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A294689 Collatz cycle of negative numbers starting with -17.

-17, -25, -37, -55, -82, -41, -61, -91, -136, -68, -34, -17, -25, -37, -55, -82, -41, -61, -91, -136, -68, -34, -17, -25, -37, -55, -82, -41, -61, -91, -136, -68, -34, -17, -25, -37, -55, -82, -41, -61, -91, -136, -68, -34

Offset: 1

Joe Slater, Nov 07 2017

This is the longest of the three known cycles of negative numbers under the reduced Collatz transformation x: x=x/2 when x is even, and x=(3x+1)/2 when x is odd. The other two cycles of negative numbers are (-1,-1,-1,...) and (-5,-7,-10,-5,-7,-10,...). Only one cycle of positive numbers is known: (1,2,1,2,...). (0,0,0,...) is also a cycle.

Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture

Positive number transforms are in A014682; the absolute value of negative number transforms can be found in A123684.

a(1) = -17, a(n+1) = -A123684(abs(a(n))).

a(n+11) = a(n).

A287319 Smallest number k which becomes a power of 2 after being transformed by the reduced Collatz function k=(3*k+1)/2 precisely n times.

Original entry on oeis.org

1, 3, 151, 26512143, 318400215865581346424671, 1240913164837493520914469575281720548839055905624577375251388717505927743

Offset: 1

Joe Slater, May 23 2017

nonn

a(7) is too large to include.

I conjecture that all members of a(n) are members of A054646 and A010120, "Smallest number to give 2^(2n) in a hailstone (3x + 1) sequence" and "Smallest start for a `3x+1' sequence containing 2^n".

			For n = 3, the reduced Collatz sequence k = (3*k+1)/2 is 151, 227, 341, 512.

Cf. A054646 and A010120.

a(n) = ((2^(3^(n-1)+n)-3^n+2^n))/3^n.

A283754 The smallest number k such that k*2^n mod 3^n = 1.

Original entry on oeis.org

2, 7, 17, 76, 38, 262, 1589, 4075, 11879, 35464, 17732, 363160, 181580, 90790, 9611333, 11980120, 92083502, 175181914, 862431935, 2174608168, 1087304084, 543652042, 271826021, 235493860078, 117746930039, 1329806379184, 664903189592, 332451594796, 166225797398, 68713490263582, 446139009321089

Offset: 1

Joe Slater, Mar 23 2017

nonn

a(n) is the coefficient "a" in the Diophantine equation with two coefficients a and b, a * 2^n - b * 3^n = 1.

			2 * 2^1 mod 3^1 = 1, 7 * 2^2 mod 3^2 =1, 17 * 2^3 mod 3^3 = 1...

Robert Israel, Table of n, a(n) for n = 1..2095

Cf. A055620.

seq(2^(-n) mod 3^n, n=1..100); # Robert Israel, Mar 28 2017

Table[ PowerMod[ (3^n +1)/2, n, 3^n], {n, 30}] (* Robert G. Wilson v, Mar 28 2017 *)

a(n)= my(z=3^n); lift( Mod((z + 1)/2, z)^n); \\ Joerg Arndt, Mar 24 2017

a(n) = ((3^n + 1)/2)^n mod 3^n (proved).

Conjecture: 2*a(n+1)-a(n) = 3^n * A055620(n). - Robert Israel, Mar 28 2017

Corrected and more terms from Joerg Arndt, Mar 24 2017

A274048 a(n) = A116640(A018900(n)) = A116623(A059893(A018900(n))).

Original entry on oeis.org

19, 23, 29, 31, 37, 49, 47, 53, 65, 89, 79, 85, 97, 121, 169, 143, 149, 161, 185, 233, 329, 271, 277, 289, 313, 361, 457, 649, 527, 533, 545, 569, 617, 713, 905, 1289, 1039, 1045, 1057, 1081, 1129, 1225, 1417, 1801

Offset: 1

Joe Slater, Jun 07 2016

A subset of A116640 containing all terms A116640(m) such that m has binary weight of 2. This sequence is related to the Collatz and Terras trajectories; specifically those trajectories that include three odd numbers besides 1.

			The first three terms of A018900 are 3,5,6. Taking these terms from A116640 gives 19,23,29, which are the first three terms of this sequence. The sequence is generated from the regular triangle
1;
1,2;
1,2,3;
etc., so the first three terms are
2^(1+1) + 3*(3+2^1) = 19;
2^(2+1) + 3*(3+2^1) = 23;
2^(2+1) + 3*(3+2^2)= 29.

Subsequence of A116640.

Cf. A116623.

a(n) = 2^(Row(n)+1) + 3*(3+2^Col(n)) where Row(n) = A002024(n) = the row position of n when the sequence is viewed as a regular triangle; and Col(n) = A002260(n) = the column position of n when the sequence is viewed as a regular triangle.

A269305 2^n mod 27 successively displaced by 0, 3, 6, 9, etc.

Original entry on oeis.org

1, 2, 4, 8, 16, 5, 10, 20, 13, 26, 25, 23, 19, 11, 22, 17, 7, 14, 4, 5, 7, 11, 19, 8, 13, 23, 16, 2, 1, 26, 22, 14, 25, 20, 10, 17, 7, 8, 10, 14, 22, 11, 16, 26, 19, 5, 4, 2, 25, 17, 1, 23, 13, 20, 10, 11, 13, 17, 25, 14, 19, 2, 22, 8, 7, 5, 1, 20

Offset: 0

Joe Slater, Feb 22 2016

Repeats with period 18*9 = 162.

Cf. A070337.

[(2^n+3*Floor(n/18)) mod 27: n in [0..80]]; // Vincenzo Librandi, Feb 23 2016

Table[Mod[(2^n + 3 Floor[n/18]), 27], {n, 0, 70}] (* Vincenzo Librandi, Feb 23 2016 *)

a(n) = (2^n + 3*(n\18)) % 27; \\ Michel Marcus, Feb 24 2016

a(n) = (2^n + 3*floor(n/18)) mod 27.

A056150 Number of combinations for each possible sum when throwing 3 (normal) dice.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1

Offset: 3

Joe Slater (joe(AT)yoyo.cc.monash.edu.au), Aug 05 2000

The 3rd row of A063260. - Michel Marcus, Mar 04 2013

			Using three normal (six-sided) dice we can produce a sum of 3 in just one way: 1,1,1. We can produce a sum of 4 in three ways: 1,1,2; 1,2,1; 2,1,1. We can produce a sum of 5 in 6 ways and so on.

A108907 gives sums for 6 dice.
A166322 gives sums for 7 dice.
A063260 gives the sums for 2 dice through to 6 dice.

Transpose[Tally[Total/@Tuples[Range[6],{3}]]][[2]] (* Harvey P. Dale, Dec 17 2014 *)

Vec(((sum(k=1,6,x^k))^3+O(x^66))) /* Joerg Arndt, Mar 04 2013 */

Corrected by Rick L. Shepherd, May 24 2002

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User: Joe Slater

A294689 Collatz cycle of negative numbers starting with -17.

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A287319 Smallest number k which becomes a power of 2 after being transformed by the reduced Collatz function k=(3*k+1)/2 precisely n times.

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A283754 The smallest number k such that k*2^n mod 3^n = 1.

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A274048 a(n) = A116640(A018900(n)) = A116623(A059893(A018900(n))).

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A269305 2^n mod 27 successively displaced by 0, 3, 6, 9, etc.

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Magma

Mathematica

PARI

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A056150 Number of combinations for each possible sum when throwing 3 (normal) dice.

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Mathematica

PARI

Extensions