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A090677
Number of ways to partition n into sums of squares of primes.
19
1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 3, 2, 2, 2, 4, 3, 2, 3, 4, 4, 2, 3, 4, 5, 3, 3, 5, 5, 4, 3, 5, 5, 5, 4, 5, 6, 5, 5, 5, 7, 6, 6, 6, 7, 7, 6, 7, 7, 8, 7, 8, 8, 8, 8, 8, 9, 8, 9, 9, 10, 9, 9, 10, 11, 11, 10, 11
OFFSET
0,26
COMMENTS
From Hieronymus Fischer, Nov 11 2007: (Start)
First statement of monotony: a(n+p^2)>=a(n) for all primes p. Proof: we restrict ourselves on a(n)>0 (the case a(n)=0 is trivial). Let T(i), 1<=i<=a(n), be the a(n) different sums of squares of primes representing n. Then, adding p^2 to those expressions, we get a(n) sums of squares of primes T(i)+p^2, obviously representing n+p^2, thus a(n+p^2) cannot be less than a(n).
Second statement of monotony: a(n+m)>=max(a(n),a(m)) for all m with a(m)>1. Proof: let T(i), 1<=i<=a(n), be the a(n) different sums of squares of primes representing n; let S(i), 1<=i<=a(m), be the a(m) different sums of squares of primes representing m. Then, adding these expressions, we get a(n) sums of squares of primes T(i)+S(1), representing n+m, further we get a(m) sums T(1)+S(i), also representing n+m. Thus a(n+m) cannot be less than the maximum of a(n) and a(m).
The minimum b(k):=min( n | a(j)>k for all j>n) exists for all k>=0. See A134755 for that sequence representing b(k). (End)
REFERENCES
R. F. Churchouse, Representation of integers as sums of squares of primes. Caribbean J. Math. 5 (1986), no. 2, 59-65.
LINKS
Roger Woodford, Bounds for the Eventual Positivity of Difference Functions of Partitions, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.3.
FORMULA
G.f.: 1/((1-x^4)*(1-x^9)*(1-x^25)*(1-x^49)*(1-x^121)*(1-x^169)*(1-x^289)...).
G.f.: 1 + Sum_{i>=1} x^(prime(i)^2) / Product_{j=1..i} (1 - x^(prime(j)^2)). - Ilya Gutkovskiy, May 07 2017
EXAMPLE
a(25)=2 because 25 = 5^2 = 4*(2^2)+3^2.
a(83)=8 because 83 = 3^2+5^2+7^2 = 4*(2^2)+2*(3^2)+7^2
= 2*(2^2)+3*(5^2) = 6*(2^2)+3^2+2*(5^2)
= 2^2+6*(3^2)+5^2 = 10*(2^2)+2*(3^2)+5^2
= 5*(2^2)+7*(3^2) = 14*(2^2)+3*(3^2).
MATHEMATICA
CoefficientList[ Series[ Product[1/(1 - x^Prime[i]^2), {i, 111}], {x, 0, 101}], x] (* Robert G. Wilson v, Sep 20 2004 *)
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Dec 19 2003
STATUS
approved