%I A090494
%S A090494 1,1,8,7776,1146617856,1289945088000000000,46798828032806092800000000000,
%T A090494 2350577043461005964030008507760640000000000000,82062624596364021632633836764627761035757255393280000000000000\
               00,
%U A090494 2746781358330240881921653545637784861521126603512175621574459373964492800000000000000000
%N A090494 Product_{j=1..n} Product_{k=1..n} lcm(j,k).
%F A090494 Let p be a prime and let ordp(n,p) denote the exponent of the largest 
               power of p which divides n. For example, ordp(48,2)=4 since 48 = 
               3*(2^4). Then the prime factorization of a(n) appears to be given 
               by the formula ordp(a(n),p)= sum_{k >= 1} [(2*(p^k)-1)*floor((n/(p^k)))^2] 
               + 2*sum_{k >= 1} [floor(n/(p^k))*mod(n,p^k)], for each prime p. See 
               the comments sections of A092143, A092287, A129365 and A129454 for 
               similar conjectural prime factorizations. - Peter Bala (pbala(AT)toucansurf.com), 
               Apr 23 2007
%p A090494 f := n->mul(mul(lcm(j,k),k=1..n),j=1..n);
%Y A090494 Cf. A018806, A090494.
%Y A090494 Cf. A092143, A092287, A129365, A129454.
%Y A090494 Sequence in context: A079189 A114133 A115442 this_sequence A079656 A114773 
               A057107
%Y A090494 Adjacent sequences: A090491 A090492 A090493 this_sequence A090495 A090496 
               A090497
%K A090494 nonn
%O A090494 0,3
%A A090494 N. J. A. Sloane (njas(AT)research.att.com), Feb 03 2004