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%I #25 Sep 21 2018 11:08:56
%S 0,0,1,4,19,177,1766,26868,421725,7692857,208699781,5420553787,
%T 180993613044,7075587523888,278356624078085,11601694011103611,
%U 552358618257458385,31520661477937912115,1750572856110551805720
%N a(n) = floor((Product_{k=1..n-1} prime(k))/prime(n)).
%C Every distinct prime dividing ((Product_{k=1..n-1} prime(k)) (mod prime(n))) also divides a(n).
%C Let Pn(n) = A002110(n) denote the primorial function. The number of natural numbers < Pn(n) that have prime(n+1) as a prime factor is equal to a(n). For example 19 numbers < Pn(4) = 210 have 11 as a prime factor. - _Jamie Morken_, Sep 18 2018
%H Muniru A Asiru, <a href="/A126147/b126147.txt">Table of n, a(n) for n = 1..185</a>
%p seq(floor(mul(ithprime(k),k=1..n-1)/ithprime(n)),n=1..20); # _Muniru A Asiru_, Sep 21 2018
%t f[n_] := Floor[ Product[ Prime@k, {k, n - 1}] / Prime@n]; Array[f, 19] (* _Robert G. Wilson v_, Mar 07 2007 *)
%Y Cf. A062347, A002110.
%K nonn
%O 1,4
%A _Leroy Quet_, Mar 07 2007
%E More terms from _Robert G. Wilson v_, Mar 07 2007
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