A346775 Starting from n!+1, the length of the longest sequence of consecutive numbers which all take the same number of steps to reach 1 in the Collatz (or '3x+1') problem.

1, 1, 1, 1, 1, 1, 3, 5, 1, 7, 1, 1, 3, 1, 3, 1, 1, 1, 30, 1, 30, 3, 7, 1, 3, 3, 7, 1, 1, 7, 15, 3, 1, 1, 3, 15, 26, 15, 1, 1, 1, 1, 7, 7, 26, 7, 1, 7, 3, 1, 1, 3, 1, 7, 3, 7, 1, 1, 26, 15, 7, 30, 1, 1, 1, 1, 3, 15, 3, 1, 1, 31, 648, 26, 26, 30, 90, 1, 1, 3, 15

Offset

0,7

Comments

The largest value known in this sequence is a(219)=78553595.

2^32 < a(238) < 11442739136455298475. - Martin Ehrenstein, Aug 21 2021

Jeremy Sawicki found that a(238) = 107150589645. - Dmitry Kamenetsky, Aug 25 2024

Links

Dmitry Kamenetsky, Table of n, a(n) for n = 0..237

Dmitry Kamenetsky, Smallest m>1 such that the number of Collatz steps needed for 238!+m to reach 1 differs from that of 238!+1., Mathematics StackExchange, Aug 2021.

Index entries for sequences related to 3x+1 (or Collatz) problem

Example

a(6) = 3, because 6!+1, 6!+2 and 6!+3 all take 46 steps to reach 1, while 6!+4 requires 20 steps to reach 1.

Mathematica

f[n_] := Length[NestWhileList[If[EvenQ@#, #/2, 3 # + 1] &, n, # != 1 &]] - 1; Table[k = 1; While[f[n! + k] == f[n! + k + 1], k++]; k, {n, 0, 100}] (* Bence Bernáth, Aug 14 2021 *)

Prog

(PARI) a6577(n0)={my(n=n0, k=0); while(n>1, k++; n=if(n%2, 3*n+1, n/2)); k};
for(n=0, 80, my(n0=n!+1, nc=a6577(n0), k=1); while(a6577(n0++)==nc, k++); print1(k, ", ")) \\ Hugo Pfoertner, Aug 04 2021

Crossrefs

Cf. A006577, A070974, A277109.

Sequence in context: A065077 A118788 A196020 * A028510 A338721 A122053

Adjacent sequences: A346772 A346773 A346774 * A346776 A346777 A346778

Keyword

nonn

Author

Dmitry Kamenetsky, Aug 03 2021

Status

approved