A204213 T(n,k) = Number of length n+1 nonnegative integer arrays starting and ending with 0 with adjacent elements differing by no more than k.

1, 1, 2, 1, 3, 4, 1, 4, 9, 9, 1, 5, 16, 32, 21, 1, 6, 25, 78, 120, 51, 1, 7, 36, 155, 404, 473, 127, 1, 8, 49, 271, 1025, 2208, 1925, 323, 1, 9, 64, 434, 2181, 7167, 12492, 8034, 835, 1, 10, 81, 652, 4116, 18583, 51945, 72589, 34188, 2188, 1, 11, 100, 933, 7120, 41363, 164255, 387000, 430569, 147787, 5798

Offset

1,3

Comments

Table starts

...1....1.....1......1.......1.......1........1........1........1.........1

...2....3.....4......5.......6.......7........8........9.......10........11

...4....9....16.....25......36......49.......64.......81......100.......121

...9...32....78....155.....271.....434......652......933.....1285......1716

..21..120...404...1025....2181....4116.....7120....11529....17725.....26136

..51..473..2208...7167...18583...41363....82440...151125...259459....422565

.127.1925.12492..51945..164255..431445...991152..2057553..3945655...7098949

.323.8034.72589.387000.1493142.4629851.12262470.28832499.61766005.122779448

Links

R. H. Hardin, Table of n, a(n) for n = 1..9999

Formula

Empirical for rows:

T(1,k) = 1

T(2,k) = k + 1

T(3,k) = k^2 + 2*k + 1

T(4,k) = (4/3)*k^3 + (7/2)*k^2 + (19/6)*k + 1

T(5,k) = (23/12)*k^4 + (37/6)*k^3 + (91/12)*k^2 + (13/3)*k + 1

T(6,k) = (44/15)*k^5 + (133/12)*k^4 + 17*k^3 + (161/12)*k^2 + (167/30)*k + 1

T(7,k) = (841/180)*k^6 + (101/5)*k^5 + (1325/36)*k^4 + (73/2)*k^3 + (946/45)*k^2 + (34/5)*k + 1

...

G.f. for column k: exp( Sum_{n>=1} L(n,k)*x^n/n ) - 1, where L(n,k) = central coefficient of (1+x+x^2+x^3+...+x^(2*k))^n. - Paul D. Hanna, Aug 01 2013

T(n,k):=Sum_{i=1..n}((Sum_{j=0..(i*k)/(2*k+1)}(binomial(i,j)*(-1)^j*binomial(-j*(2*k+1)+i*(k+1)-1,i*k-j*(2*k+1))))*T(n-i,k))/n, T(0,k)=1. - Vladimir Kruchinin, Apr 06 2017

Example

Some solutions for n=5 k=3
..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0
..3....0....1....2....2....2....1....3....0....3....1....2....2....3....2....1
..5....0....4....5....1....1....4....4....2....4....0....4....5....1....0....1
..5....1....2....4....3....0....1....1....3....1....1....2....5....3....2....2
..2....3....1....2....0....2....0....0....1....3....1....2....2....1....2....1
..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0

Mathematica

T[n_, k_] := T[n, k] = If[n == 0, 1, Sum[(Sum[Binomial[i, j]*(-1)^j* Binomial[-j*(2*k + 1) + i*(k + 1) - 1, i*k - j*(2*k + 1)], {j, 0, (i*k)/(2*k + 1)}])*T[n - i, k], {i, 1, n}]/n];
Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Sep 24 2019, after Vladimir Kruchinin *)

Prog

(PARI)
{L(n, k)=polcoeff( ( (1-x^(2*k+1))/(1-x) +x*O(x^(k*n)) )^n, k*n)}
{T(n, k)=polcoeff(exp(sum(m=1, n, L(m, k)*x^m/m)+x*O(x^n)), n)}
for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print("")) \\ Paul D. Hanna, Aug 01 2013
(Maxima)
T(n, k):=if n=0 then 1 else sum((sum(binomial(i, j)*(-1)^j*binomial(-j*(2*k+1)+i*(k+1)-1, i*k-j*(2*k+1)), j, 0, (i*k)/(2*k+1)))*T(n-i, k), i, 1, n)/n;  /* Vladimir Kruchinin, Apr 06 2017 */

Crossrefs

Column 1 is A001006; column 2 is A104184; column 3 is A204208.

Row 4 is A051662.

Sequence in context: A055208 A051128 A137614 * A143326 A327086 A186686

Adjacent sequences: A204210 A204211 A204212 * A204214 A204215 A204216

Keyword

nonn,tabl

Author

R. H. Hardin, Jan 12 2012

Status

approved