OFFSET
1,4
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
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.
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
EXAMPLE
prime(n)'s are 2, 3, 5, 7, 11, 13, 17, ... giving the sequence 1, 0, 1, 2, 3, 5, 7, ...
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Mark Underwood, Sep 05 2012
STATUS
approved