A120437 Differences of A037314 (sum of base-3 digits of n = sum of base-9 digits of n).

1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 547, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 547, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1, 61, 1, 1, 7, 1, 1, 7, 1, 1

Offset

1,3

Comments

It appears that sign(a(n+1) - a(n)) gives A102283. - Filip Zaludek, Oct 29 2016

This is clear: a(n) = 1 for n == 1 or 2 (mod 3), and a(n) >= 7 for n == 0 (mod 3): see comment by Franklin T. Adams-Watters on A037314. - Robert Israel, Nov 06 2016

Links

Michel Marcus, Table of n, a(n) for n = 1..2000

Formula

It appears that the sequence is given by the following recursion: a(n)=1 if n=1, a(n)=9a(3^(k-1))-2 if n=3^k for some k>0, a(n)=a(n-3^(k-1)) if 3^(k-1)<n<3^k for some k>0. This recursion formula has been verified for n<=2000.

a(n) = A066443(A007949(n)). (This is equivalent to the conjectured recursion above; that recursion is correct.) - Franklin T. Adams-Watters, Jul 24 2006

G.f. g(x) satisfies g(x) = 9 g(x^3) + x*(1+2*x)/(1+x+x^2). - Robert Israel, Nov 06 2016

Maple

A037314[0]:= 0;
for n from 0 to 33 do for k from 0 to 2 do
  A037314[3*n+k]:= 9*A037314[n]+k
od od:
seq(A037314[i]-A037314[i-1], i=1..100); # Robert Israel, Nov 06 2016

Mathematica

Differences@ Table[FromDigits[RealDigits[n, 3], 9], {n, 1, 100}] (* Michael De Vlieger, Nov 10 2016, after Clark Kimberling at A037314 *)

Prog

(PARI) a037314(n) = {my(d = digits(n, 3)); subst(Pol(d), x, 9); }
a(n) = a037314(n) - a037314(n-1); \\ Michel Marcus, Oct 30 2016

Crossrefs

Cf. A000695, A007949, A037314, A066443.

Sequence in context: A102421 A019620 A105395 * A336459 A367764 A174095

Adjacent sequences: A120434 A120435 A120436 * A120438 A120439 A120440

Keyword

nonn,base

Author

John W. Layman, Jul 17 2006

Status

approved