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A278081 1/12 of the number of primitive quadruples with sum = 0 and sum of squares = 2m^2 where m = 2n-1. 7
1, 2, 6, 8, 6, 10, 14, 12, 16, 18, 16, 24, 30, 18, 30, 32, 20, 48, 38, 28, 40, 42, 36, 48, 56, 32, 54, 60, 36, 58, 62, 48, 84, 66, 48, 72, 72, 60, 80, 80, 54, 82, 96, 60, 88, 112, 64, 108, 96, 60, 102, 104, 96, 106, 110, 76, 112, 144, 84, 128, 110, 80, 150, 128 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Set b(m)=a(n) for m=2n-1, b(m)=0 for m even.

Conjecture: b(m) is multiplicative: for k>=1, b(2^k)=0; for p an odd prime, b(p*k)=p^(k-1)b(p); b(p)= p+1 for p=(5,7,13,23) (mod 24); b(p)=p-1 for p=(1,11,17,19) (mod 24); b(3)=3. It would be nice to have a proof of this.

This sequence applies also to the case sum=4n, ssq=6n^2. Generally, there is a 1-to-1 correspondence between a quadruple (h,i,j,k) with sum=rn, ssq = sn^2 and another with r'n, s'n^2, if r+r'=4, s-r=s'-r', namely (h',i',j',k')=(n,n,n,n)-(h,i,j,k).

LINKS

Andrew Howroyd, Table of n, a(n) for n = 1..500

Colin Mallows, R programs for A278081-A278086

EXAMPLE

a(2) = 2 = b(3) because of (-3,-1,2,2) and (-2,-2,1,3) (12 permutations each).

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(m=2*n-1, s=2*m^2); sum(i=-sqrtint(s), sqrtint(s), sum(j=-sqrtint(s-i^2), sqrtint(s-i^2), q(i+j, s-i^2-j^2, gcd(i, j)) ))/12} \\ Andrew Howroyd, Aug 02 2018

CROSSREFS

Cf. A278082, A278083, A278084, A278085, A278086.

Cf. A005886, A046897.

Sequence in context: A074758 A228163 A029671 * A154970 A182586 A011224

Adjacent sequences:  A278078 A278079 A278080 * A278082 A278083 A278084

KEYWORD

nonn

AUTHOR

Colin Mallows, Nov 14 2016

EXTENSIONS

Terms a(51) and beyond from Andrew Howroyd, Aug 02 2018

STATUS

approved

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Last modified November 14 20:15 EST 2019. Contains 329130 sequences. (Running on oeis4.)