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A331562
Number A(n,k) of sequences with k copies each of 1,2,...,n avoiding absolute differences between adjacent elements larger than one; square array A(n,k), n>=0, k>=0, read by antidiagonals.
20
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 2, 1, 1, 1, 20, 12, 2, 1, 1, 1, 70, 92, 26, 2, 1, 1, 1, 252, 780, 506, 48, 2, 1, 1, 1, 924, 7002, 11482, 2288, 86, 2, 1, 1, 1, 3432, 65226, 284002, 135040, 10010, 148, 2, 1, 1, 1, 12870, 623576, 7426610, 8956752, 1543862, 41618, 250, 2, 1
OFFSET
0,9
COMMENTS
All columns are linear recurrences with constant coefficients and for k > 0 the order of the recurrence is bounded by 3*k-1. For k up to at least 17 this upper bound is exact. - Andrew Howroyd, May 16 2020
Row 2, the sequence of central binomial numbers A000984, satisfies the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r (see Meštrović, equation 39). This is also known to be true for row 3 (A103882) and row 4 (A177316). We conjecture that each row sequence of the table satisfies the same congruences. - Peter Bala, Oct 26 2024.
LINKS
Andrew Howroyd, Antidiagonals n = 0..50, flattened (antidiagonals 0..15 from Alois P. Heinz)
EXAMPLE
A(2,2) = 6: 1122, 1212, 1221, 2112, 2121, 2211.
A(3,2) = 12: 112233, 112323, 112332, 121233, 123321, 211233, 233211, 321123, 323211, 332112, 332121, 332211.
A(2,3) = 20: 111222, 112122, 112212, 112221, 121122, 121212, 121221, 122112, 122121, 122211, 211122, 211212, 211221, 212112, 212121, 212211, 221112, 221121, 221211, 222111.
A(3,3) = 92: 111222333, 111223233, 111223323, 111223332, ..., 333221112, 333221121, 333221211, 333222111.
Square array A(n,k) begins:
1, 1, 1, 1, 1, 1, 1, ...
1, 1, 1, 1, 1, 1, 1, ...
1, 2, 6, 20, 70, 252, 924, ...
1, 2, 12, 92, 780, 7002, 65226, ...
1, 2, 26, 506, 11482, 284002, 7426610, ...
1, 2, 48, 2288, 135040, 8956752, 640160976, ...
1, 2, 86, 10010, 1543862, 276285002, 54331653686, ...
MAPLE
b:= proc(l, q) option remember; (n-> `if`(n<2, 1, add(
`if`(l[j]=1, `if`(j in [1, n], b(subsop(j=[][], l),
`if`(j=1, 0, n)), 0), b(subsop(j=l[j]-1, l), j)), j=
`if`(q<0, 1..n, max(1, q-1)..min(n, q+1)))))(nops(l))
end:
A:= (n, k)-> `if`(k=0, 1, b([k$n], -1)):
seq(seq(A(n, d-n), n=0..d), d=0..10);
MATHEMATICA
b[l_, q_] := b[l, q] = With[{n = Length[l]}, If[n < 2, 1, Sum[
If[l[[j]] == 1, If[j == 1 || j == n, b[ReplacePart[l, j -> Nothing],
If[j == 1, 0, n]], 0], b[ReplacePart[l, j -> l[[j]] - 1], j]], {j,
If[q < 0, Range[n], Range[Max[1, q - 1], Min[n, q + 1]]]}]]];
A[n_, k_] := If[k == 0, 1, b[Table[k, {n}], -1]];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Jan 03 2021, after Alois P. Heinz *)
PROG
(PARI)
step(m, R)={my(M=matrix(3, m+1, q, p, q--; p--; sum(j=0, m-p-q, sum(i=max(p+j-#R+1, 2*p+q+j-m), p, R[1+q, 1+p+j-i] * binomial(p, i) * binomial(p+q+j-i-1, j) * binomial(m-1, 2*p+q+j-i-1))))); M[3, ]+=2*M[2, ]+M[1, ]; M[2, ]+=M[1, ]; M}
AdjPathsBySig(sig)={if(#sig<1, 1, my(R=[1; 1; 1]); for(i=1, #sig-1, R=step(sig[i], R)); my(m=sig[#sig]); sum(i=1, min(m, #R), binomial(m-1, i-1)*R[3, i]))}
T(n, k) = {if(k==0, 1, AdjPathsBySig(vector(n, i, k)))} \\ Andrew Howroyd, May 16 2020
CROSSREFS
Columns k=0-9 give: A000012, A130130 (for n>0), A177282, A177291, A177298, A177301, A177304, A177307, A177310, A177313.
Main diagonal gives A331623.
Sequence in context: A060185 A348091 A129110 * A257101 A112624 A294875
KEYWORD
nonn,tabl
AUTHOR
Alois P. Heinz, Jan 20 2020
STATUS
approved