%I #15 Nov 01 2019 15:50:10
%S 2,4,6,0,10,12,0,16,18,0,22,0,0,28,30,0,0,36,0,40,42,0,46,0,0,52,0,0,
%T 58,60,0,0,66,0,70,72,0,0,78,0,82,0,0,88,0,0,0,96,0,100,102,0,106,108,
%U 0,112,0,0,0,0,0,0,126,0,130,0,0,136,138,0,0,0,0,148,150,0,0,156,0,0,162
%N a(n) = (2n)! mod n(2n+1).
%C If the zeros are removed and a 3 is inserted at the front, the first 3000 terms (or more) of the condensed sequence coincide with A039915. - _R. J. Mathar_, Mar 02 2007
%F a(n) = A000142(2n) mod A000217(2n).
%e a(4) = 0 because 8*7*6*5*4*3*2*1 / 8+7+6+5+4+3+2+1 divides evenly (0 remainder).
%t Table[Mod[(2n)!, n*(2n + 1)], {n, 85}] (* _Ray Chandler_, Aug 23 2006 *)
%Y Cf. A005097 gives indices of nonzero terms; A047845 gives indices of zero terms.
%Y Cf. A039915, A006093.
%K nonn
%O 1,1
%A _Ben Paul Thurston_, Aug 20 2006
%E Edited, corrected and extended by _R. J. Mathar_ and _Ray Chandler_, Aug 23 2006