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a(n)=Product[p(n)-j, j=1..n]/n!=A090114(n)/n!.
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%I #4 Oct 15 2013 22:32:21

%S 1,1,4,15,252,924,11440,43758,497420,13123110,54627300,1251677700,

%T 12033222880,52860229080,511738760544,10363194502115,197548686920970,

%U 925029565741050,17302625882942400,161884603662657876

%N a(n)=Product[p(n)-j, j=1..n]/n!=A090114(n)/n!.

%C It needs proof that A090114(n) is always divisible by n!, that is, these terms are integers.

%e n=5: p(5)=11, a(5)=(11-1)()(11-2)(11-3)(11-4)(11-5)/5!= 10.9.8.7.6/120=30240=252

%t Table[Apply[Times, Table[Prime[w]-j, {j, 1, w}]]/w!, {w, 1, 15}]

%Y Cf. A000142, A090114.

%K nonn

%O 1,3

%A _Labos Elemer_, Jan 08 2004