cp's OEIS Frontend

A259276 G.f.: A(x) = exp( Sum_{n>=1} 6^n * x^n/(n*(1+x^n)) ).

%I A259276 #10 Apr 18 2024 11:28:36
%S A259276 1,6,30,186,1110,6630,39846,239010,1433910,8603790,51622446,309733890,
%T A259276 1858404990,11150428470,66902565630,401415404586,2408492418870,
%U A259276 14450954480790,86705726950470,520234361647890,3121406169699270,18728437018590366,112370622111206670,674223732666113010
%N A259276 G.f.: A(x) = exp( Sum_{n>=1} 6^n * x^n/(n*(1+x^n)) ).
%C A259276 Compare to: exp( Sum_{n>=1} x^n/(1+x^n)/n ) = Sum_{n>=0} x^(n*(n+1)/2).
%C A259276 In general, for m > 1, if g.f. = exp(Sum_{k>=1} m^k * x^k/(k*(1+x^k))) then a(n) ~ 2 * m^(n - 1/8) / EllipticTheta(2, 0, 1/sqrt(m)). - _Vaclav Kotesovec_, Apr 18 2024
%F A259276 G.f.: -1/5 + (6/5)/(1+x - 6*x/(1+x^2 - 6*x^2/(1+x^3 - 6*x^3/(1+x^4 - 6*x^4/(1+x^5 - 6*x^5/(1+x^6 - 6*x^6/(1+x^7 - 6*x^7/(1+x^8 - 6*x^8/(...))))))))), a continued fraction.
%F A259276 G.f.: A(x) = (1 + x*B(x))/(1 - 5*x*B(x)), where B(x) = (1 + x^2*C(x))/(1 - 5*x^2*C(x)), C(x) = (1 + x^3*D(x))/(1 - 5*x^3*D(x)), D(x) = (1 + x^4*E(x))/(1 - 5*x^4*E(x)), ...
%F A259276 a(n) ~ c * 6^n, where c = 2^(7/8) / (3^(1/8) * EllipticTheta(2, 0, 1/sqrt(6))) = 0.8537393061385536395511385858124987202486847622399194... - _Vaclav Kotesovec_, Oct 18 2020, updated Apr 18 2024
%e A259276 G.f.: A(x) = 1 + 6*x + 30*x^2 + 186*x^3 + 1110*x^4 + 6630*x^5 +...
%e A259276 such that
%e A259276 log(A(x)) = 6*x/(1+x) + 6^2*x^2/(2*(1+x^2)) + 6^3*x^3/(3*(1+x^3)) + 6^4*x^4/(4*(1+x^4)) + 6^5*x^5/(5*(1+x^5)) +...
%t A259276 nmax = 30; CoefficientList[Series[Exp[Sum[6^k * x^k / (1 + x^k)/k, {k, 1, nmax}]], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Oct 18 2020 *)
%o A259276 (PARI) {a(n)=if(n==0, 1, polcoeff(exp(sum(m=1, n, 6^m*x^m/(1+x^m+x*O(x^n))/m)), n))}
%o A259276 for(n=0, 30, print1(a(n), ", "))
%o A259276 (PARI) {a(n)=local(A=1+x); for(i=1, n, A=(1 + x^(n+1-i)*A)/(1 - 5*x^(n+1-i)*A+ x*O(x^n))); polcoeff(A, n)}
%o A259276 for(n=0, 30, print1(a(n), ", "))
%Y A259276 Cf. A165941, A259273, A259274, A259275.
%K A259276 nonn
%O A259276 0,2
%A A259276 _Paul D. Hanna_, Jun 23 2015