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%I #5 Mar 31 2012 13:21:04
%S 23,39,55,64,68,73,80,84,91,96,100,105,109,113,114,118,122,123,127,
%T 131,132,136,140,140,144,145,145,149,149,153,154,156,158,160,163,164,
%U 167,168,168,172,172,176,176,176,180,180,181,181,185,185,185,189,189,190
%N Minimal number such that all greater numbers can be written as sums of squares of primes in more than n ways.
%C 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).
%C 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.
%C 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).
%C 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.
%F a(n)=min( m | A090677(j)>n for all j>m).
%e a(0)=23, since numbers >23 can be written as sum of squares of primes.
%e a(1)=39, since there are at least two ways, to write a number >39 as a sum of squares of primes.
%Y Cf. A078134, A078135, A078136, A078139, A090677, A078137, A134754.
%K nonn
%O 0,1
%A _Hieronymus Fischer_, Nov 11 2007
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