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%I #27 Jan 14 2018 04:37:30
%S 0,1,5,11,35,36,126,130,264,260,715,406,1365,952,1530,1716,3876,1830,
%T 5985,3300,5796,5500,12650,5460,15075,10556,16965,12810,31465,9920,
%U 40920,24616,34650,30192,49210,26106,82251,46740,67158,47320
%N Sum of tetrahedral numbers A000292(k), 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. Cf. A038566. See also the Apostol reference p. 133, and the Wikipedia link.
%C This is the m=3 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, and 2 are A000010(n), A023896(n) and A127415(n), respectively, where in the last two the offset for n=1 should be taken as 0 (not 1).
%D T. Apostol, Introduction to Analytic Number Theory, Springer, 1986.
%H G. C. Greubel, <a href="/A189918/b189918.txt">Table of n, a(n) for n = 1..10000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Reduced_residue_system">Reduced residue system</a>
%F a(n) = Sum_{k=0..n-1, gcd(k,n)=1 } * A000292(k), n>=1.
%F a(n) = (n*(n+2)/4!) *{n*(n+2) + mu(rad(n))*rad(n)} *phi(n)/n, n>=2, with rad(n) = A007947(n) the squarefree kernel of n, mu(n)=A008683(n), and phi(n)= A000010(n).
%F Note that phi(n)/n = A076512(n)/A109395(n) = phi(rad(n))/rad(n).
%F Proof by principle of inclusion-exclusion.
%e a(6) = A000292(1) + A000292(5)= 1 + 35 = 36.
%e a(6) = (6*8/4!)*(6*8 + 1*6)*((1/2)*(2/3)) = 36.
%e a(12) = A000292(1) + A000292(5) + A000292(7) + A000292(11) = 1 + 35 + 84 + +286 = 406.
%e a(12) = (12*14/4!)*(12*14 + 1*6)*((1/2)*(2/3)) = 406.
%p A000292 := proc(n) binomial(n+2,3) ; end proc:
%p A189918 := proc(n) local a; a := 0 ; for k from 0 to n-1 do if igcd(k,n) = 1 then a := a+A000292(k); end if; end do: a ; end proc:
%p seq(A189918(n),n=1..40) ; # _R. J. Mathar_, Jun 13 2011
%t a[n_] := Sum[ Boole[GCD[k, n] == 1]*k*(k+1)*(k+2)/6, {k, 0, n-1}]; Table[a[n], {n, 1, 40}] (* _Jean-François Alcover_, Jul 12 2012 *)
%o (PARI) a(n) = sum(k=0, n-1, if (gcd(n,k)==1, k*(k+1)*(k+2)/6)); \\ _Michel Marcus_, Feb 01 2016
%Y Cf. A000010, A038566, A076512/A109395, A023896, A127415.
%K nonn
%O 1,3
%A _Wolfdieter Lang_, May 19 2011
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