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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A385911 If n = 2^b*3^c, then a(n) = (-1)^b * binomial(b+c, b), else a(n) = 0, for n >= 1.

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%I A385911 #6 Jul 20 2025 17:24:33
%S A385911 1,-1,1,1,0,-2,0,-1,1,0,0,3,0,0,0,1,0,-3,0,0,0,0,0,-4,0,0,1,0,0,0,0,
%T A385911 -1,0,0,0,6,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,-4,0,0,0,0,0,0,0,0,0,1,
%U A385911 0,0,0,0,0,0,0,-10,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-6,0
%N A385911 If n = 2^b*3^c, then a(n) = (-1)^b * binomial(b+c, b), else a(n) = 0, for n >= 1.
%C A385911 See comment by _David Wasserman_ in related sequence A061984.
%C A385911 Triangle A385910 has g.f. A(x,y) where A(x,y) = A(x^3 + 3*x*y*A(x,y)^3, y) / A(x^2 + 2*x*y*A(x,y)^2, y).
%C A385911 a(n) = A385910(n+1, 1) for n >= 1.
%F A385911 G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies the following formulas.
%F A385911 (1) A(x) = Sum_{n>=0} (-1)^n * Sum_{k>=0} binomial(n+k,k) * x^(2^n*3^k).
%F A385911 (2) If n = 2^b*3^c, then a(n) = (-1)^b * binomial(b+c, b) else a(n) = 0, for n >= 1.
%e A385911 G.f. A(x) = x - x^2 + x^3 + x^4 - 2*x^6 - x^8 + x^9 + 3*x^12 + x^16 - 3*x^18 - 4*x^24 + x^27 - x^32 + 6*x^36 + 5*x^48 - 4*x^54 + x^64 - 10*x^72 + x^81 - 6*x^96 + ...
%e A385911 where A(x) equals the sum of the following series
%e A385911 A(x) = (x + x^3 + x^9 + x^27 + ... + x^(3^k) + ...)
%e A385911   - (x^2 + 2*x^6 + 3*x^18 + 4*x^54 + ... + (k+1)*x^(2*3^k) + ...)
%e A385911   + (x^4 + 3*x^12 + 6*x^36 + 10*x^108 + ... + C(2+k,k)*x^(2^2*3^k) + ...)
%e A385911   - (x^8 + 4*x^24 + 10*x^72 + 20*x^216 + ... + C(3+k,k)*x^(2^3*3^k) + ...)
%e A385911   + (x^16 + 5*x^48 + 15*x^144 + 35*x^432 + ... + C(4+k,k)*x^(2^4*3^k) + ...)
%e A385911   + ... + (-1)^n * Sum_{k>=0} binomial(n+k,k) * x^(2^n*3^k) + ...
%o A385911 (PARI) {a(n) = my(p2,p3); if(n<1,0, p2 = valuation(n,2); p3 = valuation(n,3);
%o A385911 if(n/(2^p2*3^p3)>1,0, (-1)^p2 * binomial(p2 + p3, p2) ))}
%o A385911 for(n=1,100,print1(a(n),", "))
%Y A385911 Cf. A385910, A061984.
%K A385911 sign
%O A385911 1,6
%A A385911 _Paul D. Hanna_, Jul 12 2025