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

Author

John W. Layman, Jul 17 2006

Keywords

• nonn
• base

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

Crossrefs

Cf. A000695, A007949, A037314, A066443.

Programs

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 *)

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

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)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