|
| |
|
|
A216374
|
|
Number of ways to express the square of the n-th prime as the sum of four nonzero squares.
|
|
3
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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.
|
|
|
LINKS
|
|
|
|
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).
|
|
|
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
|
|
|
|
STATUS
|
approved
|
| |
|
|
