This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A238353 #26 Jan 06 2015 12:11:47
%S A238353 1,1,0,2,0,0,2,1,0,0,3,1,1,0,0,2,3,1,1,0,0,4,3,2,1,1,0,0,2,6,3,2,1,1,
%T A238353 0,0,4,6,6,2,2,1,1,0,0,3,10,6,5,2,2,1,1,0,0,4,11,11,6,4,2,2,1,1,0,0,2,
%U A238353 16,13,10,5,4,2,2,1,1,0,0,6,17,19,12,9,4,4,2,2,1,1,0,0,2,24,24,18,11,8,4,4,2,2,1,1,0,0
%N A238353 Triangle T(n,k) read by rows: T(n,k) is the number of partitions of n (as weakly ascending list of parts) with maximal ascent k, n >= 0, 0 <= k <= n.
%C A238353 Reversed rows and also the columns converge to A002865 (setting A002865(0)=0).
%C A238353 Column k=0 is A000005 (n>=1), column k=1 is A237665.
%C A238353 Row sums are A000041.
%C A238353 Sum_{i=0..k} T(n,i) for k=0-9 gives: A000005, A034296, A224956, A238863, A238864, A238865, A238866, A238867, A238868, A238869.
%H A238353 Joerg Arndt and Alois P. Heinz, <a href="/A238353/b238353.txt">Rows 0..140, flattened</a>
%F A238353 G.f. for column k>=1: sum(j>=1, q^j/(1-q^j) * (prod(i=1..j-1, (1-q^((k+1)*i))/(1-q^i) ) - prod(i=1..j-1, (1-q^(k*i))/(1-q^i) ) ) ), see the comment about the g.f. in A238863.
%e A238353 Triangle starts:
%e A238353 00: 1;
%e A238353 01: 1, 0;
%e A238353 02: 2, 0, 0;
%e A238353 03: 2, 1, 0, 0;
%e A238353 04: 3, 1, 1, 0, 0;
%e A238353 05: 2, 3, 1, 1, 0, 0;
%e A238353 06: 4, 3, 2, 1, 1, 0, 0;
%e A238353 07: 2, 6, 3, 2, 1, 1, 0, 0;
%e A238353 08: 4, 6, 6, 2, 2, 1, 1, 0, 0;
%e A238353 09: 3, 10, 6, 5, 2, 2, 1, 1, 0, 0;
%e A238353 10: 4, 11, 11, 6, 4, 2, 2, 1, 1, 0, 0;
%e A238353 11: 2, 16, 13, 10, 5, 4, 2, 2, 1, 1, 0, 0;
%e A238353 12: 6, 17, 19, 12, 9, 4, 4, 2, 2, 1, 1, 0, 0;
%e A238353 13: 2, 24, 24, 18, 11, 8, 4, 4, 2, 2, 1, 1, 0, 0;
%e A238353 14: 4, 27, 34, 22, 17, 10, 7, 4, 4, 2, 2, 1, 1, 0, 0;
%e A238353 15: 4, 35, 39, 33, 20, 15, 9, 7, 4, 4, 2, 2, 1, 1, 0, 0;
%e A238353 ...
%e A238353 The 7 partitions of 5 and their maximal ascents are:
%e A238353 1: [ 1 1 1 1 1 ] 0
%e A238353 2: [ 1 1 1 2 ] 1
%e A238353 3: [ 1 1 3 ] 2
%e A238353 4: [ 1 2 2 ] 1
%e A238353 5: [ 1 4 ] 3
%e A238353 6: [ 2 3 ] 1
%e A238353 7: [ 5 ] 0
%e A238353 There are 2 rows with 0 ascents, 3 with 1 ascent, 1 for ascents 2 and 3, giving row 5 of the triangle.
%p A238353 b:= proc(n, i, t) option remember; `if`(n=0, 1,
%p A238353 `if`(i<1, 0, b(n, i-1, t)+`if`(i>n, 0, (p->
%p A238353 `if`(t=0 or t-i=0, p, add(coeff(p, x, j)*x^
%p A238353 max(j, t-i), j=0..degree(p))))(b(n-i, i, i)))))
%p A238353 end:
%p A238353 T:= n-> (p-> seq(coeff(p, x, k), k=0..n))(b(n$2, 0)):
%p A238353 seq(T(n), n=0..15);
%t A238353 b[n_, i_, t_] := b[n, i, t] = If[n == 0, 1, If[i<1, 0, b[n, i-1, t] + If[i>n, 0, Function[{p}, If[t == 0 || t-i == 0, p, Sum[Coefficient[p, x, j]*x^ Max[j, t-i], {j, 0, Exponent[p, x]}]]][b[n-i, i, i]]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, k], {k, 0, n}]][b[n, n, 0]]; Table[T[n], {n, 0, 15}] // Flatten (* _Jean-François Alcover_, Jan 06 2015, translated from Maple *)
%Y A238353 Cf. A238354 (partitions by minimal ascent).
%K A238353 nonn,tabl
%O A238353 0,4
%A A238353 _Joerg Arndt_ and _Alois P. Heinz_, Feb 26 2014