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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A307219 a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 2, 5, 6, 2, 6, 3, 6, 5, 14, 8, 6, 5, 6, 15, 10, 6, 14, 24, 14, 6, 12, 12, 6, 16, 19, 18, 18, 36, 18, 10, 16, 20, 20, 12, 28, 18, 8, 24, 38, 27, 40, 20, 17, 30, 52, 18, 22, 26, 29, 21, 42, 31, 26, 26, 18, 44, 38, 40, 46, 26, 30, 44, 38, 36, 52, 28, 27, 38, 103, 22, 38, 78
Offset: 3

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Author

Marius A. Burtea, Mar 29 2019

Keywords

Comments

If p >= 5 is a prime number it can be written as p = 6m-1 or p = 6m+1. The identities ((6m-1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m-1)^2 and ((6m+1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m+1)^2 show that the number (p^2 + 2)/3 can be written as a sum of 3 squares of integers in at least one way.

Examples

			For n = 3, p = prime(3) = 5, (5^2+2)/3 = 9 = 2^2 + 2^2 + 1^2, so a(3) = 1.
For n = 9, p = prime(9) = 23, (23^2+2)/3 = 177 = 13^2 + 2^2 + 2^2 = 8^2 + 8^2 + 7^2, so a(9) = 2.
For n = 17, p = prime(17) = 59, (59^2+2)/3 = 1161 = 34^2 + 2^2 + 1^2 = 33^2 + 6^2 + 6^2 = 24^2 + 11^2 + 4^2 = 31^2 + 14^2 + 2^2 = 31^2 + 10^2 + 10^2 = 30^2 + 15^2 + 6^2 = 29^2 + 16^2 + 8^2 = 28^2 + 19^2 + 4^2 = 28^2 + 16^2 + 11^2 = 26^2 + 22^2 + 1^2 = 26^2 + 17^2 + 14^2 = 24^2 + 24^2 + 3^2 = 24^2 + 21^2 + 12^2 = 20^2 + 20^2 + 19^2, so a(17) = 14.
		

References

  • Ion Cucurezeanu, Pătrate și cuburi perfecte de numere întregi (Squares and perfect cubes of integer numbers), Ed. Gil., Zalău, 2007, ch. 1, p. 21, pr. 166.
  • Laurențiu Panaitopol, Dinu Șerbănescu, Number theory and combinatorial problems for juniors, Ed. Gil, Zalău, (2003), ch. 1, p. 5, pr. 4. (in Romanian).

Crossrefs

Programs

  • Magma
    [#RestrictedPartitions(Floor((p*p+2)/3),3,{d*d:d in [1..p]}): p in PrimesInInterval(5,500) ];
    
  • PARI
    a(n)={k=(prime(n+2)^2+2)/3;sum(a=1, k, sum(b=1, a, sum(c=1, b, a^2+b^2+c^2==k)));} \\ Jinyuan Wang, Mar 30 2019