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A090677
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Number of ways to partition n into sums of squares of primes.
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19
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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
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OFFSET
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0,26
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COMMENTS
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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)
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REFERENCES
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R. F. Churchouse, Representation of integers as sums of squares of primes. Caribbean J. Math. 5 (1986), no. 2, 59-65.
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LINKS
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FORMULA
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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
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MATHEMATICA
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CoefficientList[ Series[ Product[1/(1 - x^Prime[i]^2), {i, 111}], {x, 0, 101}], x] (* Robert G. Wilson v, Sep 20 2004 *)
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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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