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E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n * 3^n / n!.
4

%I #9 Jul 06 2019 09:28:45

%S 1,3,36,837,29592,1439775,90723564,7109399241,672900166584,

%T 75245901590187,9770338275393240,1452674820992915817,

%U 244491148094925021156,46131995287645828742727,9678693008639052537757380,2241968557540165237891804185,569848346606872473737714179056,158069419606634839915503628956051,47621655849844748263169576451111984,15515379326590122849811694557147948473,5445580659887211921286711773580373201820

%N E.g.f.: Sum_{n>=0} ((1+x)^n - 1)^n * 3^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 = 3.

%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 = 3, m = 1.

%H Paul D. Hanna, <a href="/A326273/b326273.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 * 3^n / n!.

%F (2) Sum_{n>=0} (1+x)^(n^2) * exp(-3*(1+x)^n) * 3^n / n!.

%e E.g.f: A(x) = 1 + 3*x + 36*x^2/2! + 837*x^3/3! + 29592*x^4/4! + 1439775*x^5/5! + 90723564*x^6/6! + 7109399241*x^7/7! + 672900166584*x^8/8! + 75245901590187*x^9/9! + 9770338275393240*x^10/10! +...

%e such that

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

%e also

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

%o (PARI) {a(n)=n!*polcoeff(sum(m=0, n, 3^m*((1+x+x*O(x^n))^m-1)^m/m!), n)}

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

%Y Cf. A192935, A326272, A326274.

%Y Cf. A326093.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jun 22 2019