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A307219
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a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.
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0
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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
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OFFSET
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3,3
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COMMENTS
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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.
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REFERENCES
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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).
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LINKS
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EXAMPLE
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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.
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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