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

%I A120437 #28 Nov 19 2016 02:54:58
%S A120437 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,
%T A120437 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,
%U A120437 1,7,1,1,7,1,1,61,1,1,7,1,1,7,1,1
%N A120437 Differences of A037314 (sum of base-3 digits of n = sum of base-9 digits of n).
%C A120437 It appears that sign(a(n+1) - a(n)) gives A102283. - _Filip Zaludek_, Oct 29 2016
%C A120437 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
%H A120437 Michel Marcus, <a href="/A120437/b120437.txt">Table of n, a(n) for n = 1..2000</a>
%F A120437 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.
%F A120437 a(n) = A066443(A007949(n)). (This is equivalent to the conjectured recursion above; that recursion is correct.) - _Franklin T. Adams-Watters_, Jul 24 2006
%F A120437 G.f. g(x) satisfies g(x) = 9 g(x^3) + x*(1+2*x)/(1+x+x^2). - _Robert Israel_, Nov 06 2016
%p A120437 A037314[0]:= 0;
%p A120437 for n from 0 to 33 do for k from 0 to 2 do
%p A120437   A037314[3*n+k]:= 9*A037314[n]+k
%p A120437 od od:
%p A120437 seq(A037314[i]-A037314[i-1],i=1..100); # _Robert Israel_, Nov 06 2016
%t A120437 Differences@ Table[FromDigits[RealDigits[n, 3], 9], {n, 1, 100}] (* _Michael De Vlieger_, Nov 10 2016, after _Clark Kimberling_ at A037314 *)
%o A120437 (PARI) a037314(n) = {my(d = digits(n, 3)); subst(Pol(d), x, 9); }
%o A120437 a(n) = a037314(n) - a037314(n-1); \\ _Michel Marcus_, Oct 30 2016
%Y A120437 Cf. A000695, A007949, A037314, A066443.
%K A120437 nonn,base
%O A120437 1,3
%A A120437 _John W. Layman_, Jul 17 2006