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Binomial(n,4) mod n.
0

%I #6 Mar 30 2012 17:40:22

%S 0,0,0,1,0,3,0,6,0,0,0,3,0,7,0,12,0,0,0,5,0,11,0,18,0,0,0,7,0,15,0,24,

%T 0,0,0,9,0,19,0,30,0,0,0,11,0,23,0,36,0,0,0,13,0,27,0,42,0,0,0,15,0,

%U 31,0,48,0,0,0,17,0,35,0,54,0,0,0,19,0,39,0,60,0,0,0,21,0,43,0,66,0,0,0,23,0

%N Binomial(n,4) mod n.

%C Any patterns?

%F Apparently a(n)= +2*a(n-8) -a(n-16). [_R. J. Mathar_, Dec 12 2010]

%t Table[Mod[Binomial[n,4],n],{n,150}]

%Y Cf. A000332.

%K nonn

%O 1,6

%A _Zak Seidov_, Aug 07 2010