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%I #25 Aug 21 2019 09:42:35
%S 1,2,3,4,4,6,8,5,8,9,12,16,6,10,12,16,18,24,32,7,12,15,16,20,24,27,32,
%T 36,48,64,8,14,18,20,24,30,32,36,40,48,54,64,72,96,128,9,16,21,24,25,
%U 28,36,40,45,48,48,60,64,72,81,80,96,108,128,144,192,256
%N Number of divisors of A036035(n,k).
%H Alois P. Heinz, <a href="/A074139/b074139.txt">Rows n = 0..30, flattened</a>
%H Byungchul Cha et al., <a href="https://arxiv.org/abs/1811.07451">An Investigation on Partitions with Equal Products</a>, arXiv:1811.07451 [math.NT], 2018.
%F T(n,k) = A000005(A036035(n,k)). - _R. J. Mathar_, Aug 28 2018
%e Express A036035(n,k) by its prime signature; add one to each exponent, then multiply: 180 = (2^2)*(3^2)*(5^1) therefore the number of divisors is (2+1)*(2+1)*(1+1)= 18
%e From _Michel Marcus_, Nov 11 2015: (Start)
%e As an irregular triangle, whose n-th row has A000041(n) terms, sequence begins:
%e 1;
%e 2;
%e 3, 4;
%e 4, 6, 8;
%e 5, 8, 9, 12, 16;
%e 6, 10, 12, 16, 18, 24, 32;
%e ...
%e (End)
%o (PARI) tabf(nn) = {for (n=1, nn, forpart(p=n, print1(prod(k=1, #p, (1+p[k])), ", ")); print(););} \\ _Michel Marcus_, Nov 11 2015
%Y Row sums give A074141.
%Y Cf. A036035, A074140.
%K nonn,look,tabf
%O 0,2
%A _Amarnath Murthy_, Aug 28 2002
%E More terms from _Alford Arnold_, Sep 17 2002
%E Term ordering corrected by _Alois P. Heinz_, Aug 21 2019
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