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A227578 Number A(n,k) of lattice paths from {n}^k to {0}^k using steps that decrement one component such that for each point (p_1,p_2,...,p_k) we have p_1<=p_2<=...<=p_k; square array A(n,k), n>=0, k>=0, read by antidiagonals. 25

%I #48 Jul 24 2021 03:04:59

%S 1,1,1,1,1,1,1,1,2,1,1,1,5,4,1,1,1,14,29,8,1,1,1,42,290,185,16,1,1,1,

%T 132,3532,7680,1257,32,1,1,1,429,49100,456033,238636,8925,64,1,1,1,

%U 1430,750325,34426812,77767945,8285506,65445,128,1

%N Number A(n,k) of lattice paths from {n}^k to {0}^k using steps that decrement one component such that for each point (p_1,p_2,...,p_k) we have p_1<=p_2<=...<=p_k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

%C Conjecture: column k is asymptotic to c(k) * (k+1)^(k*n)/n^((k^2-1)/2), where c(k) is a constant dependent only on k. - _Vaclav Kotesovec_, Jul 21 2013

%H Alois P. Heinz, <a href="/A227578/b227578.txt">Antidiagonals n = 0..25, flattened</a>

%H Antonio Vera López, Luis Martínez, Antonio Vera Pérez, Beatriz Vera Pérez and Olga Basova, <a href="https://doi.org/10.1016/j.laa.2017.05.027">Combinatorics related to Higman's conjecture I: Parallelogramic digraphs and dispositions</a>, Linear Algebra and its Applications, Volume 530, 1 October 2017, p. 414-444. This entry is mentioned in the Introduction, but it seems that actually they mean A181196.

%e A(4,0) = 1: [()].

%e A(3,1) = 4: [(3),(0)], [(3),(1),(0)], [(3),(2),(0)], [(3),(2),(1),(0)].

%e A(2,2) = 5: [(2,2),(0,2),(0,0)], [(2,2),(0,2),(0,1),(0,0)], [(2,2),(1,2),(0,2),(0,0)], [(2,2),(1,2),(0,2),(0,1),(0,0)], [(2,2),(1,2),(1,1),(0,1),(0,0)].

%e A(1,3) = 1: [(1,1,1),(0,1,1),(0,0,1),(0,0,0)].

%e A(0,4) = 1: [(0,0,0,0)].

%e Square array A(n,k) begins:

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 1, 1, 1, 1, 1, ...

%e 1, 2, 5, 14, 42, 132, ...

%e 1, 4, 29, 290, 3532, 49100, ...

%e 1, 8, 185, 7680, 456033, 34426812, ...

%e 1, 16, 1257, 238636, 77767945, 36470203156, ...

%p b:= proc(l) option remember; `if`(l[-1]=0, 1, add(add(b(subsop(

%p i=j, l)), j=`if`(i=1, 0, l[i-1])..l[i]-1), i=1..nops(l)))

%p end:

%p A:= (n, k)-> `if`(k=0, 1, b([n$k])):

%p seq(seq(A(n, d-n), n=0..d), d=0..10);

%t b[l_] := b[l] = If[ l[[-1]] == 0, 1, Sum[ Sum[ b[ReplacePart[l, i -> j]], {j, If[i == 1, 0, l[[i-1]]], l[[i]]-1}], {i, 1, Length[l]}]]; a[n_, k_] := If[k == 0, 1, b[Array[n&, k]]]; Table[Table[a[n, d - n], {n, 0, d}], {d, 0, 10}] // Flatten (* _Jean-François Alcover_, Dec 09 2013, translated from Maple *)

%Y Columns k=0-10 give: A000012, A011782, A059231, A227580, A227583, A227596, A227597, A227598, A227599, A227600, A227601.

%Y Rows n=0+1, 2-10 give: A000012, A000108(k+1), A181197(k+2), A227584, A227602, A227603, A227604, A227605, A227606, A227607.

%Y Main diagonal gives: A227579.

%Y Cf. A060854 (steps decrement one component by 1), A262809, A263159.

%Y A181196 is a similar but different array.

%K nonn,tabl

%O 0,9

%A _Alois P. Heinz_, Jul 16 2013

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