A288913 -id:A288913 - 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

A056914 a(n) = Lucas(4*n+1).

Original entry on oeis.org

1, 11, 76, 521, 3571, 24476, 167761, 1149851, 7881196, 54018521, 370248451, 2537720636, 17393796001, 119218851371, 817138163596, 5600748293801, 38388099893011, 263115950957276, 1803423556807921, 12360848946698171

Offset: 0

Barry E. Williams, Jul 11 2000

a(n) = (t(i+4n+1) - t(i))/(t(i+2n+1) - t(i+2n)), where (t) is any sequence of the form t(n+2) = 4t(n+1) - 4t(n) + t(n-1) or of the form t(n+1) = 3t(n) - t(n-1) without regard to initial values as long as t(i+2n+1) - t(i+2n) != 0. - Klaus Purath, Jun 24 2024

V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers, A Publication of the Fibonacci Association, Houghton Mifflin Co., 1969, pp. 27-29.

G. C. Greubel, 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. (A056914) = sqrt{5*(A033889)^2 - 4}.

Cf. quadrisection of A000032: A056854 (first), this sequence (second), A246453 (third, without 11), A288913 (fourth).

List([0..30], n-> Lucas(1,-1,4*n+1)[2] ); # G. C. Greubel, Jan 16 2020

[Lucas(4*n+1): n in [0..30]]; // G. C. Greubel, Dec 24 2017

with(combinat); seq(fibonacci(4*n+2)+fibonacci(4*n), n = 0..30); # G. C. Greubel, Jan 16 2020

LucasL[4*Range[0,30]+1] (* or *) LinearRecurrence[{7,-1}, {1,11}, 30] (* G. C. Greubel, Dec 24 2017 *)

my(x='x+O('x^30)); Vec((1+4*x)/(1-7*x+x^2)) \\ G. C. Greubel, Dec 24 2017

[lucas_number2(4*n+1,1,-1) for n in (0..30)] # G. C. Greubel, Jan 16 2020

a(n) = 7*a(n-1) - a(n-2), with a(0)=1, a(1)=11.
a(n) = (11*(((7+3*sqrt(5))/2)^n - ((7-3*sqrt(5))/2)^n) - (((7+3*sqrt(5))/2)^(n-1) - ((7-3*sqrt(5))/2)^(n-1)))/3*sqrt(5).
G.f.: (1+4*x)/(1-7*x+x^2). - Philippe Deléham, Nov 02 2008

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

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

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

A056914 a(n) = Lucas(4*n+1).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

Extensions

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

Programs

Mathematica

Formula

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).