|
%I #21 Jun 14 2021 10:04:27
%S 1,2,1,1,1,3,2,2,1,1,1,1,1,1,4,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,5,4,4,
%T 3,3,3,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,6,5,5,4,4,4,3,3,3,3,
%U 3,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%N Irregular triangle read by rows T(n,k) in which row n lists in nonincreasing order all divisors of the terms of the n-th row of triangle A110730, n >= 1, k >= 1.
%C Note that in the definition A110730 can be replaced with A333516 or with A345116 since these three triangles contain in every row the same terms but in distinct order.
%C The sum of n-th row is equal to A175254(n) equaling the volume (also the number of cubes) of the stepped pyramid with n levels described in A245092.
%e Triangle begins:
%e 1;
%e 2, 1, 1, 1;
%e 3, 2, 2, 1, 1, 1, 1, 1, 1;
%e 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
%e ...
%e For n = 3 the third row of A110730 is [1, 1, 1, 2, 2, 3], so the divisors of these terms in nonincreasing order are [3, 2, 2, 1, 1, 1, 1, 1, 1], the same as the third row of triangle.
%o (PARI) row(n) = my(v=[]); for (k=1, n, for (j=1, n-k+1, v = concat(v, divisors(k)))); vecsort(v,,4); \\ _Michel Marcus_, Jun 14 2021
%Y Cf. A110730, A175254, A237593, A245092, A333516, A345116.
%K nonn,tabf
%O 1,2
%A _Omar E. Pol_, Jun 12 2021
|
