A216374 Number of ways to express the square of the n-th prime as the sum of four nonzero squares.

1, 0, 1, 2, 3, 5, 7, 9, 13, 20, 23, 32, 38, 42, 50, 63, 77, 83, 99, 111, 117, 137, 150, 172, 204, 221, 230, 247, 257, 275, 347, 368, 402, 414, 475, 488, 527, 567, 595, 638, 682, 698, 776, 792, 825, 842, 945, 1055, 1092, 1112, 1150, 1210, 1230, 1333, 1397, 1463, 1530, 1553, 1622, 1668, 1692, 1813, 1989, 2041

Offset: 1

Author

Mark Underwood, Sep 05 2012

Keywords

• nonn

Comments

The simple counting and the conjectured first formula agree for all the primes from 3 to 997. The counting and the conjectured second formula agree for all the primes from 5 to 997. The author of this sequence would like to know whether the formulas are already known and/or how it could be proved.

I suspect Jacobi's theorem will suffice. - Charles R Greathouse IV, Sep 30 2012

Examples

			prime(n)'s are 2, 3, 5, 7, 11, 13, 17, ... giving the sequence 1, 0, 1, 2, 3, 5, 7, ...
		

Links

• Sergey Beliy and others, Pythagorean "five"tuples and "six"tuples, digest of 9 messages in Yahoo group "Unsolved Problems in Number Theory, Logic, and Cryptography", Sep 04 2012.

Crossrefs

Cf. A025428, A001248.

Programs

PARI

forprime(p=2,1000, k=0; for(s1=1,sqrt((p^2)/4),for(s2=s1,sqrt((p^2 - s1^2)/3), for(s3=s2,sqrt((p^2-s1^2 - s2^2)/2), if(issquare(p^2-s1^2-s2^2-s3^2),k++)))) ; f = floor((p^2+4*p+24)/48.) ; f2 = (p^2 + 4*p + (19*(5*(p%48)+2)^2)%48 - 24)/48 ;                 print1([p,k,f,f2]" "))
/* code above prints [p, k, f, f2] where p is the prime, k is the number of ways the square of p can be expressed as the sum of four nonzero squares, and f and f2 are the formulas derivations. f and k are observed to be the same for p from 3 to 997; f2 and k are observed to be the same for p from 5 to 997. */

PARI

A216374(n)=sum(s1=1,.5*n=prime(n+1),my(t);sum(s2=s1,sqrtint((n^2-s1^2)\3),sum(s3=s2,sqrtint((t=n^2-s1^2-s2^2)\2),issquare(t-s3^2)))) \\ M. F. Hasler, Sep 11 2012

Formula

a(n) = floor((prime(n)^2 + 4*prime(n) + 24)/48) (conjectured for n>1).

a(n) = (prime(n)^2 + 4*prime(n) + (19*(5*(prime(n) mod 48)+2)^2) mod 48 - 24)/48 (conjectured for n>2).

a(n) = A025428(A001248(n)), where A001248(n) = A000040(n)^2 = prime(n)^2. - M. F. Hasler, Sep 10 2012