A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Author

Ralf Stephan, May 31 2004

Keywords

• nonn
• tabl

Comments

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

Examples

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...
		

Links

• Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114.
• Elizabeth Wilmer, A note on Stephan's conjecture 87
• Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Crossrefs

Rows are first differences of rows in array A073134.

Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).

Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

Programs

Mathematica

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI

T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)

Formula

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).

For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).

T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.

T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).