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.
LINKS
T. L. Todorova, D. I. Tolev, On the distribution of alpha p modulo one for primes p of a special form, Nov 1, 2007.
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
KEYWORD
nonn
AUTHOR
Jonathan Vos Post and John Sokol (john.sokol(AT)gmail.com), Nov 06 2007
STATUS
approved