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%I #41 Nov 22 2023 11:04:28
%S 1,0,1,2,3,5,7,9,13,20,23,32,38,42,50,63,77,83,99,111,117,137,150,172,
%T 204,221,230,247,257,275,347,368,402,414,475,488,527,567,595,638,682,
%U 698,776,792,825,842,945,1055,1092,1112,1150,1210,1230,1333,1397,1463,1530,1553,1622,1668,1692,1813,1989,2041
%N Number of ways to express the square of the n-th prime as the sum of four nonzero squares.
%C 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.
%C I suspect Jacobi's theorem will suffice. - _Charles R Greathouse IV_, Sep 30 2012
%H Sergey Beliy and others, <a href="/A216374/a216374.txt">Pythagorean "five"tuples and "six"tuples</a>, digest of 9 messages in Yahoo group "Unsolved Problems in Number Theory, Logic, and Cryptography", Sep 04 2012.
%F a(n) = floor((prime(n)^2 + 4*prime(n) + 24)/48) (conjectured for n>1).
%F a(n) = (prime(n)^2 + 4*prime(n) + (19*(5*(prime(n) mod 48)+2)^2) mod 48 - 24)/48 (conjectured for n>2).
%F a(n) = A025428(A001248(n)), where A001248(n) = A000040(n)^2 = prime(n)^2. - _M. F. Hasler_, Sep 10 2012
%e prime(n)'s are 2, 3, 5, 7, 11, 13, 17, ... giving the sequence 1, 0, 1, 2, 3, 5, 7, ...
%o (PARI)
%o 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]" "))
%o /* 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. */
%o (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
%Y Cf. A025428, A001248.
%K nonn
%O 1,4
%A _Mark Underwood_, Sep 05 2012
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