A005248 -id:A005248 - cp's OEIS frontend

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

2, 8, 62, 488, 3842, 30248, 238142, 1874888, 14760962, 116212808, 914941502, 7203319208, 56711612162, 446489578088, 3515205012542, 27675150522248, 217885999165442, 1715412842801288, 13505416743244862, 106327921103157608

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Sep 21 2003

a(n+1)/a(n) converges to (4+sqrt(15)) = 7.872983... a(0)/a(1)=2/8; a(1)/a(2)=8/62; a(2)/a(3)=62/488; a(3)/a(4)=488/3842; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.127016... = 1/(4+sqrt(15)) = (4-sqrt(15)).
Twice A001091. - John W. Layman, Sep 25 2003
Except for the first term, positive values of x (or y) satisfying x^2 - 8xy + y^2 + 60 = 0. - Colin Barker, Feb 13 2014

			a(4) = 3842 = 8*a(3) - a(2) = 8*488 - 62 = (4+sqrt(15))^4 + (4-sqrt(15))^4 = 3841.9997397 + 0.0002603 = 3842.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A086594, A058316, A006245, A009271, A005248, A174502.

I:=[2,8]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 15 2014

a[0] = 2; a[1] = 8; a[n_] := 8a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
CoefficientList[Series[(2 - 8 x)/(1 - 8 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 15 2014 *)
LinearRecurrence[{8,-1},{2,8},30] (* Harvey P. Dale, Jan 18 2015 *)

[lucas_number2(n,8,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.
G.f.: (2-8*x)/(1-8*x+x^2). [Philippe Deléham, Nov 02 2008]
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 = 4 - sqrt(15). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.12474 84992 41370 33639 ... = 2 + 1/(8 + 1/(62 + 1/(488 + ...))).  Cf. A174502 and A005248.
Also F(-alpha) = 0.87474 74663 84045 35032 ... has the continued fraction representation 1 - 1/(8 - 1/(62 - 1/(488 - ...))) and the simple continued fraction expansion 1/(1 + 1/((8-2) + 1/(1 + 1/((62-2) + 1/(1 + 1/((488-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((8^2-4) + 1/(1 + 1/((62^2-4) + 1/(1 + 1/((488^2-4) + 1/(1 + ...))))))).
(End)

A099016 a(n) = 3L(2n)/5 - (-1)^n/5, where L = A000032.

Original entry on oeis.org

1, 2, 4, 11, 28, 74, 193, 506, 1324, 3467, 9076, 23762, 62209, 162866, 426388, 1116299, 2922508, 7651226, 20031169, 52442282, 137295676, 359444747, 941038564, 2463670946, 6449974273, 16886251874, 44208781348, 115740092171

Offset: 0

Paul Barry, Sep 22 2004

Let M = an infinite triangle with (1,2,2,3,3,4,4,...) as the left border and all other columns = (0,1,2,3,4,5,...). Then lim_{n->infinity} M^n = A099016, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 26 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..300 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032.

[3*Lucas(2*n)/5-(-1)^n/5: n in [0..35]]; // Vincenzo Librandi, Jun 09 2011

F:=Fibonacci; [F(n+1)^2+F(n)*F(n-2): n in [0..30]]; // Bruno Berselli, Feb 15 2017

with(combinat):seq(3*fibonacci(n)^2+(-1)^n, n= 0..27)

CoefficientList[Series[(1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2)), {x,0,50}], x] (* G. C. Greubel, Dec 31 2017 *)

x='x+O('x^30); Vec((1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2))) \\ G. C. Greubel, Dec 31 2017

G.f.: (1 - 2*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = 2*F(n)^2 + F(n)*F(n-1) + F(n-1)^2, where F = A000045.
a(n) = 3*((3/2 - sqrt(5)/2)^n + (3/2 + sqrt(5)/2)^n)/5 - (-1)^n/5.
a(n) = A099015(n)/F(n+1).
a(n) = 3*A005248(n)/5 - (-1)^n/5.
a(n) = 3*A000032(2*n)/5 - (-1)^n/5.
a(n) = A061646(n) + F(n)^2.
a(n) = 3*F(n)^2 + (-1)^n.
a(n) = F(n+1)^2 + F(n)*F(n-2). See also A192914, fourth formula. - Bruno Berselli, Feb 15 2017

A134571 Array T(n,k) by antidiagonals; T(n,k) = position in row n of k-th occurrence of the Fibonacci number F(2n) in A134567.

Original entry on oeis.org

1, 3, 2, 4, 7, 5, 6, 10, 18, 13, 8, 15, 26, 47, 34, 9, 20, 39, 68, 123, 89, 11, 23, 52, 102, 178, 322, 233

Offset: 1

Clark Kimberling, Nov 02 2007

(Row 1) = A000201, the lower Wythoff sequence (Row 2) = (Column 2 of Wythoff array) = A035336 (Row 3) = (Column 4 of Wythoff array) = A035338 (Row 4) = (Column 6 of Wythoff array) = A035340 (Column 1) = A001519 (bisection of Fibonacci sequence) (Column 2) = A005248 (bisection of Lucas sequence) (Column 3) = A052995 Row 1 is the ordered union of all odd-numbered columns of the Wythoff array; and A134571 is a permutation of the positive integers.

It looks like this array is A080164 transposed. - Peter Munn, Sep 02 2025

			Northwest corner:
1 3 4 6 8 9 11 12 14 16
2 7 10 15 20 23
5 18 26 39 52 60
13 47 68 102 136 157
Row 1 consists of numbers k such that 1 is the least m for which {-m*tau}<{k*tau}, where tau=(1+sqrt(5))/2 and {} denotes fractional part.

Cf. A080164, A134567, A134570.

A195720 Decimal expansion of arccos(sqrt(1/6)) and of arcsin(sqrt(5/6)) and arctan(sqrt(5)).

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Sep 23 2011

			arccos(sqrt(1/6)) = 1.150261991510931...

G. C. Greubel, Table of n, a(n) for n = 1..5000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
Index entries for transcendental numbers

Cf. A188595, A195721, A195722.

[Arccos(Sqrt(1/6))]; // G. C. Greubel, Nov 23 2017

r = Sqrt[1/6]; RealDigits[ArcCos[r], 10, 100][[1]]

atan(sqrt(5)) \\ Michel Marcus, Mar 29 2016

Equals Sum_{k >= 1} sqrt(5)/L(2n) where L=A000032. See also A005248. - Michel Marcus, Mar 29 2016

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

A236331 Positive integers n such that x^2 - 18xy + y^2 + n = 0 has integer solutions.

Original entry on oeis.org

64, 256, 320, 576, 704, 1024, 1216, 1280, 1600, 1856, 1984, 2304, 2624, 2816, 2880, 3136, 3520, 3776, 3904, 4096, 4544, 4864, 5056, 5120, 5184, 5696, 6080, 6336, 6400, 6464, 6976, 7424, 7744, 7936, 8000, 8384, 8896, 9216, 9280, 9536, 9664, 9920, 10496, 10816

Offset: 1

Colin Barker, Feb 16 2014

nonn

			64 is in the sequence because x^2 - 18xy + y^2 + 64 = 0 has integer solutions, for example (x, y) = (1, 13).

Cf. A001519 (n = 64), A052995 (n = 256), A055819 (n = 256), A005248 (n = 320), A237132 (n = 704), A237133 (n = 1216).

Cf. A031363, A084917, A237351, A237599, A237606, A237609, A237610, A236330.

A237133 Values of x in the solutions to x^2 - 3xy + y^2 + 19 = 0, where 0 < x < y.

Original entry on oeis.org

4, 5, 7, 11, 17, 28, 44, 73, 115, 191, 301, 500, 788, 1309, 2063, 3427, 5401, 8972, 14140, 23489, 37019, 61495, 96917, 160996, 253732, 421493, 664279, 1103483, 1739105, 2888956, 4553036, 7563385, 11920003, 19801199, 31206973, 51840212, 81700916, 135719437

Offset: 1

Colin Barker, Feb 04 2014

The corresponding values of y are given by a(n+2).

Positive values of x (or y) satisfying x^2 - 18xy + y^2 + 1216 = 0.

			11 is in the sequence because (x, y) = (11, 28) is a solution to x^2 - 3xy + y^2 + 19 = 0.

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

Cf. A001519, A005248, A055819, A237132, A218735.

LinearRecurrence[{0,3,0,-1},{4,5,7,11},40] (* Harvey P. Dale, Dec 15 2014 *)

Vec(-x*(x-1)*(4*x^2+9*x+4)/((x^2-x-1)*(x^2+x-1)) + O(x^100))

a(n) = 3*a(n-2)-a(n-4).
G.f.: -x*(x-1)*(4*x^2+9*x+4) / ((x^2-x-1)*(x^2+x-1)).
a(n) = (1/2) * (F(n+4) + (-1)^n*F(n-5)), n>4, with F the Fibonacci numbers (A000045). - Ralf Stephan, Feb 05 2014

A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.

Original entry on oeis.org

4, -3, 5, 0, 13, 21, 68, 165, 445, 1152, 3029, 7917, 20740, 54285, 142133, 372096, 974173, 2550405, 6677060, 17480757, 45765229, 119814912, 313679525, 821223645, 2149991428, 5628750621, 14736260453, 38580030720, 101003831725, 264431464437, 692290561604, 1812440220357

Offset: 0

Bruno Berselli, Sep 20 2017

This is the case k=3 of the identity F(n)^2 - F(k)^2*(-1)^(n+k) = F(n+k)*F(n-k), known also as Catalan's identity.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Catalan's Identity.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A001622, A005248: Lucas(2*n), A001654: F(n)*F(n+1).

Cf. A007598 (k=0), A059929 (k=1, without initial 1), A192883 (k=2, without initial -1), this sequence (k=3).

List([0..10^2],n ->Fibonacci(n)^2+4*(-1)^n); # Muniru A Asiru, Sep 26 2017

[Fibonacci(n)^2+4*(-1)^n: n in [0..40]];

with(combinat,fibonacci):  A292612:=seq(fibonacci(n)^2+4*(-1)^n, n=0..10^2); # Muniru A Asiru, Sep 26 2017

Table[Fibonacci[n]^2 + 4 (-1)^n, {n, 0, 40}]

for(n=0, 40, print1(fibonacci(n)^2+4*(-1)^n", "));

Vec((4-11*x+3*x^2)/((1+x)*(1-3*x+x^2))+O(x^30)) \\ Colin Barker, Sep 20 2017

[fibonacci(n)^2+4*(-1)^n for n in range(40)]

G.f.: (4 - 11*x + 3*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = a(-n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = 4*A001654(n+1) - 11*A001654(n) + 3*A001654(n-1) with A001654(-1)=0.
5*a(n) = Lucas(2*n) + 18*(-1)^n. Note that Lucas(2*n) + r*(-1)^n is divisible by 5 for r = -2, 3, -7, 8, -12, 13, -17, 18, -22, 23, -27, ... = (-1/4)*(3 + 5*(2*m+1)*(-1)^m) = (-1)^m*A047221(m). On the other hand, a(n) is divisible by 5 when n is a member of A047221.
a(n) = (1/5)*(18*(-1)^n + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n). - Colin Barker, Sep 20 2017
Sum_{n>=4} 1/a(n) = 143/960. - Amiram Eldar, Oct 05 2020
Sum_{n>=4} (-1)^n/a(n) = 3/(4*phi) - 407/960, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 06 2020

A335673 Composite integers m such that A003500(m) == 4 (mod m).

Original entry on oeis.org

10, 209, 230, 231, 399, 430, 455, 530, 901, 903, 923, 989, 1295, 1729, 1855, 2015, 2211, 2345, 2639, 2701, 2795, 2911, 3007, 3201, 3439, 3535, 3801, 4823, 5291, 5719, 6061, 6767, 6989, 7421, 8569, 9503, 9591, 9869, 9890, 10439, 10609, 11041, 11395, 11951, 11991

Offset: 1

Ovidiu Bagdasar, Jun 17 2020

nonn

If p is a prime, then A003500(p)==4 (mod p).
This sequence contains the composite integers for which the congruence holds.
The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.
For a=4, b=1, V(n)=A003500(n).

			m=10 is the first composite integer for which A003500(m)==4 (mod m).

D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).

Amiram Eldar, Table of n, a(n) for n = 1..10000 (first 1000 odd terms from Chai Wah Wu)
D. Andrica and O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, preprint for Mediterr. J. Math. 18, 47 (2021).

Cf. A005248, A335669 (a=3,b=-1), A335672 (a=3,b=1), A335674 (a=5,b=1).

A330206 is the subsequence of odd terms.

Select[Range[3, 20000], CompositeQ[#] && Divisible[Round@LucasL[2#, Sqrt[2]] - 4, #] &] (* Amiram Eldar, Jun 18 2020 *)

my(M=[1,2;1,3]); forcomposite(m=5, 10^5, if(trace(Mod(M,m)^m)==4, print1(m,", "))); \\ Joerg Arndt, Jun 18 2020

More terms from Joerg Arndt, Jun 18 2020

A335674 Odd composite integers m such that A003501(m) == 5 (mod m).

Original entry on oeis.org

15, 21, 35, 105, 161, 195, 255, 345, 385, 399, 465, 527, 551, 609, 741, 897, 1105, 1295, 1311, 1807, 1919, 2001, 2015, 2071, 2085, 2121, 2415, 2737, 2915, 3289, 3815, 4031, 4033, 4355, 4879, 4991, 5291, 5777, 5983, 6049, 6061, 6083, 6479, 6601, 6785, 7645, 7905, 8695, 8855, 8911, 9361, 9591, 9889

Offset: 1

Ovidiu Bagdasar, Jun 17 2020

nonn

If p is a prime, then A003501(p)==5 (mod p).
This sequence contains the odd composite integers for which the congruence holds.
The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.
For a=5, b=1, V(n) recovers A003501(n).

			15 is the first odd composite integer for which the relation A003501(15)=16098445550==5 (mod 15) holds.

D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).

Chai Wah Wu, Table of n, a(n) for n = 1..1000
D. Andrica and O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, preprint for Mediterr. J. Math. 18, 47 (2021).

Cf. A005248, A335669 (a=3,b=-1), A335672 (a=3,b=1), A335673 (a=4,b=1).

Select[Range[3, 5000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 5/2] - 5, #] &] (* Amiram Eldar, Jun 18 2020 *)

cp's OEIS Frontend

A086903 a(n) = 8*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 8.

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A099016 a(n) = 3*L(2*n)/5 - (-1)^n/5, where L = A000032.

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A134571 Array T(n,k) by antidiagonals; T(n,k) = position in row n of k-th occurrence of the Fibonacci number F(2n) in A134567.

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A195720 Decimal expansion of arccos(sqrt(1/6)) and of arcsin(sqrt(5/6)) and arctan(sqrt(5)).

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

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A236331 Positive integers n such that x^2 - 18xy + y^2 + n = 0 has integer solutions.

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A237133 Values of x in the solutions to x^2 - 3xy + y^2 + 19 = 0, where 0 < x < y.

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A292612 a(n) = F(n)^2 + 4*(-1)^n = F(n+3)*F(n-3), where F = A000045.

A099016 a(n) = 3L(2n)/5 - (-1)^n/5, where L = A000032.

A292612 a(n) = F(n)^2 + 4(-1)^n = F(n+3)F(n-3), where F = A000045.