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(n!)^2 mod Pt(n), where Pt(n) is product of first n positive triangular numbers (A000217).
0

%I #6 Dec 04 2013 17:41:30

%S 0,1,0,36,900,8100,0,25401600,514382400,12859560000,6224027040000,

%T 56016243360000,9466745127840000,1855482045056640000,0,

%U 6679735362203904000000,13513104637738497792000000,156365925093831188736000000,225792395835492236534784000000,22579239583549223653478400000000

%N (n!)^2 mod Pt(n), where Pt(n) is product of first n positive triangular numbers (A000217).

%C Indices of zeros appear to be 2^k-1.

%e a(4) = 1*4*9*16 mod 1*3*6*10 = 576 mod 90 = 36.

%o (Python)

%o s=t=1

%o for n in range(1,33):

%o s*=n*n

%o t*=n*(n+1)/2

%o print str(s%t)+',',

%Y Cf. A000142, A000217, A000290, A001044(n!^2).

%Y Cf. A006472 (triangular factorial, essentially equal to Pt(n)).

%Y Cf. A006788 (floor(n!^2/Pt)).

%K nonn

%O 1,4

%A _Alex Ratushnyak_, Dec 03 2013