A342709 -id:A342709 - cp's OEIS frontend

A033890 a(n) = Fibonacci(4*n + 2).

1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673

Offset: 0

N. J. A. Sloane

(x,y) = (a(n), a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions are in A033888. - Floor van Lamoen, Dec 10 2001
This sequence consists of the odd-indexed terms of A001906 (whose terms are the values of x such that 5*x^2 + 4 is a square). The even-indexed terms of A001906 are in A033888. Limit_{n->infinity} a(n)/a(n-1) = phi^4 = (7 + 3*sqrt(5))/2. - Gregory V. Richardson, Oct 13 2002
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4, lim_{n->infinity} a(n) = x*(k*x+1)^n, k = a(1) - 3, x = (1 + sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
Indices of square numbers which are also 12-gonal. - Sture Sjöstedt, Jun 01 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with 3's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
If we let b(0) = 0 and, for n >= 1, b(n) = A033890(n-1), then the sequence b(n) will be F(4n-2) and the first difference is L(4n) or A056854. F(4n-2) is also the ratio of golden spiral length (rounded to the nearest integer) after n rotations. L(4n) is also the pitch length ratio. See illustration in links. - Kival Ngaokrajang, Nov 03 2013
The aerated sequence (b(n))n>=1 = [1, 0, 8, 0, 55, 0, 377, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
Solutions y of Pell equation x^2 - 5*y^2 = 4; corresponding x values are in A342710 (see A342709). - Bernard Schott, Mar 19 2021

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Nathan D. Cahill and Darren A. Narayan, Fibonacci and Lucas Numbers as Tridiagonal Matrix Determinants, Fib. Quart. 42, no. 3, Aug. 2004, pp. 216-221. See p. 219.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
Kival Ngaokrajang, Illustration of golden spiral length and pitch ratio
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (7,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A000045, A001906, A100047.

Cf. A342709, A342710.

[Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

A033890 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,8]);
    else
        7*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

Table[Fibonacci[4n + 2], {n, 0, 14}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{7, -1}, {1, 8}, 50] (* G. C. Greubel, Jul 13 2017 *)
a[n_] := (GoldenRatio^(2 (1 + 2 n)) - GoldenRatio^(-2 (1 + 2 n)))/Sqrt[5]
Table[a[n] // FullSimplify, {n, 0, 21}] (* Gerry Martens, Aug 20 2025 *)

PARI
```
a(n)=fibonacci(4*n+2);
```

G.f.: (1+x)/(1-7*x+x^2).
a(n) = 7*a(n-1) - a(n-2), n > 1; a(0)=1, a(1)=8.
a(n) = S(n,7) + S(n-1,7) = S(2*n,sqrt(9) = 3), where S(n,x) = U(n,x/2) are Chebyshev's polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), S(n,3) = A001906(n+1).
a(n) = ((7+3*sqrt(5))^n - (7-3*sqrt(5))^n + 2*((7+3*sqrt(5))^(n-1) - ((7-3*sqrt(5))^(n-1)))) / (3*(2^n)*sqrt(5)). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -9). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 for L(n,+7). - Reinhard Zumkeller, Jun 01 2005
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),7/2) + f(a(n-2),7/2). - Marcos Carreira, Dec 27 2006
a(n+1) = 8*a(n) - 8*a(n-1) + a(n-2); a(1)=1, a(2)=8, a(3)=55. - Sture Sjöstedt, May 27 2009
a(n) = A167816(4*n+2). - Reinhard Zumkeller, Nov 13 2009
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*log(3). - Francesco Daddi, Aug 01 2011
a(n) = A000045(A016825(n)). - Michel Marcus, Mar 22 2015
a(n) = A001906(2*n+1). - R. J. Mathar, Apr 30 2017
E.g.f.: exp(7*x/2)*(5*cosh(3*sqrt(5)*x/2) + 3*sqrt(5)*sinh(3*sqrt(5)*x/2))/5. - Stefano Spezia, Apr 14 2025
From Peter Bala, Jun 08 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/9 [telescoping series: 3/(a(n) - 1/a(n)) = 1/Fibonacci(4*n+4) + 1/Fibonacci(4*n)].
Product_{n >= 1} (a(n) + 3)/(a(n) - 3) = 5/2 [telescoping product:
(a(n) + 3)/(a(n) - 3) = b(n)/b(n-1), where b(n) = (Lucas(4*n+4) - 3)/(Lucas(4*n+4) + 3)].
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(9/5) [telescoping product:
(a(n) + 1)/(a(n) - 1) = c(n)/c(n-1) for n >= 1, where c(n) = Fibonacci(2*n+2)/Lucas(2*n+2)]. (End)
From Gerry Martens, Aug 20 2025: (Start)
a(n) = ((3 + sqrt(5))^(1 + 2*n) - (3 - sqrt(5))^(1 + 2*n)) / (2^(1 + 2*n)*sqrt(5)).
a(n) = Sum_{k=0..2*n} binomial(2*n + k + 1, 2*k + 1). (End)

A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.

Original entry on oeis.org

1, 4, 25, 169, 1156, 7921, 54289, 372100, 2550409, 17480761, 119814916, 821223649, 5628750625, 38580030724, 264431464441, 1812440220361, 12422650078084, 85146110326225, 583600122205489, 4000054745112196, 27416783093579881

Offset: 0

R. K. Guy, Mar 04 2003

Indices of 12-gonal numbers which are also squares (A342709). - Bernard Schott, Mar 19 2021
Values of y in solutions of x^2 = 5*y^2 - 4*y in positive integers. See A360467 for how this relates to a problem regarding the subdivision of a square into four triangles of integer area. - Alexander M. Domashenko, Feb 26 2023
And the corresponding x values of x^2 = 5*y^2 - 4*y are in A033890. - Bernard Schott, Feb 26 2023

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 19.
Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (vii)).
Pridon Davlianidze, Problem B-1264, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 1 (2020), p. 82; It's All About Catalan, Solution to Problem B-1264, ibid., Vol. 59, No. 1 (2021), pp. 87-88.
Derek Jennings, On Sums of Reciprocals of Fibonacci and Lucas Numbers, The Fibonacci Quarterly, Vol. 32, No. 1 (1994), pp. 18-21.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000045 (Fibonacci numbers), A000032 (Lucas numbers), A081067.
Cf. A001622, A051324, A239798, A342709.
Cf. A033890, A360467.

I:=[1, 4, 25]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

[(Lucas(4*n+2) + 2)/5: n in [0..30]]; // G. C. Greubel, Dec 17 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 40 do printf(`%d,`,(luc(4*n+2)+2)/5) od: # James Sellers, Mar 05 2003

CoefficientList[Series[-(1-4*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,4,25},50] (* Vincenzo Librandi, Jun 26 2012 *)
Table[(LucasL[4*n+2] + 2)/5, {n,0,30}] (* G. C. Greubel, Dec 17 2017 *)

main(size)={ return(concat([1],vector(size,n,fibonacci(2*n+1)^2))) } /* Anders Hellström, Jul 11 2015 */

for(n=0,30, print1(fibonacci(2*n+1)^2, ", ")) \\ G. C. Greubel, Dec 17 2017

a(n) = A001519(n+1)^2 = A122367(n)^2 = A058038(n) + 1.
a(n) = A103433(n+1) - A103433(n).
a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
a(n) = Fibonacci(2*n)*Fibonacci(2*n+2) +1. - Gary Detlefs, Apr 01 2012
G.f.: (1-4*x+x^2)/((1-x)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
Sum_{n>=0} 1/(a(n) + 1) = 1/3*sqrt(5). - Peter Bala, Nov 30 2013
Sum_{n>=0} 1/a(n) = sqrt(5) * Sum_{n>=1} (-1)^(n+1)*n/Fibonacci(2*n) (Jennings, 1994). - Amiram Eldar, Oct 30 2020
Product_{n>=1} (1 + 1/a(n)) = phi^2/2 (A239798), where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 01 2021

More terms from James Sellers, Mar 05 2003

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

A333641 11-gonal (or hendecagonal) square numbers.

Original entry on oeis.org

0, 1, 196, 29241, 1755625, 261468900, 38941102225, 2337990844401, 348201795147556, 51858411008887561, 3113535139359330841, 463705205422871375236, 69060571958250748760481, 4146338334574433921200225, 617522713934165528806340100, 91968930524758079223806760025

Offset: 1

Bernard Schott, Mar 31 2020

The 11-gonal square numbers correspond to the nonnegative integer solutions of the Diophantine equation k*(9*k-7)/2 = m^2, equivalent to (18*k-7)^2 - 72*m^2 = 49. Substituting x = 18*k-7 and y = m gives the Pell equation x^2-72*y^2 = 49. The integer solutions (x,y) = (-7,0), (11,1), (119,14), (1451,171), (11243,1325), ... correspond to the following solutions (k,m) = (0,0), (1,1), (7,14), (81,171), (625,1325), ...

			1755625 is a term because 625*(9*625-7)/2 = 1325^2 = 1755625; that means that 1755625 is the 625th 11-gonal number and the square of 1325.

Index entries for linear recurrences with constant coefficients, signature (1,0,1331714,-1331714,0,-1,1).

Intersection of A000290 (squares) and A051682 (11-gonals).
Cf. A106525.
Cf. A001110 (square triangulars), A036353 (square pentagonals), A046177 (square hexagonals), A036354 (square heptagonals), A036428 (square octagonals), A036411 (square 9-gonals), A188896 (only {0,1} are square 10-gonals), this sequence (square 11-gonals), A342709 (square 12-gonals).

for k from 0 to 8000000 do
d:= k*(9*k-7)/2;
if issqr(d) then print(k,sqrt(d),d); else fi; od:

Last /@ Solve[(18*x - 7)^2 - 72*y^2 == 49 && x >= 0 && y >= 0 && y < 10^16, {x, y}, Integers] /. Rule -> (#2^2 &) (* Amiram Eldar, Mar 31 2020 *)

concat(0, Vec(-x*(1 + 195*x + 29045*x^2 + 394670*x^3 + 29045*x^4 + 195*x^5 + x^6)/(-1 + x + 1331714*x^3 - 1331714*x^4 - x^6 + x^7) + O(x^20))) \\ Jinyuan Wang, Mar 31 2020

a(n) = k*(9*k-7)/2 for n > 1, where k = (A106525(4*n-6) + 7)/18. - Jinyuan Wang, Mar 31 2020

More terms from Amiram Eldar, Mar 31 2020

cp's OEIS Frontend

A033890 a(n) = Fibonacci(4*n + 2).

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A081068 a(n) = (Lucas(4*n+2) + 2)/5, or Fibonacci(2*n+1)^2, or A081067(n)/5.

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

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A333641 11-gonal (or hendecagonal) square numbers.

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A378948 Numbers that are simultaneously 12-gonal and cubes.

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A081068 a(n) = (Lucas(4n+2) + 2)/5, or Fibonacci(2n+1)^2, or A081067(n)/5.