A056914 -id:A056914 - cp's OEIS frontend

A056854 a(n) = Lucas(4*n).

2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447

Offset: 0

Barry E. Williams, Aug 29 2000

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

			Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper);  n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A004187, A005248.
Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* Harvey P. Dale, Aug 08 2011 *)

a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])

a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))

[lucas_number2(n,7,1) for n in range(27)] #Zerinvary Lajos, Jun 25 2008

a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
a(n) = A000032(4*n), where A000032 = Lucas numbers.
a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
G.f.: (2-7x)/(1-7x+x^2).
a(n) = A005248(2*n); bisection of A005248.
a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - Wolfdieter Lang, Aug 27 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - Bruno Berselli, May 25 2015
a(n) = sqrt(45*(A004187(n))^2+4).
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) (mod p^k) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - Greg Dresden and Tracy Z. Wu, Sep 10 2020
a(n) = Sum_{k>=1} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 20 2025

More terms from James Sellers, Aug 31 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A246453 Lucas numbers (A000204) of the form n^2 + 2.

Original entry on oeis.org

3, 11, 18, 123, 843, 5778, 39603, 271443, 1860498, 12752043, 87403803, 599074578, 4106118243, 28143753123, 192900153618, 1322157322203, 9062201101803, 62113250390418, 425730551631123, 2918000611027443, 20000273725560978, 137083915467899403, 939587134549734843

Offset: 1

Michel Lagneau, Aug 26 2014

a(n) = {11} union {A000204(2+4*n)} for n=0,1,...

Intersection of A000204 and A059100. - Michel Marcus, Aug 26 2014

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000204 (Lucas), A059100 (n^2+2).

Cf. quadrisection of A000032: A056854 (first), A056914 (second), this sequence (third, without 11), A288913 (fourth).

I:=[3,11,18,123]; [n le 4 select I[n] else 7*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

with(combinat,fibonacci):lst:={}:lst1:={}:nn:=5000:
  for n from 1 to nn do:
    lst:=lst union {2*fibonacci(n-1)+fibonacci(n)}:
  od:
   for m from 1 to nn do:
    if {m^2+2} intersect lst = {m^2+2}
    then
    lst1:=lst1 union {m^2+2}:
    else
    fi:
   od:
   print(lst1):

CoefficientList[Series[x*(3-10*x-56*x^2+8*x^3)/(1-7*x+x^2), {x,0,50}], x] (* or *) LinearRecurrence[{7,-1}, {3, 11, 18, 123}, 30] (* G. C. Greubel, Dec 21 2017 *)
Select[LucasL[Range[100]],IntegerQ[Sqrt[#-2]]&] (* Harvey P. Dale, Dec 31 2018 *)

lista(nn) = for (n=0, nn, luc = fibonacci(n+1) + fibonacci(n-1); if (issquare(luc-2), print1(luc, ", "))); \\ Michel Marcus, Mar 29 2016

Vec(x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

From Colin Barker, Jun 20 2017: (Start)
G.f.: x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2).
a(n) = (2^(-n)*((7+3*sqrt(5))^n*(-20+9*sqrt(5)) + (7-3*sqrt(5))^n*(20+9*sqrt(5)))) / sqrt(5) for n>2.
a(n) = 7*a(n-1) - a(n-2) for n>4. (End)
E.g.f.: 2*exp(7*x/2)*(9*cosh(3*sqrt(5)*x/2) - 4*sqrt(5)*sinh(3*sqrt(5)*x/2)) + 4*x^2 - 18. - Stefano Spezia, Apr 14 2025

Corrected by Michel Marcus, Mar 29 2016

A288913 a(n) = Lucas(4*n + 3).

Original entry on oeis.org

4, 29, 199, 1364, 9349, 64079, 439204, 3010349, 20633239, 141422324, 969323029, 6643838879, 45537549124, 312119004989, 2139295485799, 14662949395604, 100501350283429, 688846502588399, 4721424167835364, 32361122672259149, 221806434537978679, 1520283919093591604

Offset: 0

Bruno Berselli, Jun 19 2017

a(n) mod 4 gives A101000.

Colin Barker, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000032, A000045, A004187, A101000.
Cf. A033891: fourth quadrisection of A000045.
Partial sums are in A081007 (after 0).
Positive terms of A098149, and subsequence of A001350, A002878, A016897, A093960, A068397.
Quadrisection of A000032: A056854 (first), A056914 (second), A246453 (third, without 11), this sequence (fourth).

[Lucas(4*n + 3): n in [0..30]]; // G. C. Greubel, Dec 22 2017

LucasL[4 Range[0, 21] + 3]
LinearRecurrence[{7,-1}, {4,29}, 30] (* G. C. Greubel, Dec 22 2017 *)

Vec((4 + x)/(1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

from sympy import lucas
def a(n):  return lucas(4*n + 3)
print([a(n) for n in range(22)]) # Michael S. Branicky, Apr 29 2021

def L():
    x, y = -1, 4
    while True:
        yield y
        x, y = y, 7*y - x
r = L(); [next(r) for  in (0..21)] # _Peter Luschny, Jun 20 2017

G.f.: (4 + x)/(1 - 7*x + x^2).
a(n) = 7*a(n-1) - a(n-2) for n>1, with a(0)=4, a(1)=29.
a(n) = ((sqrt(5) + 1)^(4*n + 3) - (sqrt(5) - 1)^(4*n + 3))/(8*16^n).
a(n) = Fibonacci(4*n+4) + Fibonacci(4*n+2).
a(n) = 4*A004187(n+1) + A004187(n).
a(n) = 5*A003482(n) + 4 = 5*A081016(n) - 1.
a(n) = A002878(2*n+1) = A093960(2*n+3) = A001350(4*n+3) = A068397(4*n+3).
a(n+1)*a(n+k) - a(n)*a(n+k+1) = 15*Fibonacci(4*k). Example: for k=6, a(n+1)*a(n+6) - a(n)*a(n+7) = 15*Fibonacci(24) = 695520.

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

Original entry on oeis.org

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

Clark Kimberling, Feb 19 2012

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

			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

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

Cf. A207606.

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

(* 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 *)

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

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

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

A354336 a(n) is the integer w such that (L(2n)^2, -L(2n-1)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

1, 11, 61, 401, 2731, 18701, 128161, 878411, 6020701, 41266481, 282844651, 1938646061, 13287677761, 91075098251, 624238009981, 4278590971601, 29325898791211, 201002700566861, 1377693005176801, 9442848335670731, 64722245344518301, 443612869075957361

Offset: 0

XU Pingya, Jun 20 2022

Subsequence of A017281.

			2*(L(4)^2)^3 + 2*(-L(3)^2)^3 + (-61)^3 = 2*(49)^3 + 2*(-1)^3 + (-61)^3 = 125, a(2) = 61.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032, A002878, A005248, A017281, A056914, A081015, A092521.

Cf. A337928, A354337.

LucasL[4*Range[22]-3] + 1 - LucasL[2*Range[22]-3]^2

a(n) = (-125 + 2*A005248(n)^6 - 2*A002878(n-1)^6)^(1/3).
a(n) = Lucas(4*n+1) - Lucas(4*n-2) + 3 = A056914(n) - 15*A092521(n-1), for n > 1.
a(n) = Lucas(4*n+1) + 1 - Lucas(2*n-1)^2.
a(n) = 2*A081015(n-1) + 1.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: (1 + 3*x - 19*x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022
a(n) = (F(2*n+1) + F(2*n-1))^2 + (F(2*n+1) + F(2*n-1)) * (F(2*n-1) + F(2*n-3)) - (F(2*n-1) + F(2*n-3))^2. - XU Pingya, Jul 17 2024

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Original entry on oeis.org

19, 149, 1039, 7139, 48949, 335519, 2299699, 15762389, 108037039, 740496899, 5075441269, 34787591999, 238437702739, 1634276327189, 11201496587599, 76776199786019, 526231901914549, 3606847113615839, 24721697893396339, 169445038140158549, 1161393569087713519

Offset: 1

XU Pingya, Jun 20 2022

Subsequence of A017377.

			2*(L(4)^2)^3 + 2*(-L(5)^2)^3 + (149)^3 = 2*(49)^3 + 2*(-121)^3 + (149)^3 = 125, a(2) = 149.

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032, A002878, A005248, A017377, A081017, A089508, A092521, A288913.

Cf. A337929, A354336.

LinearRecurrence[{7,-1},{19,149},21]-1 + LucasL[2*Range[21]-3]^2

a(n) = (125 - 2*A005248(n)^6 + 2*A002878(n)^6)^(1/3).
a(n) = Lucas(4*n+2) + Lucas(4n-1) - 3 = 2*A056914(n)-3 = 15*A092521(n) + A288913(n-1).
a(n) = 2*A081017(n) - 1.
a(n) = 10*A089508(n) + 9.
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: x*(19 - 3*x - x^2)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jun 22 2022

A081013 a(n) = Fibonacci(4n+3) - 2, or Fibonacci(2n)Lucas(2n+3).

Original entry on oeis.org

0, 11, 87, 608, 4179, 28655, 196416, 1346267, 9227463, 63245984, 433494435, 2971215071, 20365011072, 139583862443, 956722026039, 6557470319840, 44945570212851, 308061521170127, 2111485077978048, 14472334024676219

Offset: 0

R. K. Guy, Mar 01 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A322159.

List([0..40], n-> Fibonacci(4*n+3) -2); # G. C. Greubel, Jul 14 2019

[Fibonacci(4*n+3)-2: n in [0..40]]; // Vincenzo Librandi, Apr 20 2011

[Fibonacci(4*n+3)-2: n in [0..40]]; // G. C. Greubel, Jul 14 2019

with(combinat) for n from 0 to 40 do printf(`%d,`,fibonacci(4*n+3)-2) od # James Sellers, Mar 03 2003

LinearRecurrence[{8,-8,1},{0,11,87},40] (* Harvey P. Dale, Dec 05 2013 *)
Table[Fibonacci[2n] LucasL[2n+3], {n,1,40}] (* Rigoberto Florez, Apr 20 2019 *)
Table[Sum[Binomial[2n-1+i, 2n-1-i], {i, 1, 2n-1}]-1, {n, 1, 40}] (* Rigoberto Florez, Apr 20 2019 *)

my(x='x+O('x^40)); concat([0], Vec(x*(11-x)/((1-x)*(1-7*x+x^2)))) \\ G. C. Greubel, Dec 24 2017

vector(40, n, n--; fibonacci(4*n+3)-2) \\ G. C. Greubel, Jul 14 2019

[fibonacci(4*n+3)-2 for n in (0..40)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
From R. J. Mathar, Sep 03 2010: (Start)
G.f.: x*(11 - x)/((1-x)*(1-7*x+x^2)).
a(n) = A033891(n) - 2.
a(n+1) - a(n) = A056914(n+1), n>0. (End)
a(n) = 7*a(n-1) - a(n-2) + 10, n>=2. - R. J. Mathar, Nov 07 2015
From Rigoberto Florez, Apr 20 2019: (Start)
a(n) = Sum_{i=0..2n} F(i)*L(i+2), F(i) = A000045(i) and L(i) = A000032(i).
a(n) = (Sum_{i=1..2n-1} binomial(2n-1+i,2n-1-i)) - 1. (End)
Product_{n>=1} (1 + 1/a(n)) = 2*(1-1/sqrt(5)) = 2*A322159. - Amiram Eldar, Nov 28 2024

A081014 a(n) = Lucas(4n+1) + 1, or Lucas(2n)Lucas(2n+1).

Original entry on oeis.org

2, 12, 77, 522, 3572, 24477, 167762, 1149852, 7881197, 54018522, 370248452, 2537720637, 17393796002, 119218851372, 817138163597, 5600748293802, 38388099893012, 263115950957277, 1803423556807922, 12360848946698172

Offset: 0

R. K. Guy, Mar 01 2003

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers).

List([0..20], n-> Lucas(1,-1,4*n+1)[2] +1); # G. C. Greubel, Jul 14 2019

I:=[2,12,77]; [n le 3 select I[n] else 8*Self(n-1) - 8*Self(n-2) + Self(n-3): n in [0..30]]; // G. C. Greubel, Dec 24 2017

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 25 do printf(`%d,`,luc(4*n+1)+1) od: # James Sellers, Mar 03 2003

LinearRecurrence[{8,-8,1}, {2,12,77}, 20] (* G. C. Greubel, Dec 24 2017 *)
LucasL[4*Range[0,20] +1] +1 (* G. C. Greubel, Jul 14 2019 *)
CoefficientList[Series[(2-4x-3x^2)/((1-x)(1-7x+x^2)),{x,0,30}],x] (* Harvey P. Dale, Aug 27 2021 *)

vector(20, n, n--; f=fibonacci; f(4*n+2)+f(4*n)+1) \\ G. C. Greubel, Dec 24 2017

[lucas_number2(4*n+1,1,-1) +1 for n in (0..20)] # G. C. Greubel, Jul 14 2019

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
From R. J. Mathar, Sep 03 2010: (Start)
G.f.: (2 -4*x -3*x^2)/((1-x)*(1-7*x+x^2)).
a(n) = 1 + A056914(n). (End)
a(n) = 7*a(n-1) - a(n-2) - 5, n >= 2. - R. J. Mathar, Nov 07 2015

cp's OEIS Frontend

A056854 a(n) = Lucas(4*n).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

Sage

Formula

Extensions

A246453 Lucas numbers (A000204) of the form n^2 + 2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

A288913 a(n) = Lucas(4*n + 3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Sage

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Sage

Formula

A354336 a(n) is the integer w such that (L(2*n)^2, -L(2*n-1)^2, -w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A354337 a(n) is the integer w such that (L(2*n)^2, -L(2*n + 1)^2, w) is a primitive solution to the Diophantine equation 2*x^3 + 2*y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A354336 a(n) is the integer w such that (L(2n)^2, -L(2n-1)^2, -w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

A354337 a(n) is the integer w such that (L(2n)^2, -L(2n + 1)^2, w) is a primitive solution to the Diophantine equation 2x^3 + 2y^3 + z^3 = 125, where L(n) is the n-th Lucas number (A000032).

A081013 a(n) = Fibonacci(4n+3) - 2, or Fibonacci(2n)Lucas(2n+3).

A081014 a(n) = Lucas(4n+1) + 1, or Lucas(2n)Lucas(2n+1).