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A340734
Number of partitions of 2n into 4 parts that divide 2n.
0
0, 0, 1, 2, 2, 1, 7, 0, 2, 3, 3, 0, 9, 0, 1, 4, 2, 0, 8, 0, 4, 3, 1, 0, 9, 1, 1, 3, 2, 0, 10, 0, 2, 2, 1, 1, 10, 0, 1, 2, 4, 0, 8, 0, 2, 5, 1, 0, 9, 0, 3, 2, 2, 0, 8, 1, 2, 2, 1, 0, 12, 0, 1, 4, 2, 1, 7, 0, 2, 2, 3, 0, 10, 0, 1, 4, 2, 0, 7, 0, 4, 3, 1, 0, 10, 1, 1, 2, 2, 0, 11
OFFSET
0,4
FORMULA
a(n) = Sum_{k=1..floor(n/2)} Sum_{j=k..floor((2*n-k)/3)} Sum_{i=j..floor((2*n-j-k)/2)} chi(2*n/k) * chi(2*n/j) * chi(2*n/i) * chi(2*n/(n-i-j-k)), where chi(n) = 1 - ceiling(n) + floor(n).
MATHEMATICA
Table[Sum[Sum[Sum[(1 - Ceiling[2 n/k] + Floor[2 n/k]) (1 - Ceiling[2 n/j] + Floor[2 n/j]) (1 - Ceiling[2 n/i] + Floor[2 n/i]) (1 - Ceiling[2 n/(2 n - i - j - k)] + Floor[2 n/(2 n - i - j - k)]), {i, j, Floor[(2 n - j - k)/2]}], {j, k, Floor[(2 n - k)/3]}], {k, Floor[n/2]}], {n, 0, 100}]
PROG
(PARI) for(n=0, 40, my(nn=2*n, count=0); forpart(p=nn, if(#p==4, count+=sum(k=1, 4, nn%p[k]==0)==4)); print1(count, ", ")) \\ Hugo Pfoertner, Jan 18 2021
CROSSREFS
Sequence in context: A181731 A278792 A343807 * A108338 A021455 A271460
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, Jan 17 2021
STATUS
approved