OFFSET
1,5
COMMENTS
In other words, there are no "column rises", where a "column rise" means a pair of adjacent columns where each entry in the left column is strictly less than the adjacent entry in the right column.
This is R(n,k,0) in [Abramson-Promislow].
From Petros Hadjicostas, Sep 09 2019: (Start)
As stated above, in the notation of Abramson and Promislow (1978), we have T(n,k) = R(n, k, t=0).
Let P_k be the set of all lists a = (a_1, a_2, ..., a_k) of integers a_i >= 0, i = 1, ..., k, such that 1*a_1 + 2*a_2 + ... + k*a_k = k; i.e., P_k is the set all integer partitions of k. Then |P_k| = A000041(k).
From Eq. (6), p. 248, in Abramson and Promislow (1978), with t=0, we get T(n,k) = Sum_{a in P_k} (-1)^(k - Sum_{j=1..k} a_j) * (a_1 + a_2 + ... + a_k)!/(a_1! * a_2! * ... * a_k!) * (k! / ((1!)^a_1 * (2!)^a_2 * ... * (k!)^a_k))^n.
The integer partitions of k = 1..10 are listed on pp. 831-832 of Abramowitz and Stegun (1964). We see that, for k = 1..6, the corresponding multinomial coefficients k! / ((1!)^a_1 * (2!)^a_2 * ... * (k!)^a_k) are all distinct; that is, A070289(k) = A000041(k) and A309951(k,s) = A325305(k,s) for s = 0..A000041(k). For 7 <= k <= 10, this is not true anymore; i.e., A070289(k) < A000041(k) for 7 <= k <= 10 (and we conjecture that this is the case for all k >= 7).
From the theory of difference equations, we see that Abramson and Promislow's Eq. (6) on p. 248 (with t=0) implies that Sum_{s = 0..A070289(k)} (-1)^s * A325305(k,s) * T(n-s,k) = 0 for n >= A070289(k) + 1. For k = 1..5, these recurrences give R. H. Hardin's empirical recurrences shown in the Formula section below.
We also have Sum_{s = 0..A000041(k)} (-1)^s * A309951(k,s) * T(n-s,k) = 0 for n >= A000041(k) + 1, but for k >= 7, the recurrence we get (for column k) may not necessarily be minimal.
To derive the recurrence for row n, let y=0 in Eq. (8), p. 249, of Abramson and Promislow (1978). We get 1 + Sum_{k >= 1} T(n,k)*x^k/(k!)^n = 1/f_n(-x), where f_n(x) = Sum_{i >= 0} (x^i/(i!)^n). Matching coefficients, we get Sum_{s = 1..k} T(n,s) * (-1)^(s-1) * binomial(k,s)^n = 1, from which the recurrence in the Formula section follows.
(End)
LINKS
Alois P. Heinz, Antidiagonals n = 1..45 (first 20 antidiagonals from R. H. Hardin)
Milton Abramowitz and Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, National Bureau of Standards (Applied Mathematics Series, 55), 1964; see pp. 831-832 for the multinomial coefficients of integer partitions of n = 1..10.
Morton Abramson and David Promislow, Enumeration of arrays by column rises, J. Combinatorial Theory Ser. A 24(2) (1978), 247-250. MR0469773 (57 #9554); see Eqs. (6) on p. 248 and (8) on p. 249 with t=0.
Yifei Li and Sheila Sundaram, Homology of Segre products of Boolean and subspace lattices, arXiv:2408.08421 [math.CO], 2024. See p. 17.
Wikipedia, Partition (number theory).
Wikipedia, Multinomial theorem.
FORMULA
Empirical recurrence for column k:
k=1: a(n) = 1*a(n-1).
k=2: a(n) = 3*a(n-1) - 2*a(n-2).
k=3: a(n) = 10*a(n-1) - 27*a(n-2) + 18*a(n-3).
k=4: a(n) = 47*a(n-1) - 718*a(n-2) + 4416*a(n-3) - 10656*a(n-4) + 6912*a(n-5).
k=5: a(n) = 246*a(n-1) - 20545*a(n-2) + 751800*a(n-3) - 12911500*a(n-4) + 100380000*a(n-5) - 304200000*a(n-6) + 216000000*a(n-7).
[All the "empirical" recurrences above are correct. See the comments above.]
From Benoit Jubin, May 29 2012: (Start)
T(n,1) = T(1,n) = 1.
T(n,2) = 2^n - 1 since the only n X 2 matrix with rows permutations of {0,1} which has a column rise is the one where all rows are [0,1].
(k!)^n*(1 - (k-1)/2^n) <= T(n,k) <= (k!)^n (the first inequality is (11) in the Abramson-Promislow reference, the second is trivial). (End)
For r >= 1, A(n, r) = Sum_{k=0..n} |[x^k] n!^r [z^n] S(r, z)^x| where S(r, z) = Sum_{k>=0} z^k/k!^r. - Peter Luschny, Feb 27 2018
From Petros Hadjicostas, Sep 09 2019: (Start)
Recurrence for column k: Sum_{s = 0..A070289(k)} (-1)^s * A325305(k,s) * T(n-s,k) = 0 for n >= A070289(k) + 1.
Recurrence for row n: T(n,k) = (-1)^(k-1) + Sum_{s = 1..k-1} T(n,s) * (-1)^(k-s-1) * binomial(k,s)^n for k >= 1.
(End)
Sum_{k>=1} T(n,k)*z^k/(k!)^n = 1/E_n(-z) -1 where E_n(z) = Sum_{k>=0} z^k/(k!)^n. - Geoffrey Critzer, Apr 28 2023
EXAMPLE
Some solutions for n=3 and k=4:
2 1 3 0 1 3 0 2 3 0 2 1 1 3 0 2 1 3 2 0
2 0 1 3 1 3 0 2 3 1 2 0 1 0 3 2 1 3 0 2
2 3 0 1 3 0 2 1 2 3 1 0 2 0 3 1 3 1 0 2
Table starts:
1 1 1 1 1 1 1
1 3 19 211 3651 90921 3081513
1 7 163 8983 966751 179781181 53090086057
1 15 1135 271375 158408751 191740223841 429966316953825
1 31 7291 7225951 21855093751 164481310134301 2675558106868421881
1 63 45199 182199871 2801736968751 128645361626874561 14895038886845467640193
MAPLE
A212855_row := proc(m, len) proc(n, m) sum(z^k/k!^m, k = 0..infinity);
series(%^x, z=0, n+1): n!^m*coeff(%, z, n); [seq(coeff(%, x, k), k=0..n)] end;
seq(add(abs(k), k=%(j, m)), j=1..len) end:
for n from 1 to 6 do A212855_row(n, 7) od; # Peter Luschny, May 26 2017
# second Maple program:
T:= proc(n, k) option remember; `if`(k=0, 1, -add(
binomial(k, j)^n*(-1)^j*T(n, k-j), j=1..k))
end:
seq(seq(T(n, 1+d-n), n=1..d), d=1..10); # Alois P. Heinz, Apr 26 2020
MATHEMATICA
rows = 9;
row[m_, len_] := Module[{p, s0, s1, s2}, p = Function[{n, m0}, s0 = Sum[ z^k/k!^m0, {k, 0, n}]; s1 = Series[s0^x, {z, 0, n+1}] // Normal; s2 = n!^m0*Coefficient[s1, z, n]; Table[Coefficient[s2, x, k], {k, 0, n}]]; Table[Sum[Abs[k], {k, p[j, m]}], {j, 1, len}]];
T = Table[row[n, rows+1], {n, 1, rows}];
Table[T[[n-k+1, k]], {n, 1, rows}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Feb 27 2018, after Peter Luschny *)
CROSSREFS
Cf. A000012 (row 1), A000275 (row 2), A212856 (row 3), A212857 (row 4), A212858 (row 5), A212859 (row 6), A212860 (row 7).
Cf. A000012 (column 1), A000225 (column 2), A212850 (column 3), A212851 (column 4), A212852 (column 5), A212853 (column 6), A212854 (column 7).
KEYWORD
nonn,tabl
AUTHOR
R. H. Hardin, May 28 2012
STATUS
approved