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A316704 E.g.f.: Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.

%I A316704 #17 Jan 17 2024 09:47:48
%S A316704 1,2,12,108,1312,20320,381408,8420160,213813248,6139270656,
%T A316704 196691281920,6956268042240,269187901974528,11313963679948800,
%U A316704 513251989767487488,24995547184468008960,1300702468667721646080,72026879559935471124480,4229000873160355032072192,262425376836886982897958912,17161024720479004010930503680,1179556709319250468710226329600
%N A316704 E.g.f.: Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.
%C A316704 More generally, we have the following identity. Given the biexponential series
%C A316704 W(x,y) = Sum_{n>=0} 1/n! * Product_{k=1..n} (n+1-k)*x + k*y,
%C A316704 then for fixed p and q,
%C A316704 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 A316704 Further, W(x,y) satisfies the biexponential functional equation
%C A316704 ( W(x,y)/(1 + x*W(x,y)) )^x = ( W(x,y)/(1 + y*W(x,y)) )^y.
%H A316704 Paul D. Hanna, <a href="/A316704/b316704.txt">Table of n, a(n) for n = 0..300</a>
%F A316704 E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! satisfies:
%F A316704 (1) A(x) = Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k) + k*x.
%F A316704 (2) Sum_{n>=0} (2*x)^n/n! * Product_{k=1..n} (n+1-k + p) + (k + q)*x  =  A(x)^(p+q+1) / ( (1 + 2*x*A(x))^q * (1 + 2*x^2*A(x))^p ), for fixed p and q.
%F A316704 (3) A(x)/(1 + 2*x*A(x)) = ( A(x)/(1 + 2*x^2*A(x)) )^x.
%F A316704 a(n)/n! ~ c * d^n / sqrt(n), where d = 3.346513389529679772056152566067040813392... and c = 1.06774499146514892068040233... - _Vaclav Kotesovec_, Jul 18 2018
%F A316704 The constant d given above is the root of the equation d^(d-2) = 2^(d-1). - _Vaclav Kotesovec_, Jan 17 2024
%e A316704 E.g.f.: A(x) = 1 + 2*x + 12*x^2/2! + 108*x^3/3! + 1312*x^4/4! + 20320*x^5/5! + 381408*x^6/6! + 8420160*x^7/7! + 213813248*x^8/8! + 6139270656*x^9/9! + 196691281920*x^10/10! + ...
%e A316704 such that
%e A316704 A(x) = 1 + (1+x)*(2*x) + (2 + x)*(1 + 2*x)*(2*x)^2/2! + (3 + x)*(2 + 2*x)*(1 + 3*x)*(2*x)^3/3! + (4 + x)*(3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^4/4! + (5 + x)*(4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^5/5! + ...
%e A316704 Also,
%e A316704 A(x)^2/(1 + 2*x*A(x)) = 1 + (1 + 2*x)*(2*x) + (2 + 2*x)*(1 + 3*x)*(2*x)^2/2! + (3 + 2*x)*(2 + 3*x)*(1 + 4*x)*(2*x)^3/3! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(1 + 5*x)*(2*x)^4/4! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(1 + 6*x)*(2*x)^5/5! + ...
%e A316704 And,
%e A316704 A(x)^3/((1 + 2*x*A(x))*(1 + 2*x^2*A(x))) = 1 + (2 + 2*x)*(2*x) + (3 + 2*x)*(2 + 3*x)*(2*x)^2/2! + (4 + 2*x)*(3 + 3*x)*(2 + 4*x)*(2*x)^3/3! + (5 + 2*x)*(4 + 3*x)*(3 + 4*x)*(2 + 5*x)*(2*x)^4/4! + (6 + 2*x)*(5 + 3*x)*(4 + 4*x)*(3 + 5*x)*(2 + 6*x)*(2*x)^5/5! + ...
%e A316704 RELATED SERIES.
%e A316704 B(x) = sqrt( (1 + 2*x*A(x)) * (1 + 2*x^2*A(x)) ) = 1 + x + 5*x^2/2! + 45*x^3/3! + 513*x^4/4! + 7745*x^5/5! + 142485*x^6/6! + 3095421*x^7/7! + 77642145*x^8/8! + 2207145825*x^9/9! + ... + A316705(n)*x^n/n! + ...
%e A316704 where
%e A316704 B(x) = 1 + (1 + x)*x + (3 + x)*(1 + 3*x)*x^2/2! + (5 + x)*(3 + 3*x)*(1 + 5*x)*x^3/3! + (7 + x)*(5 + 3*x)*(3 + 5*x)*(1 + 7*x)*x^4/4! + (9 + x)*(7 + 3*x)*(5 + 5*x)*(3 + 7*x)*(1 + 9*x)*x^5/5! + ...
%o A316704 (PARI) /* Using the biexponential series */
%o A316704 {a(n) = my(A); A = sum(m=0,n, (2*x)^m/m! * prod(k=1,m, m+1-k + k*x +x*O(x^n))); n!*polcoeff(A,n)}
%o A316704 for(n=0,30, print1(a(n),", "))
%o A316704 (PARI) /* Using Functional Equation: */
%o A316704 {a(n) = my(A=1); for(i=1,n, A = ( (1 + 2*x*A)/(1 + 2*x^2*A +x*O(x^n))^x )^(1/(1-x)) ); n!*polcoeff(A,n)}
%o A316704 for(n=0,30, print1(a(n),", "))
%Y A316704 Cf. A316370, A316705.
%K A316704 nonn
%O A316704 0,2
%A A316704 _Paul D. Hanna_, Jul 16 2018