OFFSET
0,1
COMMENTS
The sequence is well-defined, in that a(n) exists for all n>=0. Proof by induction: a(0) exists. We set b(j):=number of ways to write j as sum of squares of primes (=A090677). If a(n) exists, then b(j)>n for all j>a(n). Setting m:=a(n)+1, we find that there are n+1 sum of squares of primes B(0,i), 1<=i<=n+1, with m=B(0,i).
Further there are n+1 such sum expressions B(1,i), B(2,i) and B(3,i), 1<=i<=n+1, representing m+1, m+2 and m+3, respectively. For all j>a(n) we have j=m+4*floor((j-m)/4)+(j-m) mod 4. Thus j=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 n+1 such representations.
Let q be the maximal prime number (to be squared) occurring as a term within those sum expressions B(r,i), 0<=r<=3,1<=i<=n+1. We select a prime number p>q and we set c:=a(n)+p^2. For j>c, we have the n+1 representations B(r(j),i)+s(j)*2^2. Additionally, for j-p^2 (which is >a(n)) there are also n+1 representations B(r_p,i)+s_p*2^2, where r_p:=r(j-p^2), s_p:=s(j-p^2).
Thus j can be written B(r(j),i)+s(j)*2^2, 1<=i<=n+1 and B(r_p,i)+s_p*2^2+p^2, 1<=i<=n+1. By choice of p all these sum representations of j are different, which implies, that there are 2n+2 such representations. It follows b(j)>2n+2>n+1 for all j>c, which implies, that a(n+1) exists.
FORMULA
a(n)=min( m | A090677(j)>n for all j>m).
EXAMPLE
a(0)=23, since numbers >23 can be written as sum of squares of primes.
a(1)=39, since there are at least two ways, to write a number >39 as a sum of squares of primes.
CROSSREFS
KEYWORD
nonn
AUTHOR
Hieronymus Fischer, Nov 11 2007
STATUS
approved