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A307219
a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.
0
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
3,3
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.
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).
EXAMPLE
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.
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius A. Burtea, Mar 29 2019
STATUS
approved