A192000 Sum of binomial numbers A000332(k+3), with k in the reduced residue system modulo n.

0, 1, 6, 16, 56, 71, 252, 296, 651, 721, 2002, 1282, 4368, 3402, 5782, 6672, 15504, 7947, 26334, 15702, 28868, 28457, 65780, 30212, 85580, 63063, 103284, 81452, 201376, 66102, 278256, 174624, 255794, 228684, 383166, 206838, 658008, 391419, 576394, 413244, 1086008

Offset

1,3

Comments

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.

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.

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).

Links

Table of n, a(n) for n=1..41.

Formula

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

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).

Example

a(6) = A000332(4) + A000292(8)= 1 + 70 = 71.
a(6) = (6/6!)*(6*3666*(1/3) + 5*137*2 - 182) = 71.
a(12) = A000332(4) + A000332(8) + A000332(10) + A000332(14) = 1 + 70 + 210 + 1001 = 1282.
a(12) = (12/6!)*(12*18258*(1/3) + 5*407*2 - 182) = 1282.

Prog

(PARI) a(n) = sum(k=0, n-1, if (gcd(n, k) == 1, binomial(k+3, 4))); \\ Michel Marcus, Feb 01 2016

Crossrefs

Cf. A189918, A023896, A127415, A189922.

Sequence in context: A175659 A221270 A316984 * A032282 A084057 A163302

Adjacent sequences: A191997 A191998 A191999 * A192001 A192002 A192003

Keyword

nonn,easy

Author

Wolfdieter Lang, Jun 22 2011

Extensions

More terms from Michel Marcus, Feb 01 2016

Status

approved