A246453 -id:A246453 - 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

A105531 Decimal expansion of arctan(1/3).

Original entry on oeis.org

3, 2, 1, 7, 5, 0, 5, 5, 4, 3, 9, 6, 6, 4, 2, 1, 9, 3, 4, 0, 1, 4, 0, 4, 6, 1, 4, 3, 5, 8, 6, 6, 1, 3, 1, 9, 0, 2, 0, 7, 5, 5, 2, 9, 5, 5, 5, 7, 6, 5, 6, 1, 9, 1, 4, 3, 2, 8, 0, 3, 0, 5, 9, 3, 5, 6, 7, 5, 6, 2, 3, 7, 4, 0, 5, 8, 1, 0, 5, 4, 4, 3, 5, 6, 4, 0, 8, 4, 2, 2, 3, 5, 0, 6, 4, 1, 3, 7, 4, 4, 3, 9, 0, 0, 7

Offset: 0

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 12 2005

arctan(1/3) + A073000 = 2*arctan(1/3) + A105533 = Pi/4.

			0.3217505543966421934014046143...

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

Peter Bala, New series for old functions
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
Eric Weisstein's World of Mathematics, Machin-Like Formulas
Index entries for transcendental numbers

Cf. A003881 (Pi/4), A073000 (arctan(1/2)), A105533 (arctan(1/7)).

RealDigits[ArcTan[1/3],10,120][[1]] (* Harvey P. Dale, Oct 28 2011 *)

PARI
```
atan(1/3) \\ Michel Marcus, Mar 29 2016
```

From Peter Bala, Feb 04 2015: (Start)
Equals (1/3)*Sum_{k >= 0} (-1)^k/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*9^k). Both sequences satisfy the same recurrence equation u(n) = (32*n + 20)*u(n-1) + 36*(2*n - 1)^2*u(n-2). From this observation we find the continued fraction expansion arctan(1/3) = (1/3)*(1 - 2/(54 + 36*3^2/(84 + 36*5^2/(116 + ... + 36*(2*n - 1)^2/((32*n + 20) + ...))))).
Equals (3/10) * Sum_{k >= 0} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ).
Define a pair of integer sequences C(n) = 10^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (2/5)^k/( (2*k + 1)*binomial(2*k,k) ). Both sequences satisfy the same recurrence equation u(n) = (44*n + 20)*u(n-1) - 80*n*(2*n - 1)*u(n-2). From this observation we obtain the continued fraction expansion arctan(1/3) = (3/10)*( 1 + 4/(60 - 480/(108 - 1200/(152 - ... - 80*n*(2*n - 1)/((44*n + 20) - ...))))). (End)
Equals Sum_{k>=1} arctan(L(4*k+2)/F(4*k+2)^2) where L=A000032 and F=A000045. See also A033890 and A246453. - Michel Marcus, Mar 29 2016 [corrected by Jason Yuen, Jan 18 2025]
From Amiram Eldar, Aug 09 2020: (Start)
Equals Sum_{k>=2} arctan(1/(2*k^2)) = Sum_{k>=2} (-1)^k arctan(2/k^2).
Equals Integral_{x=1..2} 1/(x^2 + 1) dx. (End)
Equals Sum_{n>=0} arctan(1/F(2*n+5)) = Sum_{n>=0} (-1)^n arctan(F(2*n+1)) where F=A000045. - Gleb Koloskov, Oct 01 2021

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.

A342710 Solutions x to the Pell-Fermat equation x^2 - 5*y^2 = 4.

Original entry on oeis.org

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

Offset: 0

Bernard Schott, Mar 19 2021

This Pell equation is used to find the 12-gonal square numbers (see A342709).
The corresponding solutions y are in A033890.
Essentially the same as A246453. - R. J. Mathar, Mar 24 2021

			a(1)^2 - 5 * A033890(1)^2 = 18^2 - 5 * 8^2 = 4.

Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A033890, A342709.

a(n) = 3*A049685(n). - Hugo Pfoertner, Mar 19 2021

LinearRecurrence[{7, -1}, {3, 18}, 20] (* Amiram Eldar, Mar 19 2021 *)
Table[2 ChebyshevT[2 n + 1, 3/2], {n, 0, 20}] (* Eric W. Weisstein, Sep 02 2025 *)
Table[2 Cos[(2 n + 1) ArcCos[3/2]], {n, 0, 20}] // FunctionExpand (* Eric W. Weisstein, Sep 02 2025 *)

a(n) = 7*a(n-1) - a(n-2).
a(n) = 2*T(2*n+1, 3/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 02 2022
From Stefano Spezia, Apr 14 2025: (Start)
G.f.: 3*(1 - x)/(1 - 7*x + x^2).
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2)). (End)
a(n) = 2*cos((2*n+1)*arccos(3/2)). - Eric W. Weisstein, Sep 02 2025

A351222 Decimal expansion of Sum_{k>=0} (-1)^k/Fibonacci(4*k+2).

Original entry on oeis.org

8, 9, 0, 8, 6, 7, 0, 2, 1, 9, 7, 2, 1, 1, 8, 2, 6, 0, 0, 4, 8, 8, 5, 1, 5, 2, 9, 2, 4, 1, 5, 6, 8, 0, 2, 0, 4, 3, 0, 5, 1, 2, 8, 4, 4, 1, 5, 8, 2, 0, 4, 3, 4, 5, 6, 6, 2, 0, 8, 0, 2, 7, 1, 9, 7, 5, 5, 2, 1, 5, 5, 6, 7, 2, 2, 1, 9, 9, 7, 5, 7, 6, 0, 5, 3, 1, 7, 8, 8, 3, 4, 9, 1, 6, 6, 2, 6, 7, 9, 5, 8, 5, 9, 2, 6

Offset: 0

Amiram Eldar, Feb 05 2022

			0.89086702197211826004885152924156802043051284415820...

Wray G. Brady, Problem B-319, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 13, No. 4 (1975), p. 373; Rerun, Solution to Problem B-319, ibid., Vol. 14, No. 5 (1976), p. 472.
L. Carlitz, Problem B-111, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 5, No. 1 (1967), p. 108; Another Series Equality, Solution to Problem B-110 by the proposer, ibid., Vol. 5, No. 5 (1967), pp. 470-471.

Cf. A000032, A000045, A033890, A246453.

RealDigits[NSum[(-1)^n/Fibonacci[4*n + 2], {n, 0, Infinity}, WorkingPrecision -> 1200], 10, 100][[1]]

sumpos(k=0, (-1)^k/fibonacci(4*k+2)) \\ Michel Marcus, Feb 05 2022

Equals sqrt(5) * Sum_{k>=0} 1/Lucas(4*k+2) (Carlitz, 1967).

cp's OEIS Frontend

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

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

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A105531 Decimal expansion of arctan(1/3).

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

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A342710 Solutions x to the Pell-Fermat equation x^2 - 5*y^2 = 4.

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A351222 Decimal expansion of Sum_{k>=0} (-1)^k/Fibonacci(4*k+2).

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