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a(A078135(n))=0; a(A078136(n))=1; a(A078137(n))>0;

Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078135(12)=23 and b(1)=A078136(15)=39. This is true - see comments by Hieronymus Fischer.

Also first difference of A001156 (number of partitions of n into squares). - Wouter Meeussen, Oct 22 2005

Comments from Hieronymus Fischer, Nov 11 2007 (Start): First statement of monotony: a(n+k^2)>=a(n) for all k>1. 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 sum expressions of squares >1 representing n. Then, adding k^2 to those expressions, we get a(n) sums of squares T(i)+k^2, obviously representing n+k^2, thus a(n+k^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 sum expressions of squares >1 representing n; let S(i), 1<=i<=a(m), be the a(m) different sum expressions of squares >1 representing m. Then, adding those expressions, we get a(n) sums of squares 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 author's conjecture holds true. Proof by induction: b(0) exists; if b(k) exists, then a(j)>k for all j>b(k). Setting m:=b(k)+1, we find that there are k+1 sums B(0,i) of squares >1, 1<=i<=k+1, with m=B(0,i). Further there are k+1 such sum expressions B(1,i), B(2,i) and B(3,i), 1<=i<=k+1, representing m+1, m+2 and m+3, respectively. For n>b(k) we have n=m+4*floor((n-m)/4)+(n-m) mod 4.

Thus n=m+r+s*2^2, where r=0,1,2 or 3. Hence n can be written B(r,i)+s*2^2 and there are k+1 such representations. Let q be the maximal number (to be squared) occurring as a term within those sum expressions B(r,i), 0<=r<=3,1<=i<=k+1. We select a number p>q and we set c:=b(k)+p^2. For n>c, we have the k+1 representations B(r(n),i)+s(n)*2^2.

Additionally, for n-p^2 (which is >b(k)) there are also k+1 representations B(r_p,i)+s_p*2^2, where r_p:=r(n-p^2), s_p:=s(n-p^2). Thus n can be written B(r(n),i)+s(n)*2^2, 1<=i<=k+1 and B(r_p,i)+s_p*2^2+p^2, 1<=i<=k+1.

By choice of p all these sum representations of n are different, which implies, that there are 2k+2 such representations. It follows a(n)>2k+2>k+1 for all n>c, which implies, that b(k+1) exists.

A more precise formulation of the author's conjecture is "b(k):=min( n | a(j)>k for all j>n) exists for all k>=0". (End)

A033183(n) <= a(n). [From Reinhard Zumkeller, Nov 07 2009]

From Hieronymus Fischer, Nov 11 2007: (Start)

Equivalently, prime numbers which cannot be written as sum of squares of primes (see A078137 for the proof).

Equivalently, prime numbers which cannot be written as sum of squares of 2 and 3 (see A078137 for the proof).

The sequence is finite, since numbers > 23 can be written as sums of squares >1 (see A078135).

Explicit representation as sum of squares of primes, or rather of squares of 2 and 3, for numbers m>23: we have m=c*2^2+d*3^2, where c:=(floor(m/4) - 2*(m mod 4))>=0, d:=m mod 4. For that, the finiteness of the sequence is proved. (End)

Equivalent to primes which can be written as the sum of cubes of primes; the question being "what is the minimum number of terms in such sums when they can be written in more than one way? - Jonathan Vos Post, Sep 21 2006

Mikawa and Peneva: "One of the famous and still unsettled problems in additive prime number theory is the conjecture that every sufficiently large integer satisfying some natural congruence conditions, can be written as the sum of four cubes of primes. Although the present methods lack the power to prove such a strong result, Hua... has been able to prove that every sufficiently large odd integer as the sum of nine cubes of primes. He also established that almost all integers {n == 1 mod 2, n =/= 0, +/-2 mod 9, n =/= 0 mod 7} can be expressed as the sum of five cubes of primes." - Jonathan Vos Post, Sep 21 2006