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A192000
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Sum of binomial numbers A000332(k+3), with k in the reduced residue system modulo n.
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0
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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
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OFFSET
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1,3
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
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EXAMPLE
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a(6) = (6/6!)*(6*3666*(1/3) + 5*137*2 - 182) = 71.
a(12) = (12/6!)*(12*18258*(1/3) + 5*407*2 - 182) = 1282.
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PROG
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(PARI) a(n) = sum(k=0, n-1, if (gcd(n, k) == 1, binomial(k+3, 4))); \\ Michel Marcus, Feb 01 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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