A172968 -id:A172968 - cp's OEIS frontend

A033889 a(n) = Fibonacci(4*n + 1).

1, 5, 34, 233, 1597, 10946, 75025, 514229, 3524578, 24157817, 165580141, 1134903170, 7778742049, 53316291173, 365435296162, 2504730781961, 17167680177565, 117669030460994, 806515533049393, 5527939700884757, 37889062373143906, 259695496911122585, 1779979416004714189

Offset: 0

N. J. A. Sloane

For positive n, a(n) equals (-1)^n times the permanent of the (4n) X (4n) tridiagonal matrix with sqrt(i)'s along the three central diagonals, where i is the imaginary unit. - John M. Campbell, Jul 12 2011

a(n) = 5^n*a(n; 3/5) = (16/5)^n*a(2*n; 3/4), and F(4*n) = 5^n*b(n; 3/5) = (16/5)^n*b(2*n; 3/4), where a(n; d) and b(n; d), n=0, 1, ..., d in C, denote the delta-Fibonacci numbers defined in comments to A014445. Two of these identities from the following relations follows: F(k+1)^n*a(n; F(k)/F(k+1)) = F(k*n+1) and F(k+1)^n*b(n; F(k)/F(k+1)) = F(k*n) (see also Witula's et al. papers). - Roman Witula, Jul 24 2012

Bruno Berselli, Table of n, a(n) for n = 0..500
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Kai Wan, Problem H-90, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 60, No. 3 (2022), p. 282; Solution to Problem H-90, by Ángel Plaza, ibid., Vol. 62, No. 1 (2024), pp. 95-96.
Roman Wituła, Binomials Transformation Formulae of Scaled Lucas Numbers, Demonstratio Mathematica, Vol. XLVI, No. 1 (2013), pp. 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A014445, A016813, A122070, A167816, A094567.

Cf. A003881, A049684, A081018, A081016, A081068, A172968.

[Fibonacci(4*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011

Table[Fibonacci[4*n+1], {n,0,14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)

a(n)=fibonacci(4*n+1) \\ Charles R Greathouse IV, Jul 15 2011

Vec((1-2*x)/(1-7*x+x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = 7*a(n-1) - a(n-2) for n >= 2. - Floor van Lamoen, Dec 10 2001
From R. J. Mathar, Jan 17 2008: (Start)
O.g.f.: (1 - 2*x)/(1 - 7*x + x^2).
a(n) = A004187(n+1) - 2*A004187(n). (End); corrected by Klaus Purath, Jul 29 2020
a(n) = A167816(4*n+1). - Reinhard Zumkeller, Nov 13 2009
a(n) = sqrt(1 + 2 * Fibonacci(2*n) * Fibonacci(2*n + 1) + 5 * (Fibonacci(2*n) * Fibonacci(2*n + 1))^2). - Artur Jasinski, Feb 06 2010
a(n) = Sum_{k=0..n} A122070(n,k)*2^k. - Philippe Deléham, Mar 13 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n)*Fibonacci(2*n+2) + 1. - Gary Detlefs, Apr 18 2012
a(n) = Fibonacci(2*n)^2 + Fibonacci(2*n+1)^2. - Bruno Berselli, Apr 19 2012
a(n) = Sum_{k = 0..n} A238731(n,k)*4^k. - Philippe Deléham, Mar 05 2014
a(n) = A000045(A016813(n)). - Michel Marcus, Mar 05 2014
2*a(n) = Fibonacci(4*n) + Lucas(4*n). - Bruno Berselli, Oct 13 2017
a(n) = A094567(n-1) + A094567(n), assuming A094567(-1) = 0. - Klaus Purath, Jul 29 2020
Sum_{n>=0} (-1)^n * arctan(3/a(n)) = Pi/4 (A003881) (Wan, 2022). - Amiram Eldar, Mar 01 2024
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

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

Original entry on oeis.org

1, 2, 3, 2, 4, 7, 2, 5, 16, 11, 2, 6, 30, 36, 15, 2, 7, 50, 91, 64, 19, 2, 8, 77, 196, 204, 100, 23, 2, 9, 112, 378, 540, 385, 144, 27, 2, 10, 156, 672, 1254, 1210, 650, 196, 31, 2, 11, 210, 1122, 2640, 3289, 2366, 1015, 256, 35, 2, 12, 275, 1782, 5148, 8008

Offset: 1

Clark Kimberling, Feb 19 2012

As triangle T(n,k) with 0 <= k <= n, it is (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 03 2012

			First five rows:
  1;
  2;
  3,  2;
  4,  7,  2;
  5, 16, 11,  2;
Triangle (2, -1/2, 1/2, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, ...), 0 <= k <= n, begins:
  1;
  2,   0;
  3,   2,   0;
  4,   7,   2,   0;
  5,  16,  11,   2,   0;
  6,  30,  36,  15,   2,   0;
  7,  50,  91,  64,  19,   2,   0;
  8,  77, 196, 204, 100,  23,   2,   0;

G. C. Greubel, Rows n = 1..101 of the triangle, flattened

Cf. A207607.

T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=0 then n+2
    elif k=n then 2
    else 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
      fi; end:
1, seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Mar 15 2020

(* 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 *)
T[n_, k_]:= T[n, k]= If[k<0 || k>n, 0, If[k==0, n+2, If[k==n, 2, 2*T[n-1, k] - T[n-2, k] + T[n-1, k-1] ]]]; Join[{1}, Table[T[n, k], {n, 0, 10}, {k, 0, n}]]//Flatten (* G. C. Greubel, Mar 15 2020 *)

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(u(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==1): return n+1
    elif (k==n): return 2
    else: return 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k)
[1]+[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Mar 15 2020

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.
As triangle T(n,k) with 0 <= k <= n: g.f.: (1-y*x)/(1-(2+y)*x+x^2). - Philippe Deléham, Mar 03 2012
As triangle T(n,k) with 0 <= k <= n: Sum_{k=0..n} T(n,k)*x^k = A132677(n), A000034(n)*A057077(n), A057079(n), A000027(n+1), A001519(n+1), A001075(n), A002310(n), A038725(n), A172968(n) for x = -3, -2, -1, 0, 1, 2, 3, 4, 5 respectively. - Philippe Deléham, Mar 03 2012
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k). - Philippe Deléham, Mar 03 2012
T(n,k) = C(n+k-1,2*k+1) + 2*C(n+k-1,2*k), where C is binomial. - Yuchun Ji, May 23 2019
T(n,k) = T(n-1,k) + A207607(n-1,k). - Yuchun Ji, May 28 2019

cp's OEIS Frontend

A033889 a(n) = Fibonacci(4*n + 1).

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