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%I #26 Jul 13 2018 22:54:36
%S 1,1,4,21,152,1410,15774,207984,3153632,54074952,1034749080,
%T 21858562440,505274905992,12686390177136,343815306388176,
%U 10003360314147480,311003061260534400,10289575224413883840,360967225620921712704,13383588039651073512576,522943874535097662998400,21477474848621411837159040,924978962293503284606947200
%N E.g.f.: Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x.
%C More generally, we have the following identity. Given the biexponential series
%C W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
%C then for fixed p and q,
%C Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k + p)*x + (k + q)*y = W(x,y)^(p+q+1) / ( (1 + x*W(x,y))^q * (1 + y*W(x,y))^p ).
%C Further, W(x,y) satisfies the biexponential functional equation
%C ( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.
%H Paul D. Hanna, <a href="/A316370/b316370.txt">Table of n, a(n) for n = 0..300</a>
%F E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
%F (1) A(x) = Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k) + k*x.
%F (2) Sum_{n>=0} x^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x = A(x)^(p+q+1) / ( (1 + x*A(x))^q * (1 + x^2*A(x))^p ), for fixed p and q.
%F (3) A(x)/(1 + x*A(x)) = ( A(x)/(1 + x^2*A(x)) )^x.
%F a(n) ~ 2^(n+1) * n^n / (sqrt(log(2)) * exp(n)). - _Vaclav Kotesovec_, Jul 13 2018
%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 21*x^3/3! + 152*x^4/4! + 1410*x^5/5! + 15774*x^6/6! + 207984*x^7/7! + 3153632*x^8/8! + 54074952*x^9/9! + 1034749080*x^10/10! + ...
%e such that
%e A(x) = 1 + (1+x)*x + (2 + x)*(1 + 2*x)*x^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*x^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*x^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*x^5/5! + ...
%e Also,
%e A(x)^2/(1 + x*A(x)) = 1 + (1 + 2*x)*x + (2 + 2*x)*(1 + 3*x)*x^2/2! + (3 + 2*x)*(2 + 3*x)*(1 + 4*x)*x^3/3! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*x^4/4! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(1 + 6*x)*x^5/5! + ...
%e And,
%e A(x)^3/((1 + x*A(x))*(1 + x^2*A(x))) = 1 + (2 + 2*x)*x + (3 + 2*x)*(2 + 3*x)*x^2/2! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*x^3/3! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*x^4/4! + (6 + 2*x)*(5 + 3*x)*(4 + 4*x)*(3 + 5*x)*(2 + 6*x)*x^5/5! + ...
%e RELATED SERIES.
%e A(x)/(1 + x*A(x)) = 1 + 2*x^2/2! + 3*x^3/3! + 32*x^4/4! + 190*x^5/5! + 1974*x^6/6! + 21588*x^7/7! + 289232*x^8/8! + 4387752*x^9/9! + ...
%e A(x)/(1 + x^2*A(x)) = 1 + x + 2*x^2/2! + 9*x^3/3! + 56*x^4/4! + 450*x^5/5! + 4494*x^6/6! + 53424*x^7/7! + 738464*x^8/8! + 11642184*x^9/9! + ...
%e where ( A(x)/(1 + x^2*A(x)) )^x = A(x)/(1 + x*A(x)).
%o (PARI) /* From Biexponential Series: */
%o {a(n) = my(A); A = sum(m=0,n, x^m/m! * prod(k=1,m, m+1-k + k*x +x*O(x^n))); n!*polcoeff(A,n)}
%o for(n=0,30, print1(a(n),", "))
%o (PARI) /* From Biexponential Functional Equation: */
%o {a(n) = my(A=1); for(i=0,n, A = (1 + x*A)*( A/(1 + x^2*A +x*O(x^n) ) )^x ); n!*polcoeff(A,n)}
%o for(n=0,30, print1(a(n),", "))
%Y Cf. A316700, A316701, A316702.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jul 12 2018
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