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 A278086 1/12 of the number of integer quadruples with sum = 3n and sum of squares = 7n^2. 7
 1, 1, 4, 0, 4, 4, 6, 0, 12, 4, 10, 0, 14, 6, 16, 0, 16, 12, 19, 0, 24, 10, 22, 0, 20, 14, 36, 0, 30, 16, 32, 0, 40, 16, 24, 0, 38, 19, 56, 0, 42, 24, 42, 0, 48, 22, 46, 0, 42, 20, 64, 0, 54, 36, 40, 0, 76, 30, 60, 0, 60, 32, 72, 0, 56, 40, 68, 0, 88, 24, 72, 0, 72, 38, 80, 0, 60, 56, 80, 0, 108, 42, 82, 0, 64, 42, 120, 0, 90, 48, 84, 0, 128, 46, 76, 0, 98, 42, 120, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Conjecture: a(n) is multiplicative with a(2)=1, a(2^k)=0 for k>=2, and for k>=1, p an odd prime a(p^k)=p^(k-1)*a(p) with a(p)=p+1 for p=2,3,8,10,12,13,14,15,18 (mod 19), a(p)=p-1 for p=1,4,5,6,7,9,11,16,17 (mod 19), p(19)=19.  It would be nice to have a proof of this. This sequence applies also to the case sum=n, ssq=5n^2. LINKS Andrew Howroyd, Table of n, a(n) for n = 1..500 EXAMPLE 12*a(3)=48 because of (-3,2,5,5), (-1,-1,5,6) (12 permutations each) and (-2,1,3,7) (24 permutations). PROG (PARI) q(r, s, g)={my(d=2*s - r^2); if(d<=0, d==0 && r%2==0 && gcd(g, r/2)==1, my(h); if(issquare(d, &h) && (r+h)%2==0 && gcd(g, gcd((r+h)/2, (r-h)/2))==1, 2, 0))} a(n)={my(s=7*n^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(3*n-i-j, s-i^2-j^2, gcd(i, j)) ))/12} \\ Andrew Howroyd, Aug 02 2018 CROSSREFS Cf. A278081, A278082, A278083, A278084, A278085. Cf. A046897. Sequence in context: A198414 A309721 A110854 * A021716 A258854 A258855 Adjacent sequences:  A278083 A278084 A278085 * A278087 A278088 A278089 KEYWORD nonn AUTHOR Colin Mallows, Nov 14 2016 STATUS approved

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Last modified February 22 15:34 EST 2020. Contains 332137 sequences. (Running on oeis4.)