%I #21 Jun 10 2021 16:11:52
%S 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,3,2,1,1,1,1,3,1,3,1,2,1,
%T 1,1,1,3,1,3,1,1,2,1,1,1,1,3,1,1,3,1,1,2,1,4,1,1,3,1,1,1,3,1,1,2,1,4,
%U 1,1,3,1,1,1,1,3,1,1,5,1,4,1,1,3,1,1,1,1,1,3,1
%N Triangle of numerators corresponding to A344007.
%e The triangle underlying A344007 begins:
%e 1
%e 1/2, 1/2
%e 1/6, 1/3, 1/2
%e 1/6, 1/4, 1/4, 1/3
%e 2/15, 1/6, 1/5, 1/4, 1/4
%e 1/12, 2/15, 1/6, 1/6, 1/5, 1/4
%e 1/12, 3/28, 2/15, 1/7, 1/6, 1/6, 1/5
%e 3/40, 1/12, 3/28, 1/8, 2/15, 1/7, 1/6, 1/6
%e 1/18, 3/40, 1/12, 3/28, 1/9, 1/8, 2/15, 1/7, 1/6
%e 1/18, 1/15, 3/40, 1/12, 1/10, 3/28, 1/9, 1/8, 2/15, 1/7
%e ...
%e The numerators are:
%e 1
%e 1, 1
%e 1, 1, 1
%e 1, 1, 1, 1
%e 2, 1, 1, 1, 1
%e 1, 2, 1, 1, 1, 1
%e 1, 3, 2, 1, 1, 1, 1
%e 3, 1, 3, 1, 2, 1, 1, 1
%e 1, 3, 1, 3, 1, 1, 2, 1, 1
%e 1, 1, 3, 1, 1, 3, 1, 1, 2, 1
%e ...
%o (PARI) lista(nn) = {my(row, nrow, drow); for (n=1, nn, if (n==1, row = [1], k = vecmax(row); nrow = row; nrow[n-1] = 1/n; nrow = concat(nrow, k - 1/n); row = vecsort(nrow);); drow = apply(numerator, row); for (k=1, #drow, print1(drow[k], ", ")););} \\ _Michel Marcus_, Jun 09 2021
%Y Cf. A003506, A344007.
%K nonn,tabl
%O 1,11
%A _Evan Lee_, Jun 09 2021
%E Corrected by _Hugo Pfoertner_ and _Michel Marcus_, Jun 09 2021