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%I #49 Jun 09 2023 15:32:55
%S 1,1,2,6,25,137,944,7884,77514,877002,11218428,160010244,2516742498,
%T 43260962754,806650405800,16213824084864,349441656710217,
%U 8037981040874313,196539809431339642,5090276002949080318,139202688233361310841,4008133046329085884137
%N Binomial transform = self-convolution: first column of the triangle (A084783).
%C In the triangle (A084783), the diagonal (A084785) is the self-convolution of this sequence and the row sums (A084786) gives the differences of the diagonal and this sequence.
%C Ramanujan considers the continued fraction phi(x) = 1 / (x + 1 - 1^2 / (x + 3 - 2^2 / (x + 5 - 3^2 / (x + 7 - 4^2 / ...)))) and states that phi(x+1) approaches x phi(x)^2 as x gets large. The asymptotic expansion is phi(x) = 1/x - 1/x^2 + 2/x^3 - 6/x^4 + 24/x^5 - ... + (-1)^n * n! / x^(n+1) + ... but if we replace this with f(x) = a(0)/x - a(1)/x^2 + a(2)/x^3 - a(3)/x^4 + ... then formally f(x+1) = x f(x)^2 which is similar to my Feb 16 2006 formula. - _Michael Somos_, Jun 20 2015
%C This is also the Euler transform of A060223. - _Gus Wiseman_, Oct 16 2016
%D S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 223.
%H Paul D. Hanna, <a href="/A084784/b084784.txt">Table of n, a(n) for n = 0..200</a>
%F G.f. satisfies A(n*x)^2 = n-th binomial transform of A(n*x).
%F G.f. A(x) satisfies 1 + x = A(x/(1 + x))^2 / A(x). - _Michael Somos_, Feb 16 2006
%F G.f.: A(x) = Product_{n>=1} 1/(1 - n*x)^(1/2^(n+1)). - _Paul D. Hanna_, Jun 16 2010
%F G.f.: A(x) = exp( Sum_{n>=1} A000670(n)*x^n/n ) where Sum_{n>=0} A000670(n)*x^n = Sum_{n>=0} n!*x^n/Product_{k=0..n} (1-k*x). - _Paul D. Hanna_, Sep 26 2011
%F a(n) ~ (n-1)! / (2 * (log(2))^(n+1)). - _Vaclav Kotesovec_, Nov 18 2014
%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 25*x^4 + 137*x^5 + 944*x^6 + ...
%e where
%e A(x) = (1-x)^(-1/4)*(1-2*x)^(-1/8)*(1-3*x)^(-1/16)*(1-4*x)^(-1/32)*...
%e Also,
%e log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 75*x^4/4 + 541*x^5/5 + 4683*x^6/6 + ... + A000670(n)*x^n/n + ...
%e thus, the logarithmic derivative equals the series:
%e A'(x)/A(x) = 1/(1-x) + 2!*x/((1-x)*(1-2*x)) + 3!*x^2/((1-x)*(1-2*x)*(1-3*x)) + 4!*x^3/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) + ...
%p a:= proc(n) option remember;
%p 1+add(a(j)*(binomial(n,j)-a(n-j)), j=1..n-1)
%p end:
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Jun 09 2023
%t a[ n_]:= If[n<1, Boole[n==0], Module[{A= 1/x - 1/x^2}, Do [A= 2 A - Normal @ Series[ (x A^2) /. x -> x-1, {x, Infinity, k+1}], {k,2,n}]; (-1)^n Coefficient[A, x, -n-1]]]; (* _Michael Somos_, Jun 20 2015 *)
%t nn=20;CoefficientList[Series[Exp[Sum[Times[1/k,i!,StirlingS2[k,i],x^k],{k,nn},{i,k}]],{x,0,nn}],x] (* _Gus Wiseman_, Oct 18 2016 *)
%o (PARI) {a(n) = my(A); if( n<0, 0, A=1; for(k=1, n, A = truncate(A + O(x^k)) + x * O(x^k); A += A - 1 / subst(A^-2, x, x / (1 + x)) / (1 + x);); polcoeff(A, n))}; /* _Michael Somos_, Feb 18 2006 */
%o (PARI) /* Using o.g.f. exp( Sum_{n>=1} A000670(n)*x^n/n ): */
%o {a(n)=polcoeff(exp(intformal(sum(m=1, n+1, m!*x^(m-1)/prod(k=1, m, 1-k*x+x*O(x^n))))), n)}
%o for(n=0,30,print1(a(n),", "))
%o (Magma)
%o m:=50;
%o f:= func< n,x | Exp((&+[(&+[Factorial(j)*StirlingSecond(k,j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
%o R<x>:=PowerSeriesRing(Rationals(), m+1); // A084784
%o Coefficients(R!( f(m,x) )); // _G. C. Greubel_, Jun 08 2023
%o (SageMath)
%o m=40
%o def f(n, x): return exp(sum(sum(factorial(j)*stirling_number2(k,j) *x^k/k for j in range(1,k+1)) for k in range(1,n+2)))
%o def A084784_list(prec):
%o P.<x> = PowerSeriesRing(QQ, prec)
%o return P( f(m,x) ).list()
%o A084784_list(m) # _G. C. Greubel_, Jun 08 2023
%o (Python) # after _Alois P. Heinz_
%o from functools import cache
%o from math import comb as binomial
%o @cache
%o def a(n: int) -> int:
%o return 1 + sum((binomial(n, j) - a(n - j)) * a(j) for j in range(1, n))
%o print([a(n) for n in range(22)]) # _Peter Luschny_, Jun 09 2023
%Y Cf. A000670, A060223, A084783, A084785, A084786, A195983.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jun 13 2003
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