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Sum of binomial numbers A000332(k+3), with k in the reduced residue system modulo n.
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%I #10 Feb 01 2016 11:37:56

%S 0,1,6,16,56,71,252,296,651,721,2002,1282,4368,3402,5782,6672,15504,

%T 7947,26334,15702,28868,28457,65780,30212,85580,63063,103284,81452,

%U 201376,66102,278256,174624,255794,228684,383166,206838,658008,391419,576394,413244,1086008

%N Sum of binomial numbers A000332(k+3), with k in the reduced residue system modulo n.

%C The reduced residue system modulo n used here is the set of numbers k from the set {0,1,...,n-1} which satisfy gcd(k,n)=1. There are phi(n) = A000010(n) such numbers k.

%C This is the m=4 member of a family of sequences, call them rmnS(m) (reduced mod n sum), with entries rmnS(m;n):=sum(binomial(k+m-1,m),0<=k<=n-1 with gcd(k,n)=1), m>=0, n>=1. Recall gcd(0,n)=n.

%C The members for m=0, 1, 2 and 3 are A000010, A023896, A127415, and A189918, respectively, where in the m=1 and 2 cases the offset for n=1 should be taken as 0 (not 1).

%F a(n) = sum(A000332(k+3), 0<=k<=n-1, gcd(k,n)=1), n>=1.

%F a(n) = (n/6!)*(n*(6*n^3+45*n^2+110*n+90)*P(1,n) + 5*(2*n^2+9*n+11)*P(-1,n) - P(-3,n)), n>=2, with P(k,n):= J(k,n)/n^k, where J(k,n) is the Jordan function (see A000010, A007434, A059376 - A059378, A069091 - A069095).

%e a(6) = A000332(4) + A000292(8)= 1 + 70 = 71.

%e a(6) = (6/6!)*(6*3666*(1/3) + 5*137*2 - 182) = 71.

%e a(12) = A000332(4) + A000332(8) + A000332(10) + A000332(14) = 1 + 70 + 210 + 1001 = 1282.

%e a(12) = (12/6!)*(12*18258*(1/3) + 5*407*2 - 182) = 1282.

%o (PARI) a(n) = sum(k=0, n-1, if (gcd(n,k) == 1, binomial(k+3, 4))); \\ _Michel Marcus_, Feb 01 2016

%Y Cf. A189918, A023896, A127415, A189922.

%K nonn,easy

%O 1,3

%A _Wolfdieter Lang_, Jun 22 2011

%E More terms from _Michel Marcus_, Feb 01 2016