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a(n) = denominator of (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)!.
3

%I #25 May 01 2021 08:21:18

%S 1,3,120,2520,51840,2494800,6227020800,653837184000,27360571392000,

%T 30411275102208000,51090942171709440000,1846572624206069760000,

%U 15511210043330985984000000,1361108681302294020096000000,8841761993739701954543616000000

%N a(n) = denominator of (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)!.

%D O. Furdui, Limits, Series and Fractional Part Integrals. Problems in Mathematical Analysis, Springer, New York, 2013. See Problem 3.114 and 3.118.

%H Seiichi Manyama, <a href="/A343831/b343831.txt">Table of n, a(n) for n = 1..225</a>

%H Math StackExchange, <a href="https://math.stackexchange.com/questions/2344699/compute-s-n-sum-limits-a-1-a-2-cdots-a-n-1-infty-fraca-1a-2-cdots-a-n?rq=1">Compute S_n = Sum_{a_1 a_2 ... a_n >=1} a_1 a_2 ... a_n/(a_1+a_2+...+a_n)!</a>

%F b(n) = (1/e) * Sum_{a_1>=1, a_2>=1, ... , a_n>=1} a_1 * a_2 * ... * a_n / (a_1 + a_2 + ... + a_n)! = Sum_{j=0..n} (-1)^(n+j-1) * binomial(n,j) * Sum_{k=0..n+j-1} (-1)^k/k! = Sum_{k=0..n-1} binomial(n-1,k)/(k+n)!.

%F a(n) = denominator of b(n).

%e 1, 2/3, 31/120, 179/2520, 787/51840, 6631/2494800, 2456299/6227020800, ...

%t a[n_] := Denominator @ Sum[Binomial[n - 1, k]/(k + n)!, {k, 0, n - 1}]; Array[a, 20] (* _Amiram Eldar_, May 01 2021 *)

%o (PARI) a(n) = denominator(sum(j=0, n, (-1)^(n+j-1)*binomial(n, j)*sum(k=0, n+j-1, (-1)^k/k!)));

%o (PARI) a(n) = denominator(sum(k=0, n-1, binomial(n-1, k)/(k+n)!));

%Y Cf. A343830 (numerator), A343832.

%K nonn,frac

%O 1,2

%A _Seiichi Manyama_, Apr 30 2021