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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]

Numbers such that A078134(n)=0.

"Numbers which cannot be written as sum of squares > 1" is equivalent to "Numbers which cannot be written as sum of squares of primes." Equivalently, numbers which can be written as the sum of nonzero squares can also be written as sum of the squares of primes." cf. A090677 = number of ways to partition n into sums of squares of primes. - Jonathan Vos Post, Sep 20 2006

The sequence is finite with a(12)=23 as last member. Proof: When k=a^2+b^2+..., k+4 = 2^2+a^2+b^2+... If k can be written as sum of the squares of primes, k+4 also has this property. As 24,25,26,27 have the property, by induction, all numbers > 23 can be written as sum of squares>1. - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Apr 07 2007

Also, numbers which cannot be written as sum of squares of 2 and 3 (see A078137 for the proof). 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 constructively. - Hieronymus Fischer, Nov 11 2007

Also numbers n such that every integer partition of n contains a squarefree number. For example, 21 does not belong to the sequence because there are integer partitions of 21 containing no squarefree numbers, namely: (12,9), (9,8,4), (9,4,4,4). - Gus Wiseman, Dec 14 2018

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)

A078134(a(n))>0.

Numbers which can be written as a sum of squares of primes. - Hieronymus Fischer, Nov 11 2007

Equivalently, numbers which can be written as a sum of squares of 2 and 3. Proof for numbers m>=24: if m=4*(k+6), k>=0, then m=(k+6)*2^2; if m=4*(k+6)+1 than m=(k+4)*2^2+3^2; if m=4*(k+6)+2 then m=(k+2)*2^2+2*3^2; if m=4*(k+6)+3 then m=k*2^2+3*3^2. Clearly, the numbers a(n)<24 can also be written as sums of squares of 2 and 3. Explicit representation as a sum of squares of 2 and 3 for numbers m>23: m=c*2^2+d*3^2, where c:=(floor(m/4) - 2*(m mod 4))>=0 and d:=m mod 4. - Hieronymus Fischer, Nov 11 2007

A078134(a(n))=1.

The sequence is finite with a(15)=39 as last term, since numbers m>39 can be represented as sums of squares>1 (even of squares of primes and even of squares of 2, 3 and 4 and even of squares of 2, 3 and 5) in at least two ways. Proof: if m=40+4k, k>=0, then m=(k+10)*2^2=(k+1)*2^2+4*3^2; if m=41+4k, then m=(k+8)*2^2+3^2=(k+4)*2^2+5^2; if m=42+4k, then m=(k+6)*2^2+2*3^2=(k+2)*2^2+3^2+5^2; if m=43+4k, then m=(k+4)*2^2+3*3^2=k*2^2+2*3^2+5^2. - Hieronymus Fischer, Nov 11 2007

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.

Sequence is finite, see A078129.

By Sylvester's solution to the Frobenius problem, all integers greater than 4*9 - 4 - 9 = 23 can be represented as a sum of multiples of 4 and 9. Hence all primes except 2,3,5,7,11,19,23 are in this sequence. [Charles R Greathouse IV, Apr 19 2010]