OFFSET
1,2
COMMENTS
Row n contains ceiling(n/2) terms. Row sums yield the Pell numbers (A000129). Column 1 yields A007290.
Eigenvector equals A118397, so that A118397(n) = Sum_{k=0..[n/2]} T(n+1,k)*A118397(k) for n >= 0. - Paul D. Hanna, May 08 2006
Essentially a triangle, read by rows, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 12 2011
Subtriangle of the triangle given by (1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 0, 2, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 07 2012
LINKS
G. C. Greubel, Rows n = 1..100 of the triangle, flattened
Rui Duarte and António Guedes de Oliveira, A Famous Identity of Hajós in Terms of Sets, Journal of Integer Sequences, Vol. 17 (2014), #14.9.1.
J. Ivie, Problem B-161, Fibonacci Quarterly, 8 (1970), 107-108.
FORMULA
E.g.f.: exp(x)*sinh(x*sqrt(2*y))/sqrt(2*y), cf. A034867. - Vladeta Jovovic, Apr 06 2005
From Philippe Deléham, Apr 07 2012: (Start)
As DELTA-triangle T(n,k) with 0 <= k <= n:
G.f.: (1-x+x^2-y*x^2)/(1-2*x+x^2-2*y*x^2).
T(n,k) = 2*T(n-1,k) - T(n-2,k) + 2*T(n-2,k-1), T(0,0) = T(1,0) = 1, T(1,1) = T(2,1) = T(2,2) = 0, T(2,0) = 2 and T(n,k) = 0 if k<0 or if k>n. (End)
EXAMPLE
Triangle begins:
1;
2;
3, 2;
4, 8;
5, 20, 4;
6, 40, 24;
(2, -1/2, 1/2, 0, 0, ...) DELTA (0, 1, -1, 0, 0, ...) begins:
1;
2, 0;
3, 2, 0;
4, 8, 0, 0;
5, 20, 4, 0, 0;
6, 40, 24, 0, 0, 0.
(1, 1, -1, 1, 0, 0, ...) DELTA (0, 0, 2, -2, 0, 0, ...) begins:
1;
1, 0;
2, 0, 0;
3, 2, 0, 0;
4, 8, 0, 0, 0;
5, 20, 4, 0, 0, 0;
6, 40, 24, 0, 0, 0, 0. - Philippe Deléham, Apr 07 2012
MAPLE
T:=(n, k)->binomial(n, 2*k+1)*2^k:for n from 1 to 15 do seq(T(n, k), k=0..floor((n-1)/2)) od; # yields sequence in triangular form
MATHEMATICA
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x]
v[n_, x_] := u[n - 1, x] + v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A207536 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A105070 *)
(* Clark Kimberling, Feb 18 2010 *)
Table[2^k*Binomial[n, 2*k+1], {n, 15}, {k, 0, Floor[(n-1)/2]}]//Flatten (* G. C. Greubel, Mar 15 2020 *)
PROG
(Magma) [2^k*Binomial(n, 2*k+1): k in [0..Floor((n-1)/2)], n in [1..15]]; // G. C. Greubel, Mar 15 2020
(Sage) [[2^k*binomial(n, 2*k+1) for k in (0..floor((n-1)/2))] for n in (1..15)] # G. C. Greubel, Mar 15 2020
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Apr 05 2005
STATUS
approved