dream_weaver Posted September 30, 2013 Report Share Posted September 30, 2013 (edited) Proof The Collatz Conjecture Is True Abstract Every positive integer can be reduced to 1 by the set of operational rules of the conjecture. Using an alterative approach to generating the counting, natural, or more precisely the whole numbers, mathematical demonstration can be provided for what makes the Collatz conjecture true. Every even number is divisible by 2 until it becomes odd and this will be amply exploited in this presentation of the facts. Keywords Source, destination, source series used. Basic Concepts The Collatz conjecture is also known as the 3n+1 conjecture, the Ulam conjecture, Kakutani problem, the Thwaites conjecture, Hasse’s algorithm or the Syracuse problem, etc. The Collatz conjecture asserts that taking any positive integer n, if n is an even number: divide n by 2 to obtain the integer n/2; if n is an odd number: multiply n by 3 and add 1 to obtain an even number 3n+1. Repeating this method, regardless of which positive integer is started with, will always reach a final result of 1. The Proof Organizing the whole numbers into a table. To begin with, start with the set of odd numbers couched as followed: 1(2^{0}), 3(2^{0}), 5(2^{0}), 7(2^{0}), 9(2^{0}), 11(2^{0}), 13(2^{0}), 15(2^{0}), 17(2^{0}), 19(2^{0}), 21(2^{0}),… Next add the even numbers in a similar format: 1(2^{1}), 3(2^{1}), 5(2^{1}), 7(2^{1}), 9(2^{1}), 11(2^{1}), 13(2^{1}), 15(2^{1}), 17(2^{1}), 19(2^{1}), 21(2^{1})… 1(2^{2}), 3(2^{2}), 5(2^{2}), 7(2^{2}), 9(2^{2}), 11(2^{2}), 13(2^{2}), 15(2^{2}), 17(2^{2}), 19(2^{2}), 21(2^{2})… 1(2^{3}), 3(2^{3}), 5(2^{3}), 7(2^{3}), 9(2^{3}), 11(2^{3}), 13(2^{3}), 15(2^{3}), 17(2^{3}), 19(2^{3}), 21(2^{3})… 1(2^{4}), 3(2^{4}), 5(2^{4}), 7(2^{4}), 9(2^{4}), 11(2^{4}), 13(2^{4}), 15(2^{4}), 17(2^{4}), 19(2^{4}), 21(2^{4})… … … … This accounts of all the even numbers, albeit in varyingly incremented sequences. Writing these sequences in conventional symbols reveals: ƒ odd n(2^{0}): 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21… increasing by increments of 2 ƒ odd n(2^{1}): 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42… increasing by increments of 4 ƒ odd n(2^{2}): 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84… increasing by increments of 8 ƒ odd n(2^{3}): 8, 24, 40, 56, 72, 88, 104, 120, 136, 152, 168…increasing by increments of 16 ƒ odd n(2^{4}): 16, 48, 80, 112, 144, 176, 208, 240, 272, 304, 336… increasing by increments of 32 … … … Development of the ‘destination’. 3n+1 produces an even number. The odd number which produced an even number can be revealed by taking the even product, ‘p’ and subjecting it to (p-1)/3. · Apply this to the applicable product numbers derived from ƒ odd n(2^{1}): 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42… (10-1)/3=3, (22-1)/3=7, (34-1)/3=11, revealing the ‘source’ number of every 3^{rd} even product in the sequence thus far, also increasing by increments of 4. 3 led to 10 or 5(2^{1}), 7 led to 22 or 11(2^{1}), 11 led to 17(2^{1}), and that 5, 11, 17 increment by 6 When the exponent is odd, the first number that satisfies (p-1)/3, as a whole number with no remainder, is 5(2^{odd}) · Examining ƒ odd n(2^{2}) by the same method: 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84… (4-1)/3=1, (28-1)/3=9, (52-1)/3=17, (76-1)/3=25, again revealing the ‘source’ number of every 3^{rd} even product in the sequence this far, also increasing by increments of 8. 1 led to 4 or 1(2^{2}), 9 led to 28 or 7(2^{2}), 17 led to 52 or 13(2^{2}), 25 led to 76 or 19(2^{2}), and that 1, 7, 13, 19 increment by 6 When the exponent is even, the first number that satisfies (p-1)/3, as a whole number with no remainder, is 1(2^{even}) · Examining ƒ odd n(2^{3}) by the same method: 8, 24, 40, 56, 72, 88, 104, 120, 136, 152, 168… (40-1)/3=13, (88-1)/3=29, (136-1)=45, again revealing the ‘source’ number of every 3^{rd} even product in the sequence this far, also increasing by increments of 16. 13 led to 40 or 5(2^{3}), 29 led to 88 or 11(2^{3}), 45 led to 136 or 17(2^{3}), and that 5, 11, 17… is encountered again. · Examining ƒ odd n(2^{4}) by the same method: 16, 48, 80, 112, 144, 176, 208, 240, 272, 304, 336… (16-1)/3=5, (112-1)=37, (208-1)/3=69, (304-1)/3=101, again revealing the ‘source’ number of every 3^{rd} even product in the sequence this far, also increasing by increments of 32… 5 led to 16 or 1(2^{4}), 37 led to 112 or 7(2^{4}), 69 led to 208 or 13(2^{4}), 101 led to 304 or 19(2^{4}), and that 1, 7, 13, 19… is encountered again. Incremental increments. The odd n(2^{0}) steps by 2 which can also be expressed as 2(2^{0}) The odd n(2^{1}) steps by 4 or 2(2^{1}) The odd n(2^{2}) steps by 8 or 2(2^{2}) The odd n(2^{3}) steps by 16 or 2(2^{3}) The odd n(2^{4}) steps by 32 or 2(2^{4}) By substituting n with 2, the increments by which any particular sequence steps can be ascertained. Exponentially determined ‘destinations’. With n(2^{1}), 3 led to 10, 7 led to 22, 11 led to 34… This exhausts the ‘source’ numbers 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67… by increments of 2(2^{1}) or 4, as destinations With n(2^{2}), 1 led to 4, 9 led to 28, 17 led to 52, 25 led to 76… This exhausts the ‘source’ numbers 1, 9, 17, 25, 33, 41, 49, 57, 65… by increments of 2(2^{2}) or 8, as destinations With n(2^{3}), 13 led to 40, 29 led to 88, 45 led to 136… This exhausts the ‘source’ numbers 13, 29, 45, 61… by increments of 2(2^{3}) or 16 as destinations. With n(2^{4}), 5 led to 16, 37 led to 112, 69 led to 208, 101 led to 304… This exhausts the ‘source’ numbers 5, 37, 69… by increments of 2(2^{4}) or 32, as destinations Arranging the ‘source series used’ as ‘destinations’ for exponents 1 thru 4: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, _, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, _, 55, 57, 59, 61, 63, 65, 67, 69… Applying the known to the unknown. · ƒ n(2^{5}) The exponent is odd. ƒ(5)=5(2^{5}) =5(32) = 160. (160-1)/3=53 As a ‘destination’, 53 fills a gap in among the ‘source series used’ developed in the previous section. 2(2^{5})=64 establishes the incremental step. 53+64=117, 117+64=181, 181+64=245… 117*3+1=352 = 11(2^{5}), 181*3+1=544 = 17(2^{5})… 53 led to 160 or 5(2^{5}), 117 led to 352 or 11(2^{5}), 181 led to 544 or 17(2^{5})… This exhausts 53, 117, 181… by increments of 2(2^{5}) from the ‘source series used’. · ƒ n(2^{6}) The exponent is even. ƒ(1)=1(2^{6})=1(64) =64 (64-1)/3=21 As a ‘destination’, 21 fills a gap among the ‘source series used’ developed in the previous section.. 2(2^{6})=128 establishes the incremental step. 21+128=149, 149+128=277, 277+128=405, 405+128=533… 149*3+1=448 = 7(2^{6}), 277*3+1=832 = 13(2^{6}), 533*3+1=1600 = 25(2^{6}) 21 led to 64 or 1(2^{6}), 149 led to 448 or 7(2^{6}), 277 led to 832 or 13(2^{26}), 533 led to 1600 or 25(2^{6})… This exhausts 21, 149, 277… by increments of 2(2^{6}) from the ‘source series used’. It is by this method that the gaps in the ‘source series used’ as ‘destinations’ continue to be filled by exponentially increasing determinable starting ‘destinations’ sequenced at exponentially increasing intervals. How do we know there is a single path to 1 for each number? n(2^{w}) can be divided by 2 a total of w times before it becomes n. 1(2^{6}) or 64 can be divided 6 times by 2. 64 serves as the ‘destination’ of 3(21)+1 21(2^{w}) hosts no ‘destinations’. 21(2^{w}) can be divided w times by 2 and the next step will position it 6 steps away from 1. 1(2^{12}) or 4096 can be divided 12 times by 2 before it becomes 1. 4096 reveals a ‘source’, thru (p-1)/3, as (4096-1)3=1365. 1365(2^{w}) hosts no ‘destinations’ either. In fact, 3(2^{0}), 9(2^{0}), 15(2^{0}), 21(2^{0})…, continuing in increments of 6, host no ‘destinations’. · First level of exhaustion 1(2^{even}) subjected to (p-1)/3 would reveal all the ‘source’ number leading to 1(2^{even}) as their ‘destination’. 1(2^{2}) is 4. (4-1)/3=1 1(2^{4}) is 16. (16-1)/3=5 1(2^{6}) is 64. (64-1)/3=21 Interestingly enough, 4+1=5, 16+5=21, 64+21=85. 3(85+1)=256=1(2^{8}) These exhaust the list of ‘source’ numbers having 1(2^{even}) as a ‘destination’. · Second level of exhaustion 5(2^{odd}) subjected to (p-1)/3 would reveal all the ‘source’ numbers leading to 5(2^{odd}) as their ‘destination. 5(2^{1}) is 10. (10-1)/3=3 5(2^{3}) is 40. (40-1)/3=13 5(2^{5}) is 160. (160-1)/3=53 Again, 10+3=13, 40+13=53, 160+53=213 3(213)+1=640=5(2^{7}) Repeat this process for 21(2^{w}), 85(2^{odd}), 341(2^{even}), 1365(2^{w}), 5461(2^{even})… This would exhaust all the ‘sources’ with 5(2^{odd}), 21(2^{w}), 85(2^{odd}) 341(2^{even})… as a ‘destination. · Further levels of exhaustion This method can be repeated on the resulting set, further exhausting the ‘sources’ ‘destined’ for this resulting set. Repeat again, to exhaust further layers. Putting it all together. Arranging the whole numbers into the table outlined exhausts the whole numbers by a method of incremental exponential notation. The odd n(2^{0}) exhausts the odd whole numbers serving as a ‘source’. 3n(2^{0})+1 exhausts the even whole numbers serving as ‘destinations’ convertible to the form of an odd n(2^{1 or greater}). Between 1(2^{even}) and 5(2^{odd}), the ‘destinations’ of the ‘source’ numbers can be mapped out. 3(2^{w}), 9(2^{w}), 15(2^{w}), 21(2^{w})…host no ‘destinations’. Every unique ‘source’ number has an equally unique ‘destination’. Every ‘destination’ provides a new ‘source’ number to work with, with one exception. There is but only one ‘source’, 3(n(2^{0}))+1, where n is the same for the ‘destination’, n(2^{1 or greater}), thus creating an infinite loop. ƒ(1(2^{0})à(3(2^{0}))(1(2^{0}))+1(2^{0}) = 3(2^{0})+1(2^{0}) = 4(2^{0}) = 2(2^{1}) = 1(2^{2}) The path any other whole number arrives there is determined by the whole number with which it begins. Edited September 30, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 5, 2013 Report Share Posted October 5, 2013 (edited) For those who are interested, post #26 has been posted in its entirety on MyMathForum in a thread entitled Old King Collatz at the end of page 5. The Old King Collatz link points to the beginning of the discussion on page 6 that started from the page 5 posting, if this constitutes a valid proof. Edited October 5, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

John Link Posted October 6, 2013 Report Share Posted October 6, 2013 dream_weaver, do your posts contain your own work or the work of someone else? Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 6, 2013 Report Share Posted October 6, 2013 This thread outlined how I've been trying to flesh it out. Quote Link to comment Share on other sites More sharing options...

John Link Posted October 6, 2013 Report Share Posted October 6, 2013 This thread outlined how I've been trying to flesh it out. Yes but are you presenting your own work, or are you summarizing and explaining the work of someone else? Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 6, 2013 Report Share Posted October 6, 2013 (edited) To expand on that a bit more, starting with post #9 on Sept. 12, it took until post #19 on Sept. 21 to see it as the table laid out in that format when the "keys" to the conjecture began to coalesce. To answer your question more directly: This is my own work. The proof presented in post #26, I copied a proof that was out there for the main section headings, reworded the Abstract and Basic Concepts sections, but everything else are my own observations. FWIW, the conversation on the other thread has led me to a different way of expressing the section: How do we know there is a single path to 1 for each number? Edited October 6, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

John Link Posted October 7, 2013 Report Share Posted October 7, 2013 This is my own work. In that case I suggest you submit your proof to a mathematics journal. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 8, 2013 Report Share Posted October 8, 2013 Thanks John. I also recieved a reply from Dr. Math today. He suggested puting it before a trusted math professor prior to submitting it a mathematics journal, and suggested that if the paper were not too long, he might consider reviewing it. I've been trying to look at my draft as an "outsider' over the last few days, and currently I'm making a few modifications, basically for clarification. The structrure is sound, the presentation of that structure is what is being improved. I do appreciate the suggestion. Quote Link to comment Share on other sites More sharing options...

John Link Posted October 9, 2013 Report Share Posted October 9, 2013 (edited) I also recieved a reply from Dr. Math today. He suggested puting it before a trusted math professor prior to submitting it a mathematics journal Good idea! Edited October 9, 2013 by John Link Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 12, 2013 Report Share Posted October 12, 2013 CollatzConjectureFormalWeb.pdf Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 13, 2013 Report Share Posted October 13, 2013 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 19, 2013 Report Share Posted October 19, 2013 Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 22, 2013 Report Share Posted October 22, 2013 It bears noting that the 3, 9, 15, 21... which are derived from the 4n-1 and 8n+! rows of tables 7a, 7b and 7c in post 37, represent the start of every Collatz sequence of significance. When w is a whole integer, ((3+6w)(2^{w})-1)/3 does not divide into a whole integer without a remainder. The numbers along the 6n-1 and 6n+1 rows, shown in post #37 are highlighted under the heading "Identifying the crucial differences:" in post #35, bear a strong correlation with the beginning [(2^{1}), (2^{0}), (2^{2}) rows] of the table under the heading "The integration:" It is the way in which 3, 9, 15, 21... trace to highlighted 6n-1 & 6n+1 numbers that require some further elaboration. Also to be considered is the way that the highlighted 6n+1 numbers follow the sequence 3, 9, 15, 21...along the (r-1)/4 row as well as how the 1, 7, 13, 19..., along the same row but from the 6n-1 side, repeat the sequence shown on the 6n+1 side that should finish putting this interesting little puzzle together. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 27, 2013 Report Share Posted October 27, 2013 Quote Link to comment Share on other sites More sharing options...

Plasmatic Posted October 28, 2013 Report Share Posted October 28, 2013 Weaver, when I open the paper it come out as blurry on my iPhone. Maybe its just my phone though. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 28, 2013 Report Share Posted October 28, 2013 It is 11" by 88" in size. I wasn't thinking of the iPhone access community on the presentation end of it. Quote Link to comment Share on other sites More sharing options...

Plasmatic Posted October 28, 2013 Report Share Posted October 28, 2013 You mean you haven't heard of the Journal Of Mathematical iPone Scholars??...(JOMIS). Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 28, 2013 Report Share Posted October 28, 2013 I'll try to keep them in mind for future productions. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted December 14, 2013 Report Share Posted December 14, 2013 The Collatz Conjecture Supplement. 1) 3(27)+1=82 Set 1. The odd numbers. {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47…} 2) 82/2 =41 Multiply Set 1 by 3 and add 1 to yield Set 2: {4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100, 106, 112, 118, 124…} 3) 3(41)+1=124 Divide Set 2 by 2 to yield Set 3: {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68…} 4) 124/2=62 Dividing the remaining odd numbers of Set 3 by 2 to yield Set 4: {1, 5, 4, 11, 7, 17, 10, 23, 13, 29, 16, 35, 19, 41, 22, 47, 25, 53, 28, 59, 31, 65, 34…} Note: The remaining even numbers can be isolated replicating Set 2. 5) 62/2=31 Set 5 {3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99. 105, 111, 117, 123, 129…} Note: Elements of Set 5 are absent from Set 4 6) 3(31)+1=94 4n+1 applied to Set 1 to yield set 6: {5, 13, 21, 29, 37, 45, 53, 61, 69, 77, 85, 93, 101, 109, 117, 125, 131, 139, 147, 155…} 7) 94/2=47 Multiply Set 6 by 3 and add 1 to yield Set 7: {16, 40, 64, 88, 112, 136, 160, 184, 208, 232, 256, 280, 304, 328, 352, 376, 400, 424…} 8) 3(47)+1=142 Divide Set 7 by 2 to yield Set 8: {8, 20, 32, 44, 56, 68, 80, 92, 104, 116, 128, 140, 152, 164, 176, 188, 200, 212, 224…} 9) 142/2=71 Divide Set 8 by 2 to generates Set 2 (again): {4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100, 106, 112, 118, 124…} 10) 3(71)+1=214 (3n+1)/2 = p 2(3n+1)/2 = 2p multiply both sides by 2 3n+1 = 2p 3n+1-1 = 2p-1 subtract 1 from both sides 3n = 2p-1 3n/3 = (2p-1)/3 divide both sides by 3 n = (2p-1)/3 11) 214/2=107 (3n+1)/4 = p 4(3n+1)/4 = 4p multiply both sides by 4 3n+1 = 4p 3n+1-1 = 4p-1 subtract 1 from both sides 3n = 4p-1 3n/3 = (4p-1)/3 divide both sides by 3 n = (4p-1)/3 12) 3(107)+1=322 (4(1)-1)/3 = 1 (2(3)-1)/3 = 1.666… (4(3)-1)/3 = 3.666… (2(5)-1)/3 = 3 (4(7)-1)/3 = 9 (2(9)-1)/3 = 5.666… (4(9)-1)/3 = 11.666… (2(11)-1)/3=7 (4(13)-1)/3=17 (2(15)-1)/3= 9.666… (4(15)-1)/3= 19.666… (2(17)-1)/3=11 (4(19)-1)/3=25 (2(21)-1)/3= 13.666… (4(21)-1)/3= 27.666… (2(23)-1)/3=15 (4(25)-1)/3=33 13) 322/2=161 Take every 6^{th} number starting with 1 to yield Set 9: {1, 7, 13, 19, 25, 31, 37, 43, 49, 55, 61, 67, 73, 79, 85, 91, 97, 103, 109, 115, 121, 127…} 14) 3(161)+1=484 Multiply Set 9 time 4, subtract 1, and then divide by 3 to yield Set 10: {1, 9, 17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97, 105, 113, 121, 129, 137, 145, 153, 161…} 15) 484/2=242 Take every 6^{th} number starting with 5 to yield Set 11: {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125…} 16) 242/2=121 Multiply Set 11 time 2, subtract 1, and then divide by 3 to yield Set 12: {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91…} 17) 3(121)+1=364 Combine Set 10 and Set 12 to yield Set 13: {1, 3, 7, 9, 11, 15, 17, 19, 23, 25, 27, 31, 33, 35, 39, 41, 43, 47, 49, 51, 55, 57, 59, 63…} Note: Elements of Set 6 ({5, 13, 21, 29, 37, 45, 53, 61…} are absent from Set 13. 18) 364/2=182 Apply 4n+1 to the first element of Set 1. If the result is an element of Set 11, apply (2n-1)/3 If the result is an element of Set 9, apply (4n-1)/3 If the result was an element of either Set 11 or Set 9, repeat check with new result. If the result is an element of Set 5, apply 4n+1 to the next element of Set 1 and repeat checking procedure. ƒ(elements of Set 1) = Set 1 rearranged into subsets of Set 1 of one or more elements relating each element of Set 6 to its unique element of Set 5 (tier explanation begins in section 28) 5^{th} tier ──────────┐ 4^{th} tier ────────┐ │ 3^{rd} tier ──────┐ │ │ 2^{nd} tier ───┐ │ │ │ 1^{st} tier ─┐ │ │ │ │ ƒ(1) = {[5], [3]} │ │ │ ƒ(3) = {[13], 17, 11, 7, [9]} ƒ(5) = {[[21]]} ƒ(7) = {[29], 19, 25, [33]} ƒ(9) = {[37], 49, 65, 43, [57]} ƒ(11) = {[[45]]} ƒ(13) = {[53], 35, 23, [15]} ƒ(15) = {[61], [81]} ƒ(17) = {[[69]]} ƒ(19) = {[77], [51]} ƒ(21) = {[85], 113, [75]} ƒ(23) = {[[93]]} ƒ(25) = {[101], 67, 89, 59, [39]} *ƒ(27) = {[109], 145, 193, 257, [171]} ƒ(29) = {[[117]]} *ƒ(31) = {[125], 83, 55, 73, 97, [129]} ƒ(33) = {[133], [177]} ƒ(35) = {[[141]]} ƒ(37) = {[149], [199]} ƒ(39) = {[157], 209, 139, 185, [123]} *ƒ(41) = {[[165]]} ƒ(43) = {[173], 115, [153]} ƒ(45) = {[181], 241, [321]} *ƒ(47) = {[[189]]} *… ƒ(81) = {[325), 433, 577, 769, 1025, 683, 455, [303]} … ƒ(111) = {[445], 593, 395, 263, 175, 233, 155, 103, 137, 91, 121, 161, 107, 71, 47, 31, 41, [27]} … ƒ(607) = {[2429], 1619, 1079, 719, 479, 319, 425, 283, 377, 251, 167, [111]} … ƒ(769) = {[3077], 2051, 1367, 911, 607, 809, 539, 359, 239, [159]} 19) 182/2=91 Subsets of 1 consisting of 1 element, Set 14: {21, 45, 69, 93, 117, 141, 165, 189, 213, 237, 261, 285, 309, 333, 357, 381, 405, 429…} Every 24^{th} integer starting with 21 satisfy both 4n+1 and 3p 4(n+2)+1-(4n+1) 4n+8+1-4n-1 8 3(n+2)-3n 3n+6-3n 6 This recognizes the facts that 4n+1 numbers occur at intervals of 8 while 3n numbers occur at intervals of 6, that every 24^{th} number is both a 4n+1 and 3n number. 6(8)/2=24 20) 3(91)+1=274 Subsets of 1 consisting of 2 elements {[5], [3]} based on (4n+1), (2p-1)/3 {[61], [81]} based on (4n+1), (4p-1)/3 consist of 2 permutations repeating at intervals of 72 {[77], [51]} based on (4n+1), (2p-1)/3 {[149], [99]} {[221], [147]} {[133], [177]} based on (4n+1), (4p-1)/3 {[205], [273]} {[277], [369]} (2n-1)/3 applies to every 6^{th} number starting with 5 (4n-1)/3 applies to every 6^{th} number starting with 1 Every 8^{th} number is a 4n+1 number starting with 5 6(8)/2=24 (2n-1)/3 applies to every 24^{th} number starting with 5 (4n-1)/3 applies to every 24^{th} number starting with 13 3n numbers occur at intervals of 6 6(24)/2=72 Substitute (4n+1) for p in (2p-1)/3 (2(4n+1)-1)/3=3p (8n+2-1)/3=3p (8n+1)/3=3p 8/3n+1/3= 3p Substitute (4n+1) for p in (4p-1)/3 (4(4n+1)-1)/3=3p (16n+4-1)/3=3p (16n-3)/3=3p 16/3n-1=3p Both 8/3n+1/3=3p & 16/3n+1=3p occur at intervals of 72 21) 274/2=137 Subsets of 1 consisting of 3 elements {[85], 113, [75]} based on (4n+1), (4p-1)/3, (2p-1)/3 {[173], 115, [153]} based on (4n+1), (2p-1)/3, (4p-1)/3 {[181], 241, [321]} based on (4n+1), (4p-1)/3, (4p-1)/3 {[197], 131, [87]} based on (4n+1), (2p-1)/3, (2p-1)/3 consist of 4 permutations repeating at intervals of 216 {[301], 401, [267]} {[389], 259, [345]} {[397], 529, [705]} {[413], 275, [183]} Substitute (16/3n+1) for p in (2p-1)/3 (2(16/3n+1)-1)/3=3p (32/3n+2-1)/3=3p (32/3n+1)/3=3p 32/9n+1/3=3p Substitute (16/3n+1) for p in (4p-1)/3 (4(16/3+1)-1)/3=3p (64/3+4-1)/3=3p (64/3+3)/3=3p 64/9+1=3p Substitute (8/3n+1) for p in (2p-1)/3 (2(8/3n+1)-1)/3=3p (16/3n+2-1)/3=3p (16/3n+1)/3=3p 16/9n+1/3=3p Substitute (8/3n+1) for p in (4p-1)/3 (4(8/3n+1/3)-1)/3=3p (32/3n+4/3-1)/3=3p (32/3n+1/3)/3=3p 32/9n+1/9=3p 32/9n+1/3=3p, 64/9+1=3p, 16/9n+1/3=3p & 32/9n+1/9=3p occur at intervals of 216 22) 3(137)+1=412 Subsets of 1 consisting of 4 elements {[29], 19, 25, [33]} based on (4n+1), (2p-1)/3, (4p-1)/3, (4p-1)/3 {[53], 35, 23, [15]} based on (4n+1), (2p-1)/3, (2p-1)/3, (2p-1)/3 {[229], 305, 203, [135]} based on (4n+1), (4p-1)/3, (2p-1)/3, (2p-1)/3 {[341], 227, 151, [201]} based on (4n+1), (2p-1)/3, (2p-1)/3, (4p-1)/3 {[373], 497, 331, [441]} based on (4n+1), (4p-1)/3, (2p-1)/2, (4p-1)/3 {[469]}, 625, 833, [555]} based on (4n+1), (4p-1)/3, (4p-1)/3, (2p-1)/3 {[533], 355, 473, [315]} based on (4n+1), (2p-1)/3, (4p-1)/3, (2p-1)/3 {[541], 721, 961, [1281]} based on (4n+1), (4p-1)/3, (4p-1)/3, (4p-1)/3 consist of 8 permutations repeating at intervals of 648 {[677], 451, 601, [801]} {[701], 467, 311, [207]} {[877], 1169, 779, [519]} {[989], 659, 439, [585]} {[1021], 1361, 907, [1209]} {[1117], 1489, 1985, [1323]} {[1181], 787, 1049, [669]} {[1189], 1585, 2133, [2817]} 23) 412/2=206 Subsets of 1 consisting of 5 elements {[13], 17, 11, 7, [9]} {[37], 49, 65, 43, [57]} {[101], 67, 89, 59, [39]} {[109], 145, 193, 257, [171]} {[157], 209, 139, 185, [123]} {[269], 179, 119, 79, [105]} {[317], 211, 281, 187, [249]} {[461], 307, 409, 545, [363]} {[557], 371, 247, 329, [219]} {[773], 515, 343, 457, [609]} {[901], 1201, 1601, 1067, [711]} {[1237], 1649, 1099, 1465, [1953]} {[1541], 1027, 1369, 1825, [2433]} {[1621], 2161, 2881, 3841, [5121]} {[1741], 2321, 1547, 1031, [687]} {[1781], 1187, 791, 527, [351]} consist of 16 permutations repeating at intervals of 1944 {[1957], 2609, 1739, 1159, [1545]} {[1981], 2641, 3521, 23247, [3129]} {[2045], 1363, 1817, 1211, [807]} {[2053], 2737, 3649, 4865, [3243]} {[2101], 2801, 1867, 2489, [1659]} {[2213], 1475, 983, 655, [873]} {[2261], 1507, 2009, 1339, [1785]} {[2405], 1603, 2137, 2849, [1899]} {[2501], 1667, 1111, 1481, [987]} {[2717], 1811, 1207, 1609, [2145]} {[2845], 3793, 5057, 3371, [2247]} {[3181], 4241, 2827, 3769, [5025]} {[3485], 2323, 3097, 4129, [5505]} {[3565], 4753, 6337, 8449, [11265]} {[3685], 4913, 3275, 2183, [1455]} {[3725], 2483, 1655, 1103, [735]} 24) 206/2=103 A subset consisting of 6 elements would contain 32 permutations repeating at intervals of 5,832. A subset consisting of 7 elements would contain 64 permutations repeating at intervals of 17,496. A subset consisting of 8 elements would contain 128 permutations repeating at intervals of 52,488. Each additions element doubles the permutations and triples the repeating interval. (Incidentally, the subset for [445]…[27] contains 18 elements, equating to 131,072 permutations repeating at intervals of 3,099,363,912. This would equate to a different permutation interspersed at 23,646 apart from one another among the first 3,099,363,912 numbers - were they to be spread equally.) 25) 3(103)+1=310 {[125], 83, 55, 73, 97, [129]} (4n+1), (2p-1)/3, (2p-1)/3, (4p-1)/3, (4p-1)/3, (4p-1)/3 26) 310/2=155 (4n+1), (2p-1)/3, (2p-1)/3, (2p-1)/3, (2p-1)/3, (2p-1)/3 (2(4n+1)-1)/3 substitute (4n+1) for p in (2p-1)/3 (8n+2-1)/3 multiply (8n+1)/3 combine terms 8/3n+1/3 result (2(8/3n+1/3)-1)/3 substitute previous result for p in next (2p-1)/3 (16/3n+2/3-1)/3 multiply (16/3n-1/3)/3 combine terms 16/9n-1/9 result (2(16/9n-1/9)-1)/3 substitute previous result for p in next (2p-1)/3 (32/9n-2/9-1)/3 multiply (32/9n-11/9)/3 combine terms 32/27n-11/27 result (2(32/27n-11/27)-1)/3 substitute previous result for p in next (2p-1)/3 (64/27n-22/27-1)/3 multiply (64/27n-49/27)/3 combine terms 64//81n-49/81 result (2(64/81n-49/81)-1)/3 substitute previous result for p in final (2p-1)/3 (128/81n-98/81-1)/3 combine terms (128/81n-179/81)/3 multiply 128/243n-179/243 result 5, which is (4(1)+1) and 29, which is (4(7)+1) are subject to (2p-1)/3 128/243(1)-179/243=-0.20987654 128/243(7)-179/243=2.95061728 2.95061728-(-0.20987654)=3.16049383 add 3.16049838 to –0.20987654 until remainder gone results in 63 (63+0.20987654))/3.16049383=20 6(20)+1=121 128/243(121)-179/243=63 4(121)+1=[485] (2(485)-1)/3=323 (2(323)-1)/3=215 (2(215)-1)/3=143 (2(143)-1)/3=95 (2(95)-1)/3=[63] 27) 3(155)+1=466 (4n+1), (4p-1)/3, (4p-1)/3, (4p-1)/3, (4p-1)/3, (4p-1)/3 (4(4n+1)-1)/3 = (16n+4-1)/3 = (16n+3)/3 = (16/3n+1) (4(16/3n+1)-1)/3 = (64/3n+4-1)/3 = (64/3n+3)/3 = (64/9n+1) (4(64/9n+1)-1)/3 = (256/9n+4-1)/3 = (256/9+3)/3 = (256/27n+1) (4(256/27n+1)-1)/3 = (1024/27n+4-1)/3 = (1024/27n+3)/3 = (1024/81n+1) (4(1024/81n+1)-1)/3 = (4096/81n+4-1)/3 = (4096/81n+3)/3 = (4096/243n+1) 13, which is (4(3)+1), and 37, which is (4(9)+1) are subject to (4p-1)/3 4096/243(3)+1=51.567901234567901234567901234568 4096/243(9)+1=152.703703703… 152.703703703…-51.567901234567901234567901234568 = 101.13580246913580246913580246914 51.567901234567901234567901234568…+101.13580246913580246913580246914…+101.13580246913580246913580246914… … = 4097, 12289, 20481 4097/3 = 1365.666… and 12289/3 = 4096.333… generate false positives. 20481/3 = 6827 satisfies the equation by leaving no remainder. (20481-51.567901234567901234567901234568)/101.13580246913580246913580246914 = 202 6(202)+3= 1215 4(1215)+1 = [4861] (4(4861)-1)/3 = 6481 (4(6481)-1)/3 = 8641 (4(8641)-1)/3 = 11521 (4(11521)-1)/3 = 15361 (4(15361)-1)/3 = [20481] 28) 466/2=233 1^{st} tier (1 permutation) Increments of +8 starting with 5 {5, 13, 21, 29, 37…} The base line of this approach begins with Set 6. 29) 3(233)+1=700 2^{nd} tier (2 permutations) Increments of +16 starting with 3 {3, 19, 35, 51, 67…} Which relate to the number located at increments of +24 starting with 5 Increments of +32 starting with 17 {17, 49, 81, 113, 145…} Which related to the number located at increments of +24 starting with 13 This line adds the next number to all the strands, skipping the single element strands. 30) 700/2=350 3^{rd} tier (4 permutations) Increments of +64 starting with 11 {11, 75, 139, 203, 267…} Increments of +72 starting with 13 Increments of +64 starting with 25 {25, 89, 153, 217, 281…} Increments of +72 starting with 29 Increments of +128 starting with 65 {65, 193, 321, 449, 577…} Increments of +72 starting with 37 Increments of +32 starting with 23 {23, 55, 87, 119, 151…} Increments of +72 starting with 53 (72-8, 72+56, 72-40) This line adds the next number to all the strands, skipping the single and double element strands. 31) 350/2=175 4^{th} tier (8 permutations) Increments of +128 starting with 7 {7, 135, 263, 391, 519…} Increments of +216 starting with 13 (229) Increments of +256 starting with 33 (289) Increments of +216 starting with 29 (245) Increments of +256 starting 43 (299) Increments of +216 starting with 37 (253) Increments of +64 starting with 15 (79) Increments of +216 starting with 53 (269) Increments of +128 starting with 59 (187) Increments of +216 starting with 101 (317) Increments of +512 starting with 257 (769) Increments of +216 starting with 109 (325) Increments of +128 starting with 73 (201) Increments of +216 starting with 125 (341) Increments of +256 starting with 185 (441) Increments of +216 starting with 157 (373) This line adds the next element to all the strands consisting of 4 elements or more. 32) 3(175)+1=526 5^{th} tier (16 permutations) Increments of +512 starting with 9 (521) Increments of +648 starting with 13 (661) Increments of +1024 starting with 57 (1081) Increments of +648 starting with 37 (685) Increments of +256 starting with 39 (295) Increments of +648 starting with 101 (749) Increments of +1024 starting with 171 (1195) Increments of +648 starting with 109 (757) Increments of +512 starting with 97 (609) Increments of +648 starting with 125 (773) Increments of +512 starting with 123 (635) Increments of +648 starting with 157 (805) Increments of +1024 starting with 385 (1409) Increments of +648 starting with 245 (893) Increments of +512 starting with 199 (711) Increments of +648 starting with 253 (901) Increments of +256 starting with 105 (361) Increments of +648 starting with 269 (917) Increments of +512 starting with 249 (761) Increments of +648 starting with 317 (965) Increments of +2048 starting with 1025 (3073) Increments of +648 starting with 325 (973) Increments of +256 starting with 175 (431) Increments of +648 starting with 445 (1093) Increments of +512 starting with 363 (875) Increments of +648 starting with 461 (1109) Increments of +128 starting with 95 (223) Increments of +648 starting with 485 (1133) Increments of +256 starting with 219 (475) Increments of +648 starting with 557 (1205) Increments of +1024 starting with 929 (1953) Increments of +648 starting with 589 (1237) This line adds the next element to all the strands consisting of 5 elements or more. 33) 526/2=263 6^{th} tier (32 permutations) Varying increments of varying doubles of 16 and 32 with varying starting numbers Increments of +3(648)=1944 starting with 32 different starting points under 1944 (Next 32 starting points: {125, 245, 253, 325, 445, 485, 589, 661, 685, 749, 757, 805, 893, 917, 965, 973, 1093, 1109, 1133, 1205, 1309, 1333, 1397, 1405, 1421, 1453, 1541, 1549, 1565, 1613, 1757, 1853}) 7^{th} tier (64 permutations) Varying increments of varying doubles of 16 and 32 with varying starting numbers Increments of +3(1944)=5832 starting with 62 different starting points under 5832 8^{th} tier (128 permutations) Varying increments of varying doubles of 16 and 32 with varying starting numbers Increments of +3(5832)=17496 starting with 128 different starting points under 17496 18^{th} tier (131072 permutations) Varying increments of varying doubles of 16 and 32 with varying starting numbers Increments of +1033121304 starting with 131072 different starting points fewer than 1033121304 34) 3(263)+1=790 1 3(1)+1=4 4/2=2 2/2=1 35) 790/2=395 Substitute 4(n)+1 for (n) in 3(n)+1, 3(4(n)+1)+1 = (12(n)+3)+1 = 12(n)+4 = 4(3(n)+1) 4(3(n)+1)/2 = 2(3(n)+1) 2(3(n)+1)/2 = 3(n)+1 36) 3(395)+1=1186 4(1)+1=5, (2(5)-1)/3=3 Leads to 2 permutations {[5], [3]} 37) 1186/2=593 Leads to 6 permutations 4(5)+1={[[21]]} 4(3)+1={[13], 17, 11, 7, [9]} 38) 3(593)+1=1780 Leads to 18 permutations 4(21)+1={[85], 113, [75]} 4(13)+1={[53], 35, 23, [15]} 4(17)+1={[[69]]} 4(11)+1={[[45]]} 4(7)+1={[29], 19, 25, [33]} 4(9)+1={[37], 49, 65, 43, [57]} 39) 1780/2=890 Leads to 41 permutations 4(85)+1=341… 4(113)+1=453… 4(75)+1=301… 4(53)+1=213… 4(35)+1=141… 4(23)+1=93… 4(15)+1=61… 4(69)+1=277… 4(45)+1=181… 4(29)+1=117… 4(19)+1=77… 4(25)+1=101… 4(33)+1=133… 4(37)+1=149… 4(49)+1=197… 4(65)+1=261… 4(43)+1=173… 4(57)+1=229… 40) 890/2=445 The (4n+1)/3p strands, as shown in 18), consist of a individually unique subsets of the set of odd numbers. No element of one subset is an element of another as confirmed by the sections referenced in the next two summations. The strands come in subsets of varying lengths, as shown in 19) through 27), that double in complexity with each increment of length, and occur at increasingly spaced intervals based on multiples of 3 along the number line with each increment in length. Furthermore, examining the first number (or tier) of each strand shows a single relationship between them, as shown in 28). Each subsequent tier doubles in its complexity, and occurs at increasingly spaced intervals along the number line, based on multiples of 3, with each incremental tier of the remaining strands to be considered, as shown in 29) through 33). Starting with 1, as shown in 34), each subsequent 4n+1 level increases in complexity while retaining its {n to 4n+1}, as shown in 35), to the previous level, as shown in 36) through 39). This fact {of the (n) to 4(n)+1) relationship} is further shown in steps 1) through 9) while steps 10) through 12) show the 4(n)+1 relationship terminates as a 3p using the converses of (3(n)+1)/2 and (3(n)+1)/4. Steps 13 through 17 show Set 6, 4(n)+1, as being the other end of the relationship to Set 5, 3p, as previously noted. Chart to graphically illustrate increasing complexity of the subset strands. set of 1, set of 2, set of 3, set of 4, set of 5… ┌(2p-1)3… ┌(2p-1)/3┤ │ └(4p-1)/3… ┌(2p-1)/3┤ │ │ ┌(2p-1)/3… │ └(4p-1)/3┤ │ └(4p-1)/3… ┌(2p-1)/3┤ │ │ ┌(2p-1)/3… │ │ ┌(2p-1)/3┤ │ │ │ └(4p-1)/3… │ └(4p-1)/3┤ │ │ ┌(2p-1)/3… │ └(4p-1)/3┤ │ └(4p-1)/3… (4n+1)┤ │ ┌(2p-1)/3… │ ┌(2p-1)/3┤ │ │ └(4p-1)/3… │ ┌(2p-1)/3┤ │ │ │ ┌(2p-1)/3… │ │ └(4p-1)/3┤ │ │ └(4p-1)/3… └(4p-1)/3┤ │ ┌(2p-1)/3… │ ┌(2p-1)/3┤ │ │ └(4p-1)/3… └(4p-1)/3┤ │ ┌(2p-1)/3… └(4p-1)/3┤ └(4p-1)/3... freq. 24, freq. 72, freq. 216, freq. 648, freq. 1944… Quote Link to comment Share on other sites More sharing options...

Cody T. Dianopoulos Posted December 22, 2013 Report Share Posted December 22, 2013 Cody T. Dianopoulos here. My proof was extremely flawed as natural, but my research has shown up some results. Considering the even numbers is just a bump in the road, completely eliminating them from the problem by modifying the algorithm simplifies the problem vastly. Secondly, my progress has shown that CC is not a problem in number theory, but instead a string transformation on binary strings. I'm working on a form of strong induction (in turn finding an upper bound) in order to prove CC. A sketch: assume CC holds for strings of length 0<=L<=k. Then show that given iteration of the algorithm in my most recent paper will eventually end up reducing the length to some 0<=L<=k. Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted December 22, 2013 Report Share Posted December 22, 2013 (edited) Cody, Welcome to O.O. The doubling of complexity observed in both the length of a (4n+1)/(3p) strand as well as working it tier by tier using (4n+1) as the base tier, shows a definate binary pattern is present. Edited December 22, 2013 by dream_weaver Quote Link to comment Share on other sites More sharing options...

Cody T. Dianopoulos Posted December 23, 2013 Report Share Posted December 23, 2013 dream_weaver, we should work together. Find me on Facebook as codercody. Quote Link to comment Share on other sites More sharing options...

Ernst Posted October 3, 2014 Report Share Posted October 3, 2014 Hello As someone who has spent many man hours on this I thought to share an observation I made yesterday on my lunch break and to ask if it has been reported before. If not reported then an effort must be made. We are familuar with the Collatz where we multiply by three then add one for when the state of the integer is odd. If we regroup this action and add one to an odd integer then multiply by three the cycle is attracted to the multiplier and not the addend. { 1 ,6 ,3 ,12 ,6 ,3 } ----------- { 2 ,1 ,6 ,3 ,12 ,6 ,3 } ----------- { 3 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 4 ,2 ,1 ,6 ,3 ,12 ,6 ,3 } ----------- { 5 ,18 ,9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 6 ,3 ,12 ,6 ,3 } ----------- { 7 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 8 ,4 ,2 ,1 ,6 ,3 ,12 ,6 ,3 } ----------- { 9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 10 ,5 ,18 ,9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 11 ,36 ,18 ,9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 12 ,6 ,3 ,12 ,6 ,3 } ----------- { 13 ,42 ,21 ,66 ,33 ,102 ,51 ,156 ,78 ,39 ,120 ,60 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 14 ,7 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 16 ,8 ,4 ,2 ,1 ,6 ,3 ,12 ,6 ,3 } ----------- { 17 ,54 ,27 ,84 ,42 ,21 ,66 ,33 ,102 ,51 ,156 ,78 ,39 ,120 ,60 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 18 ,9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 19 ,60 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- { 20 ,10 ,5 ,18 ,9 ,30 ,15 ,48 ,24 ,12 ,6 ,3 ,12 ,6 ,3 } ----------- So: If Odd do x+1 then multiply by three. If Even then divide by two. This thread looked really good to me as I searched and it seems to have faded a bit so I decided to chare here. Has anyone seen this before? I'll write it up if not with a short program attached. I tested this with a 3321928 bit number. It has been tested with multiplier of two and three. The "attractor" then is the multiplier and not the addend as is seen in the Collatz form. -------- mpz_set_ui( A, 3 ); // Set A to the multiplier you wish for A > 0 mpz_set_str( Y, "1\0",10); // Example C language code snippit using GMP bignum laibrary for(x=1;x<18446744073709551615u;x++) { mpz_set_ui(Z,x); w=0; printf("{ "); for(; { gmp_printf("%Zu ,",Z); if( mpz_odd_p(Z) ) { mpz_add_ui(Z,Z,1); mpz_mul(Z,Z,A); } else {mpz_divexact_ui(Z,Z,2);} if( mpz_cmp(Z,A) == 0 )w++; if(w==2){ gmp_printf("%Zu }",Z); printf("\n-----------\n");break;} } } I have enjoyed the Collatz conjecture since being introduced to it in 1992. I have worked out a great deal about it so I'm open to sharing what I know. In General I adore the cycle in mathematics. Always have. Ernst Berg [email protected] Quote Link to comment Share on other sites More sharing options...

dream_weaver Posted October 4, 2014 Report Share Posted October 4, 2014 Welcome to OO, Ernst. As to the thread fading here, it also faded on MyMathForum as well. Applying 3x+1 to the odd numbers kicks off a series of {4, 10, 16, 22, . . .} incrementing by +6 Applying 3(x+1) to the odd numbers kicks off a series of {6, 12, 18, 24, . . .}, also incrementing by +6. The latter, though, is not the Collatz. Are you suggesting there is a correlation? Quote Link to comment Share on other sites More sharing options...

Ernst Posted October 4, 2014 Report Share Posted October 4, 2014 I have not gotten that far into it yet. This variant, which doesn't follow our conventional algebraic order of operations, was a "doodle" at lunch time at work on October 3 2014. I had time to access the Internet on Friday the 3rd so I began looking for prior art. Nothing was obvious yet it was pointed out by a member of mathstackexchange.com a"one HardMaths" that it may be covered in a more generalized form in some paper. So the first thing that came to my mind was "did I discover this form" and second look to see if there is prior art. I just ran 9(X+1) and it simply goes on and on with no cycle. What do we call that "Divergent?" At this point I am simply sharing the observation I made Thursday. I am looking at 7(X+1) now { 1 ,14 ,7 ,56 ,28 ,14 ,7 } ----------- { 2 ,1 ,14 ,7 ,56 ,28 ,14 ,7 } ----------- { 3 ,28 ,14 ,7 ,56 ,28 ,14 ,7 } ----------- { 4 ,2 ,1 ,14 ,7 ,56 ,28 ,14 ,7 } After this it seems to diverge. Just changed the A(X+1) to A(X+3) looking at 9 again.. Nope seems to go on forever. Changed back to 3(X+3) and now the attractor is A*3 { 1 ,12 ,6 ,3 ,18 ,9 ,36 ,18 ,9 } for 3(X+9) { 1 ,30 ,15 ,72 ,36 ,18 ,9 ,54 ,27 ,108 ,54 ,27 ,108 } 3*9 = 27 So Perhaps then it is multiplier times addend that is the attractor for a multiplier of 2(X+9) { 2 ,1 ,20 ,10 ,5 ,28 ,14 ,7 ,32 ,16 ,8 ,4 ,2 } the multiplier times the addend for attractor fails. hmmmmmm.... The thing is to do at this point is to simply point out the alternate realm and let folks experiment. I don't know if this will help us find a solution for the Collatz Conjecture. From what little I have done this afternoon and last night it looks like it's more of a variety then what we are used to. Convergent and divergent it would seem depending on the multiplier perhaps. It's more fun if you like this sort of thing. Ernst Quote Link to comment Share on other sites More sharing options...

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