cp's OEIS Frontend

Number of partitions of n such that number of parts equal to k is multiple of k for all k. - Vladeta Jovovic, Aug 01 2004

Of course p_{4*square}(n)>0. In fact p_{4*square}(32n+28)=3 times p_{4*square}(8n+7) and p_{4*square}(72n+69) is even. These seem to be the only arithmetic properties the function p_{4*square(n)} possesses. Similar results hold for partitions into positive squares, distinct squares and distinct positive squares. - Michael David Hirschhorn, May 05 2005

The Heinz numbers of these partitions are given by A324588. - Gus Wiseman, Mar 09 2019

"WEIGH" transform of squares A000290.

a(n) = 0 for n in {A001422}, a(n) > 0 for n in {A003995}. - Alois P. Heinz, May 14 2014

Number of partitions of n in which each part i has multiplicity i. Example: a(50)=3 because we have [1,2,2,3,3,3,6,6,6,6,6,6], [1,7,7,7,7,7,7,7], and [3,3,3,4,4,4,4,5,5,5,5,5]. - Emeric Deutsch, Jan 26 2016

The Heinz numbers of integer partitions into distinct pairs are given by A324587. - Gus Wiseman, Mar 09 2019

From Gus Wiseman, Mar 09 2019: (Start)

Equivalent to Emeric Deutsch's comment, a(n) is the number of integer partitions of n where the multiplicities (where if x < y the multiplicity of x is counted prior to the multiplicity of y) are equal to the distinct parts in increasing order. The Heinz numbers of these partitions are given by A109298. For example, the first 30 terms count the following integer partitions:

1: (1)

4: (22)

5: (221)

9: (333)

10: (3331)

13: (33322)

14: (333221)

16: (4444)

17: (44441)

20: (444422)

21: (4444221)

25: (55555)

25: (4444333)

26: (555551)

26: (44443331)

29: (5555522)

29: (444433322)

30: (55555221)

30: (4444333221)

The case where the distinct parts are taken in decreasing order is A324572, with Heinz numbers given by A324571.

(End)

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]

a(A078135(n)) = 0; a(A078137(n)) > 0.

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

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Also products of elements of A011757.

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

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)

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Also products of distinct elements of A011757.