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
Examples
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
Links
- R. H. Hardin, Table of n, a(n) for n = 1..9999
Programs
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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 *) -
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 */
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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
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
Comments