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Number of partitions of n into triangular numbers dividing n.
1

%I #10 Aug 02 2019 12:58:17

%S 1,1,1,2,1,1,4,1,1,4,2,1,9,1,1,7,1,1,16,1,3,9,1,1,25,1,1,10,2,1,74,1,

%T 1,12,1,1,50,1,1,14,5,1,85,1,1,35,1,1,81,1,6,18,1,1,100,2,3,20,1,1,

%U 544,1,1,46,1,1,145,1,1,24,8,1,219,1,1,81,1,1,197,1,9,28,1,1,628,1,1,30,1,1,2264,2,1,32,1,1

%N Number of partitions of n into triangular numbers dividing n.

%H Amiram Eldar, <a href="/A294334/b294334.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>

%H <a href="/index/Pol#polygonal_numbers">Index to sequences related to polygonal numbers</a>

%F a(n) = 1 if n in A112886.

%e a(6) = 4 because 6 has 4 divisors {1, 2, 3, 6} among which 3 are triangular numbers {1, 3, 6} therefore we have [6], [3, 3], [3, 1, 1, 1] and [1, 1, 1, 1, 1, 1].

%t Table[SeriesCoefficient[Product[1/(1 - Boole[Mod[n, k] == 0 && IntegerQ[Sqrt[8 k + 1]]] x^k), {k, 1, n}], {x, 0, n}], {n, 0, 95}]

%Y Cf. A000217, A007294, A007862, A018818, A112886, A284345.

%K nonn,look

%O 0,4

%A _Ilya Gutkovskiy_, Oct 28 2017