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 A120437 Differences of A037314 (sum of base-3 digits of n = sum of base-9 digits of n). 2
 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 (list; graph; refs; listen; history; text; internal format)
 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)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 A174095 A305607 Adjacent sequences: A120434 A120435 A120436 * A120438 A120439 A120440 KEYWORD nonn,base AUTHOR John W. Layman, Jul 17 2006 STATUS approved

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Last modified January 27 20:13 EST 2023. Contains 359849 sequences. (Running on oeis4.)