A090677 - OEIS

%I #36 May 22 2024 11:33:47

%S 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,

%T 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,

%U 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

%N Number of ways to partition n into sums of squares of primes.

%C From _Hieronymus Fischer_, Nov 11 2007: (Start)

%C 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).

%C 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).

%C 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)

%D R. F. Churchouse, Representation of integers as sums of squares of primes. Caribbean J. Math. 5 (1986), no. 2, 59-65.

%H Vincenzo Librandi, <a href="/A090677/b090677.txt">Table of n, a(n) for n = 0..1000</a>

%H Christian N. K. Anderson, <a href="/A090677/a090677.txt">Table of coefficients b_i solving n=sum(b_i * prime(i)^2), for n=0..500</a>

%H Roger Woodford, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Woodford/woodford82.html">Bounds for the Eventual Positivity of Difference Functions of Partitions</a>, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.3.

%F G.f.: 1/((1-x^4)*(1-x^9)*(1-x^25)*(1-x^49)*(1-x^121)*(1-x^169)*(1-x^289)...).

%F G.f.: 1 + Sum_{i>=1} x^(prime(i)^2) / Product_{j=1..i} (1 - x^(prime(j)^2)). - _Ilya Gutkovskiy_, May 07 2017

%e a(25)=2 because 25 = 5^2 = 4*(2^2)+3^2.

%e a(83)=8 because 83 = 3^2+5^2+7^2 = 4*(2^2)+2*(3^2)+7^2

%e                    = 2*(2^2)+3*(5^2) = 6*(2^2)+3^2+2*(5^2)

%e                    = 2^2+6*(3^2)+5^2 = 10*(2^2)+2*(3^2)+5^2

%e                    = 5*(2^2)+7*(3^2) = 14*(2^2)+3*(3^2).

%t CoefficientList[ Series[ Product[1/(1 - x^Prime[i]^2), {i, 111}], {x, 0, 101}], x] (* _Robert G. Wilson v_, Sep 20 2004 *)

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

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

%K nonn

%O 0,26

%A _N. J. A. Sloane_, Dec 19 2003