cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A052918 a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).

Original entry on oeis.org

1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - Gary W. Adamson, Jul 01 2003, corrected Oct 07 2008, corrected by Jianing Song, Feb 01 2019
a(p-1) == 29^((p-1)/2) (mod p), for odd primes p. - Gary W. Adamson, Feb 22 2009 [See A087475 for more info about this congruence. - Jason Yuen, Apr 05 2025]
For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A015523. - Johannes W. Meijer, Aug 01 2010
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n on alphabet {0,1,...,5} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 5-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 5 kinds of squares available. (End)

Crossrefs

Row 5 of A073133, A172236, and A352361.
Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

Programs

  • GAP
    a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Oct 16 2019
  • Magma
    I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 23 2013
    
  • Magma
    R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019
    
  • Maple
    spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
    a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # Zerinvary Lajos, Jul 26 2006
    with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # Zerinvary Lajos, Dec 07 2008
  • Mathematica
    LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
    Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)
  • PARI
    Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011
    
  • Sage
    [lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009
    

Formula

G.f.: 1/(1 - 5*x - x^2).
a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - Henry Bottomley, May 10 2000
a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - Gary W. Adamson, Jul 01 2003
a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - Roger L. Bagula, May 29 2005
a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - T. D. Noe, Jan 19 2006
a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - Gary W. Adamson, Dec 21 2007
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit_{k->oo} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
Limit_{n->oo} A087130(n)/a(n-1) = sqrt(29). (End)
From L. Edson Jeffery, Jan 07 2012: (Start)
Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
a(n) is the upper-right term of A^(n+1);
a(n) is the lower-left term of (1/5)*A^(n+1);
a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - Vladimir Shevelev, Feb 23 2013
G.f.: x/(1 - 5*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 5 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024

A099372 a(n) = A099371(n)^2.

Original entry on oeis.org

0, 1, 81, 6724, 558009, 46308025, 3843008064, 318923361289, 26466795978921, 2196425142889156, 182276820063821025, 15126779640154255921, 1255340433312739420416, 104178129185317217638609, 8645529381948016324584129, 717474760572500037722844100, 59541759598135555114671476169
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=9.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 9 kinds of half-square available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 9 kinds of (1/4,1/4)-fence available. - Michael A. Allen, Mar 21 2024

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, A099369, this sequence, A099374.

Programs

  • Mathematica
    LinearRecurrence[{82,82,-1},{0,1,81},17] (* Stefano Spezia, Apr 06 2023 *)

Formula

a(n) = A099371(n)^2.
a(n) = 82*a(n-1) + 82*a(n-2) - a(n-3), n>=3; a(0)=0, a(1)=1, a(2)=81.
a(n) = 83*a(n-1) - a(n-2) - 2*(-1)^n, n>=2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 83/2)-(-1)^n)/85 with twice the Chebyshev polynomials of the first kind: 2*T(n, 83/2) = A099373(n).
G.f.: x*(1-x)/((1-83*x+x^2)*(1+x)) = x*(1-x)/(1-82*x-82*x^2+x^3).
E.g.f.: 2*exp(-x)*(exp(85*x/2)*cosh(9*sqrt(85)*x/2) - 1)/85. - Stefano Spezia, Apr 06 2023
a(n) = (1 - (-1)^n)/2 + 81*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 21 2024
Product_{n>=2} (1 + (-1)^n/a(n)) = (9 + sqrt(85))/18 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099279 Squares of A001076.

Original entry on oeis.org

0, 1, 16, 289, 5184, 93025, 1669264, 29953729, 537497856, 9645007681, 173072640400, 3105662519521, 55728852710976, 1000013686278049, 17944517500293904, 322001301319012225, 5778078906241926144, 103683419011035658369, 1860523463292399924496, 33385738920252162982561
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

For the generalized Fibonacci sequences U(n-1;a) = (ap(a)^n - am(a)^n)/(ap(a) - am(a)) with ap(a) = (a + sqrt(a^2+4))/2, am(a) = (a - sqrt(a^2+4))/2, a from the integers, one has for the squared sequences U(n-1;a)^2 = (2*T(n,(a^2+2)/2) - 2*(-1)^n)/(a^2+4). Here T(n,x) are Chebyshev's polynomials of the first kind (see A053120). Therefore the o.g.f. for the squared sequence is x*(1-x)/((1+x)*(1-(a^2+2)*x+x^2)) = x*(1-x)/(1 - (a^2+1)*x - (a^2+1)*x^2 + x^3). For this example a=4.
Unsigned member r=-16 of the family of Chebyshev sequences S_r(n) defined in A092184.
(-1)^(n+1)*a(n) = S_{-16}(n), n >= 0, defined in A092184.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 4 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 4 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 12 2023

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, this sequence, A099365, A099366, A099367, A099369, A099372, A099374.

Programs

  • Magma
    [Fibonacci(3*n)^2/4: n in [0..30]]; // G. C. Greubel, Aug 18 2022
  • Maple
    with (combinat):seq(fibonacci(n,4)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008
    nmax:=48: with(combinat): for n from 0 to nmax do A001654(n):=fibonacci(n) * fibonacci(n+1) od: a(0):=0: for n from 1 to nmax/3 do a(n):=a(n-1)+A001654(3*n-2) od: seq(a(n),n=0..nmax/3); # Johannes W. Meijer, Sep 22 2010
  • Mathematica
    LinearRecurrence[{17,17,-1},{0,1,16},30] (* Harvey P. Dale, Mar 26 2012 *)
    Fibonacci[3*Range[0, 30]]^2/4 (* G. C. Greubel, Aug 18 2022 *)
  • MuPAD
    numlib::fibonacci(3*n)^2/4 $ n = 0..35; // Zerinvary Lajos, May 13 2008
    
  • PARI
    my(x='x+O('x^99)); concat([0], Vec(x*(1-x)/((1-18*x+x^2)*(1+x)))) \\ Altug Alkan, Dec 17 2017
    
  • Sage
    [(fibonacci(3*n))^2/4 for n in range(0, 17)] # Zerinvary Lajos, May 15 2009
    

Formula

a(n) = A001076(n)^2.
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=16.
a(n) = 18*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2, a(0)=0, a(1)=1.
a(n) = (T(n, 9) - (-1)^n)/10 with Chebyshev's T(n, x) polynomials of the first kind. T(n, 9) = A023039(n).
G.f.: x*(1-x)/((1+x)*(1-18*x+x^2)) = x*(1-x)/(1-17*x-17*x^2+x^3).
a(n) = a(n-1) + A001654(3*n-2) with a(0)=0, where A001654 are the golden rectangle numbers. - Johannes W. Meijer, Sep 22 2010
a(n+1) = (1 + (-1)^n)/2 + 16*Sum_{r=1..n} ( r*a(n+1-r) ). - Michael A. Allen, Mar 12 2023
E.g.f.: exp(-x)*(exp(10*x)*cosh(4*sqrt(5)*x) - 1)/10. - Stefano Spezia, Apr 06 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (2 + sqrt(5))/4 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099369 Squares of A041025(n-1), n>=1, (generalized Fibonacci).

Original entry on oeis.org

0, 1, 64, 4225, 278784, 18395521, 1213825600, 80094094081, 5284996383744, 348729667233025, 23010873040995904, 1518368891038496641, 100189335935499782400, 6610977802851947141761, 436224345652293011573824
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=8.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 8 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 8 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Apr 30 2023

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, this sequence, A099372, A099374.

Programs

  • Mathematica
    LinearRecurrence[{65,65,-1},{0,1,64},20] (* Harvey P. Dale, Oct 05 2021 *)

Formula

a(n) = A041025(n-1)^2, n >= 1, a(0)=0.
a(n) = 65*a(n-1) + 65*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=64.
a(n) = 66*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 33) - (-1)^n)/34 with the Chebyshev polynomials of the first kind: T(n, 33) = A099370(n).
G.f.: x*(1-x)/((1-66*x+x^2)*(1+x)) = x*(1-x)/(1-65*x-65*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 64*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Apr 30 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (4 + sqrt(17))/8 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099374 a(n) = A041041(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 100, 10201, 1040400, 106110601, 10822240900, 1103762461201, 112572948801600, 11481337015302001, 1170983802612002500, 119428866529408953001, 12180573402197101203600
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=10.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 10 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 10 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 21 2024

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, A099367, A099369, A099372, this sequence.

Programs

  • Mathematica
    LinearRecurrence[{101,101,-1},{0,1,100},20] (* Harvey P. Dale, Nov 10 2021 *)

Formula

a(n) = A041041(n-1)^2, n >= 1, a(0)=0.
a(n) = 101*a(n-1) + 101*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=100.
a(n) = 102*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 51) - (-1)^n)/52 with the Chebyshev polynomials of the first kind: T(n, 51) = (n).
G.f.: x*(1-x)/((1-102*x+x^2)*(1+x)) = x*(1-x)/(1-101*x-101*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 100*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 21 2024
Product_{n>=2} (1 + (-1)^n/a(n)) = (5 + sqrt(26))/10 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099367 a(n) = A054413(n-1)^2, n >= 1.

Original entry on oeis.org

0, 1, 49, 2500, 127449, 6497401, 331240000, 16886742601, 860892632649, 43888637522500, 2237459621014849, 114066552034234801, 5815156694124960000, 296458924848338725201, 15113590010571150025249, 770496631614280312562500
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=7.

Crossrefs

Cf. A054413.
Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, A099366, this sequence, A099369, A099372, A099374.

Programs

  • Mathematica
    LinearRecurrence[{50,50,-1},{0,1,49},20] (* Harvey P. Dale, Jul 27 2023 *)

Formula

a(n) = A054413(n-1)^2, n >= 1. a(0)=0.
a(n) = 50*a(n-1) + 50*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=49.
a(n) = 51*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 51/2) - (-1)^n)/53 with twice the Chebyshev polynomials of the first kind: 2*T(n, 51/2) = A099368(n).
G.f.: x*(1-x)/((1-51*x+x^2)*(1+x)) = x*(1-x)/(1-50*x-50*x^2+x^3).
a(n+1) = (1 + (-1)^n)/2 + 49*Sum_{k=1..n} k*a(n+1-k). - Michael A. Allen, Feb 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (7 + sqrt(53))/14 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A099366 Squares of A005668.

Original entry on oeis.org

0, 1, 36, 1369, 51984, 1974025, 74960964, 2846542609, 108093658176, 4104712468081, 155870980128900, 5918992532430121, 224765845252215696, 8535183127051766329, 324112192982714904804, 12307728150216114616225
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=6.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 6 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 6 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Apr 21 2023

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, this sequence, A099367, A099369, A099372, A099374.

Programs

  • Maple
    with (combinat):seq(fibonacci(n,6)^2,n=0..15); # Zerinvary Lajos, Apr 09 2008
  • Mathematica
    LinearRecurrence[{37,37,-1},{0,1,36},20] (* Harvey P. Dale, Sep 23 2018 *)

Formula

a(n) = A005668(n)^2.
a(n) = 37*a(n-1) + 37*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=36.
a(n) = 38*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 19) - (-1)^n)/20 with the Chebyshev polynomials of the first kind: T(n, 19) = A078986(n).
G.f.: x*(1-x)/((1 - 38*x + x^2)*(1+x)) = x*(1-x)/(1 - 37*x - 37*x^2 + x^3).
a(n) = (1 - (-1)^n)/2 + 36*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Apr 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(10))/6 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024
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