A090677 Number of ways to partition n into sums of squares of primes.

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Christian N. K. Anderson, Table of coefficients b_i solving n=sum(b_i * prime(i)^2), for n=0..500

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 *)

Crossrefs

Cf. A111900, A001248, A001156, A023893, A111901.

Cf. A078134, A078135, A078136, A078139, A134600, A078137, A134754, A134755.

Sequence in context: A353835 A033183 A351742 * A161097 A105240 A327499

Adjacent sequences: A090674 A090675 A090676 * A090678 A090679 A090680

Keyword

nonn

Author

N. J. A. Sloane, Dec 19 2003

Status

approved