A273622 -id:A273622 - cp's OEIS frontend

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1

Offset: 0

Eric W. Weisstein, Dec 06 2005

Unsigned version of A108045.

The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016

			2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  0:    2
  1:    0  1
  2:    2  0  1
  3:    0  3  0  1
  4:    2  0  4  0  1
  5:    0  5  0  5  0  1
  6:    2  0  9  0  6  0  1
  7:    0  7  0 14  0  7  0  1
  8:    2  0 16  0 20  0  8  0  1
  9:    0  9  0 30  0 27  0  9  0  1
  10:   2  0 25  0 50  0 35  0 10  0  1
  n\k   0  1  2  3  4  5  6  7  8  9 10 ...
  .... reformatted by _Wolfdieter Lang_, Feb 10 2023

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
B. Johnson, Fibonacci numbers and matrices, 2009.
Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4.
M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Eric Weisstein's World of Mathematics, Lucas Polynomial

Cf. A108045 (signed version).
Cf. A034807, A162514.
Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199).

Lucas := proc(n,x)
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
end proc:
A114525 := proc(n,k)
    coeftayl(Lucas(n,x),x=0,k) ;
end proc:
seq(seq(A114525(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Aug 16 2019

row[n_] := CoefficientList[LucasL[n, x], x];
Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *)

From Peter Bala, Mar 18 2015: (Start)
The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x.
O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t + (x^2 + 2)*t^2 + (3*x + x^3)*t^3 + ....
L(n,x) = trace( [ x, 1; 1, 0 ]^n ).
exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson).
exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n.
exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.
L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End)
From Jeremy Dover, Jun 10 2016: (Start)
Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2.
T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n.
T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End)
From Peter Bala, Sep 03 2022: (Start)
L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.
d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.
Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].
P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).
L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.
Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and
Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)
From Peter Bala, Nov 18 2022: (Start)
L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.
For odd m, L(n, L(m,x)) = L(n*m, x).
For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.
Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)
The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - Wolfdieter Lang, Feb 10 2023

A215465 a(n) = (Lucas(4n) - Lucas(2n))/4.

Original entry on oeis.org

0, 1, 10, 76, 540, 3751, 25840, 177451, 1217160, 8344876, 57202750, 392089501, 2687463360, 18420257701, 126254611990, 865362736876, 5931286406640, 40653646980451, 278644255208560, 1909856172864751, 13090349042248500

Offset: 0

R. J. Mathar, Aug 11 2012

This is a divisibility sequence, that is, if n | m then a(n) | a(m). However, it is not a strong divisibility sequence. It is the case k = 3 of a 1-parameter family of 4th-order linear divisibility sequences with o.g.f. x*(1 - x^2)/( (1 - k*x + x^2)*(1 - (k^2 - 2)*x + x^2) ). Compare with A000290 (case k = 2) and A085695 (case k = -3). - Peter Bala, Jan 17 2014

In general, for distinct integers r and s with r = s (mod 2), the sequence Lucas(r*n) - Lucas(s*n) is a fourth-order divisibility sequence. See A273622 for the case r = 3, s = 1. - Peter Bala, May 27 2016

			a(3) = 76 because the 12th (4 * 3rd) Lucas number is 22, the 6th (2 * 3rd) Lucas number is 18, and (322 - 18)/4 = 304/4 = 76.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Peter Bala, Lucas sequences and divisibility sequences
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Hugh Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory vol. 7 (5) (2011) 1255-1277
Index entries for linear recurrences with constant coefficients, signature (10,-23,10,-1).

Cf. A085695, A215466, A273622.

[(Lucas(4*n) - Lucas(2*n))/4: n in [0..20]]; // Vincenzo Librandi, Dec 23 2012

A215465 := proc(n)
    (A000032(4*n)-A000032(2*n))/4 ;
end proc:

Table[(LucasL[4n] - LucasL[2n])/4, {n, 0, 19}] (* Alonso del Arte, Aug 11 2012 *)
CoefficientList[Series[-x*(x-1)*(1+x)/((x^2 - 7*x + 1)* (x^2 - 3*x + 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 23 2012 *)
LinearRecurrence[{10,-23,10,-1},{0,1,10,76},50] (* G. C. Greubel, Jun 02 2016 *)

{a(n) = my(w = quadgen(5)^(2*n)); (2*real(w^2-w) + imag(w^2-w))/4}; /* Michael Somos, Dec 29 2022 */

4*a(n) = A000032(4*n) - A000032(2*n).
a(n) = A056854(n)/4 - A005248(n)/4.
G.f.: -x*(x-1)*(1+x) / ( (x^2-7*x+1)*(x^2-3*x+1) ).
a(n) = 10*a(n-1) - 23*a(n-2) + 10*a(n-3) - a(n-4). - G. C. Greubel, Jun 02 2016
a(n) = 2^(-2-n)*((7-3*sqrt(5))^n-(3-sqrt(5))^n-(3+sqrt(5))^n+(7+3*sqrt(5))^n). - Colin Barker, Jun 02 2016
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 29 2022

A273623 a(n) = Fibonacci(3n) - (2 + (-1)^n)Fibonacci(n).

Original entry on oeis.org

1, 5, 32, 135, 605, 2560, 10933, 46305, 196384, 831875, 3524489, 14929920, 63245753, 267913165, 1134902560, 4807524015, 20365009477, 86267563520, 365435291981, 1548008735625, 6557470308896, 27777889982155, 117669030432337, 498454011740160, 2111485077903025

Offset: 1

Peter Bala, May 29 2016

This is a divisibility sequence: if n divides m then a(n) divides a(m). The sequence satisfies a linear recurrence of order 6. In general, for integers r and s, the sequence Fibonacci(r*n) - 2*Fibonacci((r - 2*s)*n) + Fibonacci((r - 4*s)*n) is a divisibility sequence of the sixth order. This is the case r = 3, s = 1. See A127595 (case r = 4, s = 1).

G. C. Greubel, Table of n, a(n) for n = 1..1000
P. Bala, Lucas sequences and divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (4,4,-12,-4,4,1).

Cf. A000032, A000045, A127595, A273622, A273624, A273625.

[Fibonacci(3*n)-(2+(-1)^n)*Fibonacci(n): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016

#A273623
with(combinat):
seq(fibonacci(3*n) - (2 + (-1)^n)*fibonacci(n), n = 1..25);

LinearRecurrence[{4, 4, -12, -4, 4, 1}, {1, 5, 32, 135, 605, 2560}, 100] (* G. C. Greubel, Jun 02 2016 *)
Table[Fibonacci[3 n] - (2 + (-1)^n) Fibonacci[n], {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)

a(n)=fibonacci(3*n) - (2 + (-1)^n)*fibonacci(n) \\ Charles R Greathouse IV, Jun 08 2016

a(n) = Fibonacci(3*n) - 2*Fibonacci(n) + Fibonacci(-n).
a(2*n) = 5*Fibonacci(2*n)^3;
a(2n+1) = Fibonacci(2*n+1)*(5*Fibonacci(2*n+1)^2 - 4) = Fibonacci(2*n+1)*Lucas(2*n+1)^2.
O.g.f. x*(x^4 - x^3 + 8*x^2 + x + 1)/( (1 + x - x^2 )*(1 - x - x^2)*(1 - 4*x - x^2 ) ).
a(n) = 4*a(n-1) + 4*a(n-2) - 12*a(n-3) - 4*a(n-4) + 4*a(n-5) + a(n-6). - G. C. Greubel, Jun 02 2016

A273624 a(n) = (1/11)(Fibonacci(4n) + Fibonacci(6*n)).

Original entry on oeis.org

1, 15, 248, 4305, 76255, 1361520, 24384737, 437245935, 7843863784, 140737371825, 2525326494911, 45314438127840, 813129752279233, 14590988151618255, 261824431125415640, 4698247224097107345, 84306614992412658847, 1512820749915870503760, 27146466385039244529569

Offset: 1

Peter Bala, May 29 2016

This is a divisibility sequence: if n divides m then a(n) divides a(m). More generally, if r is an even integer then the sequence Fibonacci(r*n) + Fibonacci((r + 2)*n) is a divisibility sequence. See A215466 for the case r = 2.

Also, the sequence s(n) := Fibonacci(4*n) + Fibonacci(6*n) + ... + Fibonacci((2*k + 2)*n) is a divisibility sequence when k is a positive integer that is not a multiple of 3.

G. C. Greubel, Table of n, a(n) for n = 1..795
P. Bala, Lucas sequences and divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (25,-128,25,-1).

Cf. A000045, A049673, A215466, A273623, A273625.

[1/11*(Fibonacci(4*n)+Fibonacci(6*n)): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016

#A273624
with(combinat):
seq(1/11*(fibonacci(4n) + fibonacci(6n)), n = 1..20);

LinearRecurrence[{25,-128,25,-1},{1, 15, 248, 4305},100] (* G. C. Greubel, Jun 02 2016 *)
Table[1/11 (Fibonacci[4 n] + Fibonacci[6 n]), {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)

a(n)=(fibonacci(4*n) + fibonacci(6*n))/11 \\ Charles R Greathouse IV, Jun 08 2016

a(n) = -a(-n).
a(n) = 25*a(n-1) - 128*a(n-2) + 25*a(n-3) - a(n-4).
O.g.f. (x^2 - 10*x + 1)/((x^2 - 7*x + 1)*(x^2 - 18*x + 1)).

A273625 a(n) = (1/12)(Fibonacci(2n) + Fibonacci(4n) + Fibonacci(6n)).

Original entry on oeis.org

1, 14, 228, 3948, 69905, 1248072, 22352707, 400808856, 7190208684, 129009258070, 2314882621811, 41538234954384, 745368939599413, 13375072472343218, 240005728531700340, 4306726622089196592, 77281063743045412517, 1386752354089549205976, 24884260852952644076119

Offset: 1

Peter Bala, May 31 2016

This is a divisibility sequence, that is, if n divides m then a(n) divides a(m). The sequence satisfies a sixth-order linear recurrence. More generally, the sequence s(n) := Fibonacci(2*n) + Fibonacci(4*n) + ... + Fibonacci(2*k*n) is a divisibility sequence for k = 1,2,3,.... See A215466 for the case k = 2. Cf. A273623, A273624.
From Peter Bala, Aug 05 2019: (Start)
Let  U(n;P,Q), where P and Q are integer parameters, denote the Lucas sequence of the first kind. Then, excluding the cases P = -1 and P = 0, the sequence ( U(n;P,1) + U(2*n;P,1) + U(3*n;P,1))/(P^2 + P) is a sixth-order linear divisibility sequence with o.g.f. x*(1 - 2*(P^2 - 2)*x + (3*P^3 - 3*P^2 - 8*P + 10)*x^2 - 2*(P^2 - 2)*x^3 + x^4)/((1 - P*x + x^2)*(1 - (P^2 - 2)*x + x^2)*(1 - P*(P^2 - 3)*x + x^2)). This is the case P = 3.
More generally, the sequence U(n;P,1) + U(2*n;P,1) + ... + U(k*n;P,1) is a linear divisibility sequence of order 2*k.  See, for example, A215466 with P = 3, k = 2. (End)

G. C. Greubel, Table of n, a(n) for n = 1..795
P. Bala, Lucas sequences and divisibility sequences
P. Bala, Divisibility sequences from strong divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (28,-204,434,-204,28,-1).

Cf. A000045, A001906, A215466, A273623, A273624, A215466.

[1/12*(Fibonacci(2*n)+Fibonacci(4*n)+Fibonacci(6*n)): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016

#A273625
with(combinat):
seq(1/12*(fibonacci(2*n) + fibonacci(4*n) + fibonacci(6*n)), n = 1..20);

LinearRecurrence[{28, -204, 434, -204, 28, -1},{1, 14, 228, 3948, 69905, 1248072}, 100] (* G. C. Greubel, Jun 02 2016 *)
Table[1/12 (Fibonacci[2 n] + Fibonacci[4 n] + Fibonacci[6 n]), {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)

A001906(n)=fibonacci(2*n)
a(n)=(A001906(n)+A001906(2*n)+A001906(3*n))/12 \\ Charles R Greathouse IV, Jun 08 2016

a(n) = -a(-n).
O.g.f.: x*(x^4 - 14*x^3 + 40*x^2 - 14*x + 1)/((x^2 - 3*x + 1)*(x^2 - 7*x + 1)*(x^2 - 18*x + 1)).
a(n) = 28*a(n-1) - 204*a(n-2) + 434*a(n-3) - 204*a(n-4) + 28*a(n-5) - a(n-6). - G. C. Greubel, Jun 02 2016

A037451 a(n) = Fibonacci(n) * Fibonacci(2*n).

Original entry on oeis.org

0, 1, 3, 16, 63, 275, 1152, 4901, 20727, 87856, 372075, 1576279, 6676992, 28284569, 119814747, 507544400, 2149990983, 9107510539, 38580029568, 163427634589, 692290558575, 2932589884016, 12422650070163, 52623190204271, 222915410823168, 944284833600625, 4000054745057907, 16944503814103696, 71778070001033487

Offset: 0

Gary W. Adamson, Feb 01 2000

Let F(n) = Fibonacci(n), then abs(det([F(n), F(n+k); F(n+2k), F(n+3k)])) = a(k), independent of n. - R. M. Welukar, Aug 26 2014
From Joerg Arndt, Aug 26 2014: (Start)
This is a special case of Johnson's identity (relation 32 in the Mathworld link).
F(a)*F(b) - F(c)*F(d) = (-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)), where a+b = c+d and r arbitrary.
Here a = n, b = n+3*k, c = n+k, d = n+2*k, and r = c, so that
(-1)^r*(F(a-r)*F(b-r) - F(c-r)*F(d-r)) =
(-1)^c*(F(a-c)*F(b-c) - F(c-c)*F(d-c)) =
(-1)^c*(F(a-c)*F(b-c) - 0) =
(-1)^c*(F(-k)*F(-2*k)), taking the absolute value gives a(k).
(End)
Let L(n) = A000032(n), then abs(det([L(n), L(n+k); L(n+2k), L(n+3k)])) = 5*a(k), independent of n. - M. N. Deshpande and R. M. Welukar, Aug 30 2014

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein, Fibonacci Number (MathWorld).
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000032, A000045, A098317, A273622.

[Fibonacci(n)*Fibonacci(2*n): n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

seq((fibonacci(2*n)*fibonacci(n)), n=0..25); # Zerinvary Lajos, Jun 24 2006

Table[Fibonacci[n]Fibonacci[2n],{n,0,40}] (* Harvey P. Dale, Mar 13 2011 *)

concat([0], Vec( x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)) + O(x^66) ) ) \\ Joerg Arndt, Aug 26 2014

From Emanuele Munarini, Jul 18 2003: (Start)
G.f.: ( x + x^3 )/( 1 - 3 x - 6 x^2 + 3 x^3 + x^4 ).
a(n+4) = 3*a(n+3) + 6*a(n+2) - 3*a(n+1) - a(n).
(End)
G.f.: x*(1+x^2) / ((1+x-x^2)*(1-4*x-x^2)). - Joerg Arndt, Aug 26 2014
a(n) = (1/5)*(Lucas(3*n) - (-1)^n*Lucas(n)) = (1/5)*(Lucas(3*n) - Lucas(-n)). In general, for r = s (mod 2) the sequence Lucas(r*n) - Lucas(s*n) is a divisibility sequence. Cf. A273622. - Peter Bala, May 27 2016
Lim_{n->infinity} a(n+1)/a(n) = 2 + sqrt(5) = A098317. - Ilya Gutkovskiy, Jun 01 2016
a(n) = (-(1/2*(-1-sqrt(5)))^n+(2-sqrt(5))^n-(1/2*(-1+sqrt(5)))^n+(2+sqrt(5))^n)/5. - Colin Barker, Jun 03 2016

A049673 a(n) = (F(3n) + F(n))/3, where F = A000045 (the Fibonacci sequence).

Original entry on oeis.org

0, 1, 3, 12, 49, 205, 864, 3653, 15463, 65484, 277365, 1174889, 4976832, 21082073, 89304891, 378301260, 1602509321, 6788337557, 28755857952, 121811766781, 516002920895, 2185823443596, 9259296684333, 39223010163217, 166151337308544, 703828359351025

Offset: 0

Clark Kimberling

This is an odd divisibility sequence, that is, if n divides m and n/m is odd then a(n) divides a(m). More generally, if r and s are positive integers such that r = s (mod 2) then the sequence Fibonacci(r*n) + Fibonacci(s*n) is an odd divisibility sequence. In the particular case that r is even and s = r + 2 then Fibonacci(r*n) + Fibonacci(s*n) is, in fact, a divisibility sequence. See for example A215466 and A273624. - Peter Bala, May 29 2016

Colin Barker, Table of n, a(n) for n = 0..1000
P. Bala, Lucas sequences and divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (5,-2,-5,-1).

Cf. A000045, A000129, A001045, A215466, A273624.

[(Fibonacci(3*n)+Fibonacci(n))/3: n in [0..30]]; // Vincenzo Librandi, Jun 04 2016

with(combinat): A049673:=n->(fibonacci(3*n)+fibonacci(n))/3: seq(A049673(n), n=0..30); # Wesley Ivan Hurt, Jun 01 2016

Table[(Fibonacci[3 n] + Fibonacci[n])/3, {n, 0, 30}] (* Wesley Ivan Hurt, Jun 01 2016 *)
LinearRecurrence[{5,-2,-5,-1},{0,1,3,12},30] (* Harvey P. Dale, Sep 21 2022 *)

concat(0, Vec(x*(1-2*x-x^2)/((x^2+4*x-1)*(x^2+x-1)) + O(x^30))) \\ Colin Barker, Jun 02 2016

G.f.: x*(1-2*x-x^2) / ((x^2+4*x-1)*(x^2+x-1)). - R. J. Mathar, Oct 26 2015
a(n) = 5*a(n-1) - 2*a(n-2) - 5*a(n-3) - a(n-4) for n>3. - Wesley Ivan Hurt, Jun 01 2016
a(n) = ((-(1/2*(1-sqrt(5)))^n-(2-sqrt(5))^n+(1/2*(1+sqrt(5)))^n+(2+sqrt(5))^n))/(3*sqrt(5)). - Colin Barker, Jun 02 2016
G.f.: G(F(t)), where G(t) is g.f. of A001045 and F(t) is g.f. of A000129. - Oboifeng Dira, Dec 07 2016

A273626 A fourth-order divisibility sequence: a(n) = (1/14)(Pell(4n) + Pell(2*n)).

Original entry on oeis.org

1, 30, 995, 33660, 1142629, 38810970, 1318402631, 44786716920, 1521429030985, 51683794848150, 1755727563817259, 59643053188493940, 2026108079758297261, 68828031652259981010, 2338126968060165944975, 79427488882178225107440, 2698196495024745460575889

Offset: 1

Peter Bala, May 31 2016

This is a divisibility sequence, that is, a(n) divides a(m) whenever n divides m. The sequence satisfies a linear recurrence of order 4.

A000129(n) = Pell(n) is the Lucas sequence U_n(2,-1). In general, if U(n) = U_n(P,Q) is the Lucas sequence with integer parameters P and Q then when Q = 1 or Q = -1 both U(4*n) + U(2*n) and U(4*n) - 2*U(2*n) are divisibility sequences of the fourth order. Cf. A127595, A215466 and A273627.

Colin Barker, Table of n, a(n) for n = 1..600
P. Bala, Lucas sequences and divisibility sequences
Wikipedia, Lucas Sequence
Index entries for linear recurrences with constant coefficients, signature (40,-206,40,-1).

Cf. A000129, A001109, A029547, A081555, A127595, A215466, A273627.

I:=[1,30,995,33660]; [n le 4 select I[n] else 40*Self(n-1)-206*Self(n-2)+40*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 04 2016

#A273626
A000129 := proc (n) option remember;
if n <= 1 then n else 2*A000129(n-1) + A000129(n-2) end if
end proc:
seq(1/14*(A000129(4*n) + A000129(2*n)), n = 1..20);

LinearRecurrence[{40,-206,40,-1},{1,30,995,33660},100] (* G. C. Greubel, Jun 02 2016 *)

a(n) = sqrt(2)/56*( (sqrt(2) + 1)^(4*n) - (sqrt(2) - 1)^(4*n) + (sqrt(2) + 1)^(2*n) - (sqrt(2) - 1)^(2*n) ).
a(n) = (A082405(n) + A001109(n))/7 .
a(n) = 1/14*Pell(2*n)*A081555(n).
a(n) = -a(-n).
a(n) = 40*a(n-1) - 206*a(n-2) + 40*a(n-3) - a(n-4) for n>4.
O.g.f.: x*(x^2 - 10*x + 1)/((x^2 - 6*x + 1)*(x^2 - 34*x + 1)).

A273627 A divisibility sequence: (1/8)(Pell(4n) - 2Pell(2n)).

Original entry on oeis.org

1, 48, 1715, 58752, 1998709, 67914000, 2307174311, 78376578048, 2662499775145, 90446634986352, 3072523201721819, 104375342876112000, 3545689138389464221, 120449055384533383248, 4091722194064948458575, 138998105543576763850752, 4721843866291934117429329

Offset: 1

Peter Bala, May 30 2016

The sequence of Pell numbers A000129 is a well-known divisibility sequence of order 2, that is, the sequence satisfies a linear recurrence of order 2 and Pell(n) divides Pell(m) whenever n divides m. The linear combinations Pell(4*n) - 2*Pell(2*n) and Pell(4*n) + Pell(2*n) are also divisibility sequences, this time of order 4. Cf. A127595 and A273626.

Colin Barker, Table of n, a(n) for n = 1..600
P. Bala, Lucas sequences and divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (40,-206,40,-1).

Cf. A000129, A002203, A127595, A273626.

I:=[1,48,1715,58752]; [n le 4 select I[n] else 40*Self(n-1)-206*Self(n-2)+40*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 02 2016

#A273627
A000129 := proc (n) option remember;
if n <= 1 then n else 2*A000129(n-1)+A000129(n-2) end if
end proc:
seq(1/8*(A000129(4*n)-2*A000129(2*n)), n = 1..20);

CoefficientList[Series[(x^2 + 8*x + 1)/((x^2 - 6*x + 1)*(x^2 - 34*x + 1)), {x, 0, 20}], x] (* Wesley Ivan Hurt, Jun 01 2016 *)
LinearRecurrence[{40, -206, 40, -1}, {1, 48, 1715, 58752}, 20] (* Vincenzo Librandi, Jun 02 2016 *)

a(n) = A000129(4*n) - 2*A000129(2*n).
a(n) = A000129(2*n)*(A002203(2*n) - 2).
a(n) = -a(-n).
a(n) = 40*a(n-1) - 206*a(n-2) + 40*a(n-3) - a(n-4) for n>4.
O.g.f.: x*(x^2 + 8*x + 1)/((x^2 - 6*x + 1)*(x^2 - 34*x + 1)).
a(n) = ((17+12*sqrt(2))^(1-n)*(-1+2*(3+2*sqrt(2))^n+(17+12*sqrt(2))^(2*n)-2*(99+70*sqrt(2))^n))/(384+272*sqrt(2)). - Colin Barker, Jun 02 2016

cp's OEIS Frontend

A114525 Triangle of coefficients of the Lucas (w-)polynomials.

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A215465 a(n) = (Lucas(4n) - Lucas(2n))/4.

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A273623 a(n) = Fibonacci(3*n) - (2 + (-1)^n)*Fibonacci(n).

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A273624 a(n) = (1/11)*(Fibonacci(4*n) + Fibonacci(6*n)).

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A273625 a(n) = (1/12)*(Fibonacci(2*n) + Fibonacci(4*n) + Fibonacci(6*n)).

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A037451 a(n) = Fibonacci(n) * Fibonacci(2*n).

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A049673 a(n) = (F(3n) + F(n))/3, where F = A000045 (the Fibonacci sequence).

A273623 a(n) = Fibonacci(3n) - (2 + (-1)^n)Fibonacci(n).

A273624 a(n) = (1/11)(Fibonacci(4n) + Fibonacci(6*n)).

A273625 a(n) = (1/12)(Fibonacci(2n) + Fibonacci(4n) + Fibonacci(6n)).

A273626 A fourth-order divisibility sequence: a(n) = (1/14)(Pell(4n) + Pell(2*n)).

A273627 A divisibility sequence: (1/8)(Pell(4n) - 2Pell(2n)).