A207607 Triangle of coefficients of polynomials v(n,x) jointly generated with A207606; see Formula section.

1, 1, 2, 1, 5, 2, 1, 9, 9, 2, 1, 14, 25, 13, 2, 1, 20, 55, 49, 17, 2, 1, 27, 105, 140, 81, 21, 2, 1, 35, 182, 336, 285, 121, 25, 2, 1, 44, 294, 714, 825, 506, 169, 29, 2, 1, 54, 450, 1386, 2079, 1716, 819, 225, 33, 2, 1, 65, 660, 2508, 4719, 5005, 3185, 1240, 289, 37, 2

Offset

1,3

Comments

Subtriangle of the triangle T(n,k) given by (1, 0, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

Links

G. C. Greubel, Rows n = 1..100 of the triangle, flattened

Formula

u(n,x) = u(n-1,x) + v(n-1,x), v(n,x) = x*u(n-1,x) + (x+1)v(n-1,x), where u(1,x)=1, v(1,x)=1.

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012

G.f.: (1-x+y*x)/(1-(y+2)*x+x^2). - Philippe Deléham, Mar 03 2012

For n >= 1, Sum{k=0..n} T(n,k)*x^k = A000012(n), A001906(n), A001834(n-1), A055271(n-1), A038761(n-1), A056914(n-1) for x = 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012

T(n,k) = C(n+k-1,2*k) + 2*C(n+k-1,2*k-1). where C is binomial. - Yuchun Ji, May 23 2019

T(n,k) = T(n-1,k) + A207606(n,k-1). - Yuchun Ji, May 28 2019

Sum_{k=1..n} T(n, k)*x^k = { 4*(-1)^(n-1)*A016921(n-1) (x=-4), 3*(-1)^(n-1) * A130815(n-1) (x=-3), 2*(-1)^(n-1)*A010684(n-1) (x=-2), A057079(n+1) (x=-1), 0 (x=0), A001906(n) = Fibonacci(2*n) (x=1), 2*A001834(n-1) (x=2), 3*A055271(n-1) (x=3), 4*A038761(n-1) (x=4) }. - G. C. Greubel, Mar 15 2020

Example

First five rows:
  1;
  1,  2;
  1,  5,  2;
  1,  9,  9,  2;
  1, 14, 25, 13,  2;
Triangle (1, 0, 1/2, 1/2, 0, 0, 0, ...) DELTA (0, 2, -1, 0, 0, 0, 0, ...) begins:
  1;
  1,  0;
  1,  2,  0;
  1,  5,  2,  0;
  1,  9,  9,  2,  0;
  1, 14, 25, 13,  2,  0;
  1, 20, 55, 49, 17,  2,  0;
  ...
1 = 2*1 - 1, 20 = 2*14 + 1 - 9, 55 = 2*25 + 14 - 9, 49 = 2*13 + 25 - 2, 17 = 2*2 + 1 - 0, 2 = 2*0 + 2 - 0. - Philippe Deléham, Mar 03 2012

Maple

A207607:= (n, k) -> `if`(k=1, 1, binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3) ); seq(seq(A207607(n, k), k = 1..n), n = 1..10); # G. C. Greubel, Mar 15 2020

Mathematica

(* First program *)
u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := x*u[n - 1, x] + (x + 1)*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[%]  (* A207606 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A207607 *)
(* Second program *)
Table[If[k==1, 1, Binomial[n+k-3, 2*k-2] + 2*Binomial[n+k-3, 2*k-3]], {n, 10}, {k, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)

Prog

(Python)
from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else x*u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017
(Sage)
def T(n, k):
    if k == 1: return 1
    else: return binomial(n+k-3, 2*k-2) + 2*binomial(n+k-3, 2*k-3)
[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 15 2020

Crossrefs

Cf. A207606.

Sequence in context: A259447 A228823 A249756 * A146024 A146023 A104766

Adjacent sequences: A207604 A207605 A207606 * A207608 A207609 A207610

Keyword

nonn,tabl

Author

Clark Kimberling, Feb 19 2012

Status

approved