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A347943
Number of ordered pairs (m,k) such that A000009(m) + A000009(k) = n.
2
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, 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, 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
OFFSET
0,3
COMMENTS
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).
a(n) is odd if and only if n != 4 and n/2 is in A000009.
Conjecture: a(n) = 0, 1, 2, 4, 6 for all n > 29696.
LINKS
FORMULA
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.
a(2+A000009(m)) >= 4. It seems that for n > 729124, all terms with a(n) = 4 are of the form 2 + A000009(m).
a(2*A000009(m)) >= 1. It seems that for m > 63 (2*A000009(m) > 29296) we have a(2*A000009(m)) = 1.
a(A000009(m)+A000009(k)) >= 2 for distinct m,k.
EXAMPLE
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).
a(63) = 0 since no two terms in A000009 sum up to 63.
MATHEMATICA
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 *)
PROG
(PARI) leng(n) = for(l=0, oo, if(A000009(l)>n-1, return(l))) \\ See A000009 for its program; A000009(0), A000009(1), ..., A000009(l-1) <= n-1
v(n) = my(l=leng(n), v=[]); for(i=0, l-1, v=concat(v, vector(l, j, A000009(i)+A000009(j-1)))); v=vecsort(v); v
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
CROSSREFS
Cf. A000009, A347944 (indices of 0).
Sequence in context: A269026 A124606 A110647 * A335168 A295486 A032687
KEYWORD
nonn,easy
AUTHOR
Jianing Song, Sep 20 2021
STATUS
approved