cp's OEIS Frontend

The g.f. 1/(z+1)/(z**2+1)/(z**4+1)/(z-1)**2 conjectured by Simon Plouffe in his 1992 dissertation is wrong.

Finite idempotent functions are identity maps on finite subsets, counting the empty function as the idempotent on the empty set.

From Gus Wiseman, Mar 09 2019: (Start)

Also numbers whose ordered prime signature is equal to the distinct prime indices in increasing order. A prime index of n is a number m such that prime(m) divides n. The ordered prime signature (A124010) is the sequence of multiplicities (or exponents) in a number's prime factorization, taken in order of the prime base. The case where the prime indices are taken in decreasing order is A324571.

Also numbers divisible by prime(k) exactly k times for each prime index k. These are a kind of self-describing numbers (cf. A001462, A304679).

Also Heinz numbers of integer partitions where the multiplicity of m is m for all m in the support (counted by A033461). The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Also products of distinct elements of A062457. For example, 43218 = prime(1)^1 * prime(2)^2 * prime(4)^4.

(End)

Is lim_{n->inf} a(n)^(1/n) > 1? - Paul D. Hanna, Aug 20 2002

The limit above is equal to 1 (see formula by Hardy & Ramanujan for A001156). - Vaclav Kotesovec, Dec 29 2016

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]

Also, number of partitions of n where there are fewer than k parts equal to k for all k. - Jon Perry and Vladeta Jovovic, Aug 04 2004. E.g. a(8)=5 because we have 8=6+2=5+3=4+4=3+3+2.

Convolution of A276516 and A000041. - Vaclav Kotesovec, Dec 30 2016

From Gus Wiseman, Apr 02 2019: (Start)

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). The Heinz numbers of the integer partitions described in Perry and Jovovic's comment are given by A325128, while the Heinz numbers of the integer partitions described in the name are given by A325129. In the former case, the first 10 terms count the following integer partitions:

() (2) (3) (4) (5) (6) (7) (8) (9)

(32) (33) (43) (44) (54)

(42) (52) (53) (63)

(62) (72)

(332) (432)

while in the latter case they count the following:

() (2) (3) (22) (5) (6) (7) (8) (63)

(32) (33) (52) (53) (72)

(222) (322) (62) (333)

(332) (522)

(2222) (3222)

(End)

The difference between the number of partitions of n into an even number of distinct squares and the number of partitions of n into an odd number of distinct squares. - Ilya Gutkovskiy, Jan 15 2018

These are a kind of self-describing partitions (cf. A001462, A304679).

The Heinz numbers of these partitions are given by A324571.

The case where the distinct parts are taken in increasing order is counted by A033461, with Heinz numbers given by A109298.

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)