%I #21 Oct 06 2021 03:26:49
%S 0,0,9,12,10,10,11,12,7,10,7,10,7,10,8,4,11,6,6,8,9,4,4,10,7,4,4,4,10,
%T 4,7,2,4,10,6,4,3,6,2,8,8,2,6,2,5,2,2,8,6,4,6,2,2,2,5,6,8,2,4,4,4,2,2,
%U 0,5,8,6,2,4,4,4,0,4,2,2,0,5,6,6,4,2,4,4,0,4,0,6,0,2,0
%N Number of ordered pairs (m,k) such that A000009(m) + A000009(k) = n.
%C Although a(n) != 0 for 2 <= n <= 62, it seems that most terms are zero. There are only 1351 nonzero terms among a(0) through a(10000).
%C a(n) is odd if and only if n != 4 and n/2 is in A000009.
%C Conjecture: a(n) = 0, 1, 2, 4, 6 for all n > 29696.
%H Jianing Song, <a href="/A347943/b347943.txt">Table of n, a(n) for n = 0..30000</a>
%F a(1+A000009(m)) >= 6. It seems that for n > 24578, all terms with a(n) >= 6 are of the form 1 + A000009(m), and in which case we have a(n) = 6.
%F a(2+A000009(m)) >= 4. It seems that for n > 729124, all terms with a(n) = 4 are of the form 2 + A000009(m).
%F a(2*A000009(m)) >= 1. It seems that for m > 63 (2*A000009(m) > 29296) we have a(2*A000009(m)) = 1.
%F a(A000009(m)+A000009(k)) >= 2 for distinct m,k.
%e a(6) = 11 since 6 = A000009(m) + A000009(k) for (m,k) = (0,7), (1,7), (2,7), (3,6), (4,6), (5,5), (6,3), (6,4), (7,0), (7,1), (7,2).
%e a(63) = 0 since no two terms in A000009 sum up to 63.
%t Table[Length@Select[Tuples[k=1;While[Max[p=PartitionsQ/@Range[0,k++]]<n];Most@p,{2}],Total@#==n&],{n,0,100}] (* _Giorgos Kalogeropoulos_, Sep 21 2021 *)
%o (PARI) leng(n) = for(l=0, oo, if(A000009(l)>n1, return(l))) \\ See A000009 for its program; A000009(0), A000009(1), ..., A000009(l1) <= n1
%o v(n) = my(l=leng(n), v=[]); for(i=0, l1, v=concat(v, vector(l, j, A000009(i)+A000009(j1)))); v=vecsort(v); v
%o list(n) = my(v=v(n), w=vector(n), size=#v); for(i=1, size, if(v[i]<=n, w[v[i]]++, break())); w=concat([0], w); w
%Y Cf. A000009, A347944 (indices of 0).
%K nonn,easy
%O 0,3
%A _Jianing Song_, Sep 20 2021
