

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
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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^(k1)*a(p) with a(p)=p+1 for p=2,3,8,10,12,13,14,15,18 (mod 19), a(p)=p1 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, (rh)/2))==1, 2, 0))}
a(n)={my(s=7*n^2); sum(i=sqrtint(s), sqrtint(s), sum(j=sqrtint(si^2), sqrtint(si^2), q(3*nij, si^2j^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



