

A253676


Irregular triangle T read by rows in which row n is the result of iterating the function S defined in A257480 and terminating at the first occurrence of 1, assuming the 3x+1 (or Collatz) conjecture.


4



1, 2, 1, 3, 5, 4, 2, 1, 4, 2, 1, 5, 4, 2, 1, 6, 1, 7, 8, 5, 4, 2, 1, 8, 5, 4, 2, 1, 9, 7, 8, 5, 4, 2, 1, 10, 5, 4, 2, 1, 11, 41, 31, 35, 59, 149, 112, 95, 107, 608, 770, 145, 109, 82, 16, 14, 2, 1, 12, 5, 4, 2, 1, 13, 10, 5, 4, 2, 1, 14, 2, 1, 15, 17, 13, 10, 5, 4, 2, 1
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OFFSET

1,2


LINKS

Table of n, a(n) for n=1..80.


EXAMPLE

T begins:
. 1
. 2 1
. 3 5 4 2 1
. 4 2 1
. 5 4 2 1
. 6 1
. 7 8 5 4 2 1
. 8 5 4 2 1
. 9 7 8 5 4 2 1
. 10 5 4 2 1
. 11 41 31 35 59 149 112 95 107 608 770 145 109 82 16 14 2 1
. 12 5 4 2 1
. 13 10 5 4 2 1
. 14 2 1
. 15 17 13 10 5 4 2 1


MATHEMATICA

v[n_] := IntegerExponent[n, 2]; f[x_] := (3*x + 1)/2^v[3*x + 1]; s[n_] := (3 + (3/2)^v[1 + f[4*n  3]]*(1 + f[4*n  3]))/6; t[n_] := NestWhileList[s[#] &, n, # > 1 &]; Flatten[Table[t[n], {n, 15}]] (* Replace Flatten with Grid to get the irregular triangle. *)


CROSSREFS

Cf. A257480 and cross references therein.
Sequence in context: A258248 A209139 A257161 * A182939 A056943 A258654
Adjacent sequences: A253673 A253674 A253675 * A253677 A253678 A253679


KEYWORD

nonn,tabf


AUTHOR

L. Edson Jeffery, May 02 2015


STATUS

approved



