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E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n * 4^n / n!.
4

%I #9 Jul 06 2019 09:29:51

%S 1,4,64,1920,86464,5304320,418131456,40727959552,4765747597312,

%T 655794545577984,104360850604687360,18948720298674028544,

%U 3882059495694122090496,889053986706845142876160,225799026538694916941283328,63163063632830911303738982400,19344290761718462120859544846336,6452149866509553556278434299117568,2332867461867950308492384248149311488,910538103145382496893587688740637114368,382208425560563535419125500691963382333440

%N E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n * 4^n / n!.

%C More generally, the following sums are equal:

%C (1) Sum_{n>=0} (q^n + p)^n * r^n/n!,

%C (2) Sum_{n>=0} q^(n^2) * exp(p*q^n*r) * r^n/n!;

%C here, q = (1+x) and p = -1, r = 4.

%C In general, let F(x) be a formal power series in x such that F(0)=1, then

%C Sum_{n>=0} m^n * F(q^n*r)^p * log( F(q^n*r) )^n / n! =

%C Sum_{n>=0} r^n * [y^n] F(y)^(m*q^n + p);

%C here, F(x) = exp(x), q = 1+x, p = -1, r = 4, m = 1.

%H Paul D. Hanna, <a href="/A326274/b326274.txt">Table of n, a(n) for n = 0..300</a>

%F E.g.f. may be expressed by the following sums.

%F (1) Sum_{n>=0} ((1+x)^n - 1)^n * 4^n / n!.

%F (2) Sum_{n>=0} (1+x)^(n^2) * exp(-4*(1+x)^n) * 4^n / n!.

%e E.g.f: A(x) = 1 + 4*x + 64*x^2/2! + 1920*x^3/3! + 86464*x^4/4! + 5304320*x^5/5! + 418131456*x^6/6! + 40727959552*x^7/7! + 4765747597312*x^8/8! + 655794545577984*x^9/9! + 104360850604687360*x^10/10! +...

%e such that

%e A(x) = 1 + 4*((1+x) - 1) + 4^2*((1+x)^2 - 1)^2/2! + 4^3*((1+x)^3 - 1)^3/3! + 4^4*((1+x)^4 - 1)^4/4! + 4^5*((1+x)^5 - 1)^5/5! + 4^6*((1+x)^6 - 1)^6/6! + 4^7*((1+x)^7 - 1)^7/7! + ...

%e also

%e A(x) = 1 + 4*(1+x)*exp(-4*(1+x)) + 4^2*(1+x)^4*exp(-4*(1+x)^2)/2! + 4^3*(1+x)^9*exp(-4*(1+x)^3)/3! + 4^4*(1+x)^16*exp(-4*(1+x)^4)/4! + 4^5*(1+x)^25*exp(-4*(1+x)^5)/5! + 4^6*(1+x)^36*exp(-4*(1+x)^6)/6! + 4^7*(1+x)^49*exp(-4*(1+x)^7)/7! + ...

%o (PARI) {a(n)=n!*polcoeff(sum(m=0, n, 4^m*((1+x+x*O(x^n))^m-1)^m/m!), n)}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A192935, A326272, A326273.

%Y Cf. A326094.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jun 22 2019