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a(n) = n!*Sum_{k=1..n} (k-1)^k/k!.
3

%I #16 Jul 15 2023 10:35:38

%S 0,0,1,11,125,1649,25519,458569,9433353,219117905,5677963451,

%T 162457597961,5087919552253,173136159558361,6361282619516343,

%U 250987334850557369,10584205713321808529,475079402305823570849,22614513693572549266291,1137911105533216112417161

%N a(n) = n!*Sum_{k=1..n} (k-1)^k/k!.

%C Perhaps the largest possible number of ways of choosing (v1, v2, ..., vn), possibly with repetition, from {b1, b2, ..., bn} with b1 < b2 < ... < bn, such that v1 + v2 + ... + vn < b1 + b2 + ... + bn. Clearly the actual number of ways depends on the particular values of {b1, b2, ..., bn}, but {1, n, n^2, ..., n^(n-1)} produces this result for the number of sums strictly less than (n^n-1)/(n-1) = A023037(n).

%H Seiichi Manyama, <a href="/A076483/b076483.txt">Table of n, a(n) for n = 0..386</a>

%F Limit_{n->oo} a(n)/(e*a(n-1)) - n = -1/2.

%F Limit_{n->oo} a(n)/n^n = 1/(e-1).

%e a(4) = 4!*(0^1/1! + 1^2/2! + 2^3/3! + 3^4/4!) = 0 + 12 + 32 + 81 = 125.

%t Table[n! Sum[(k-1)^k/k!, {k,n}], {n,0,17}] (* _Stefano Spezia_, Sep 11 2022 *)

%o (PARI) a(n) = n!*sum(k=1, n, (k-1)^k/k!); \\ _Seiichi Manyama_, Jul 15 2023

%Y Row sums of A076482.

%Y Cf. A023037, A075473.

%K nonn

%O 0,4

%A _Henry Bottomley_, Oct 14 2002