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A132203
Number of order-independent ways to represent 24*n+5 as the sum of squares of exactly 5 primes.
0
0, 0, 0, 1, 1, 2, 1, 2, 2, 3, 3, 3, 2, 4, 3, 5, 3, 5, 6, 5, 6, 6, 5, 8, 6, 9, 6, 7, 10, 8, 9, 9, 8, 10, 8, 11, 8, 8, 13, 11, 10, 11, 11, 14, 10, 14, 13, 9, 17, 13, 12, 15, 13, 17, 11, 15, 17, 10, 17, 17, 14, 17, 16, 19, 12, 17, 19, 13, 18, 17, 14, 17, 17, 23, 16
OFFSET
0,6
COMMENTS
Hua proved in 1938 that every sufficiently large integer n congruent to 5 mod 24 can be written as the sum of the squares of exactly 5 primes.
REFERENCES
L. K. Hua, Some results in the additive prime number theory, Quart J. Math., Oxford, 9 (1938) 68-80.
EXAMPLE
a(3) = 1 because the only way, up to permutation, to represent 24*n+5 as the sum of squares of exactly 5 primes is 77 = 5 + 24*3 = 5^2 + 5^2 + 3^2 + 3^2 + 3^2.
a(5) = 2 because 125 = 5 + 24*5 = 5^2 + 5^2 + 5^2 + 5^2 + 5^2 = 7^2 + 7^2 + 3^2 + 3^2 + 3^2.
a(9) = 3 because 221 = 5 + 24*9 = 11^2 + 5^2 + 5^2 + 5^2 + 5^2 = 13^2 + 5^2 + 3^2 + 3^2 + 3^2 = 7^2 + 7^2 + 7^2 + 7^2 + 5^2.
a(13) = 4 because 317 = 5 + 24*13 = 11^2 + 11^2 + 5^2 + 5^2 + 5^2 = 11^2 + 7^2 + 7^2 + 7^2 + 7^2 = 13^2 + 11^2 + 3^2 + 3^2 + 3^2 = 13^2 + 7^2 + 7^2 + 5^2 + 5^2.
a(15) = 5 because 365 = 5 + 24*15 = 11^2 + 11^2 + 7^2 + 7^2 + 5^2 = 13^2 + 11^2 + 5^2 + 5^2 + 5^2 = 13^2 + 13^2 + 3^2 + 3^2 + 3^2 = 13^2 + 7^2 + 7^2 + 7^2 + 7^2 = 17^2 + 7^2 + 3^2 + 3^2 + 3^2.
a(18) = 6 because 437 = 5 + 24*18 = 11^2 + 11^2 + 11^2 + 7^2 + 5^2 = 13^2 + 11^2 + 7^2 +7^2 + 7^2 = 13^2 + 13^2 + 7^2 + 5^2 + 5^2 = 17^2 + 11^2 + 3^2 + 3^2 + 3^2 = 17^2 + 7^2 + 7^2 + 5^2 + 5^2 = 19^2 + 7^2 + 3^2 + 3^2 + 3^2 = 19^2 + 7^2 + 3^2 + 3^2 + 3^2.
a(23) = 8 because 557 = 5 + 24*23 = 13^2 + 11^2 + 11^2 + 11^2 + 5^2 = 13^2 + 13^2 + 11^2 + 7^2 + 7^2 = 13^2 + 13^2 + 13^2 + 5^2 + 5^2 = 17^2 + 11^2 + 7^2 + 7^2 + 7^2 = 17^2 + 13^2 + 7^2 + 5^2 + 5^2 = 19^2 + 11^2 + 5^2 + 5^2 + 5^2 = 19^2 + 13^2 + 3^2 + 3^2 + 3^2 = 19^2 + 7^2 + 7^2 + 7^2 + 7^2.
CROSSREFS
Sequence in context: A126237 A243164 A333708 * A158925 A262868 A342097
KEYWORD
nonn
AUTHOR
Jonathan Vos Post and John Sokol (john.sokol(AT)gmail.com), Nov 06 2007
STATUS
approved