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 1-8 of 8 results.

A099371 Expansion of g.f.: x/(1 - 9*x - x^2).

Original entry on oeis.org

0, 1, 9, 82, 747, 6805, 61992, 564733, 5144589, 46866034, 426938895, 3889316089, 35430783696, 322766369353, 2940328107873, 26785719340210, 244011802169763, 2222891938868077, 20250039251982456, 184473245206710181, 1680509246112374085, 15309056460218076946
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 9's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,9} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Mar 10 2023: (Start)
Also called the 9-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) 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 9 kinds of squares available. (End)

Crossrefs

Programs

  • Magma
    I:=[0,1]; [n le 2 select I[n] else 9*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
  • Maple
    F:= gfun:-rectoproc({a(n)=9*a(n-1)+a(n-2),a(0)=0,a(1)=1},a(n),remember):
    seq(F(n),n=0..30); # Robert Israel, Feb 01 2015
  • Mathematica
    CoefficientList[Series[x/(1-9*x-x^2), {x,0,30}], x] (* G. C. Greubel, Apr 16 2017 *)
    LinearRecurrence[{9,1}, {0,1}, 30] (* G. C. Greubel, Jan 24 2018 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(1/(1-9*x-x^2)) ) \\ Charles R Greathouse IV, Feb 03 2014
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen3
    it = recur_gen3(0,1,9,9,1,0)
    [next(it) for i in range(1,22)] # Zerinvary Lajos, Jul 09 2008
    
  • Sage
    [lucas_number1(n,9,-1) for n in range(0, 20)] # Zerinvary Lajos, Apr 26 2009
    

Formula

G.f.: x/(1 - 9*x - x^2).
a(n) = 9*a(n-1) + a(n-2), n >= 2, a(0)=0, a(1)=1.
a(n) = (-i)^(n-1)*S(n-1, 9*i) with S(n, x) Chebyshev's polynomials of the second kind (see A049310) and i^2=-1.
a(n) = (ap^n - am^p)/(ap-am) with ap:= (9+sqrt(85))/2 and am:= (9-sqrt(85))/2 = -1/ap (Binet form).
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-1-k, k)*9^(n-1-2*k) n >= 1.
a(n) = F(n, 9), the n-th Fibonacci polynomial evaluated at x=9. - T. D. Noe, Jan 19 2006
a(n) = ((9+sqrt(85))^n - (9-sqrt(85))^n)/(2^n*sqrt(85)). Offset 1. a(3)=82. - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 85^((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]
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+2) = 9*A097839(n), a(2n+1) = A097841(n).
a(3n+1) = A041151(5n), a(3n+2) = A041151(5n+3), a(3n+3) = 2*A041151(5n+4).
Limit_{k -> infinity} (a(n+k)/a(k)) = (A087798(n) + A099371(n)*sqrt(85))/2.
Lim_{n->infinity} A087798(n)/A099371(n) = sqrt(85). (End)
a(n) ~ 1/sqrt(85)*((9+sqrt(85))/2)^n. - Jean-François Alcover, Dec 04 2013
a(n) = [1,0] (M^n) [0,1]^T where M is the matrix [9,1; 1,0]. - Robert Israel, Feb 01 2015
E.g.f.: 2*exp(9*x/2)*sinh(sqrt(85)*x/2)/sqrt(85). - Stefano Spezia, Apr 06 2023

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

A099365 Squares of A052918(n-1) (generalized Fibonacci).

Original entry on oeis.org

0, 1, 25, 676, 18225, 491401, 13249600, 357247801, 9632441025, 259718659876, 7002771375625, 188815108482001, 5091005157638400, 137268324147754801, 3701153746831741225, 99793882840309258276, 2690733682941518232225
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

See the comment in A099279. This is example a=5.
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 5 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 5 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 30 2023

Crossrefs

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, this sequence, A099366, A099367, A099369, A099372, A099374.

Programs

  • Magma
    [(2/29)*(Evaluate(ChebyshevFirst(n), 27/2) -(-1)^n): n in [0..30]]; // G. C. Greubel, Aug 21 2022
    
  • Maple
    with (combinat):seq(fibonacci(n,5)^2,n=0..16); # Zerinvary Lajos, Apr 09 2008
  • Mathematica
    LinearRecurrence[{26,26,-1},{0,1,25},30] (* Harvey P. Dale, Sep 25 2019 *)
  • SageMath
    def A099365(n): return (2/29)*(chebyshev_T(n, 27/2) - (-1)^n)
    [A099365(n) for n in (0..30)] # G. C. Greubel, Aug 21 2022

Formula

a(n) = A052918(n-1)^2, n >= 1, a(0) = 0.
a(n) = 26*a(n-1) + 26*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=25.
a(n) = 27*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 27/2) - (-1)^n)/29 with twice the Chebyshev's T(n, x) polynomials of the first kind. 2*T(n, 27/2) = A090248(n).
G.f.: x*(1-x)/((1-27*x+x^2)*(1+x)) = x*(1-x)/(1-26*x-26*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 25*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 30 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (5 + sqrt(29))/10 (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

A099373 Twice Chebyshev polynomials of the first kind, T(n,x), evaluated at 83/2.

Original entry on oeis.org

2, 83, 6887, 571538, 47430767, 3936182123, 326655685442, 27108485709563, 2249677658208287, 186696137145578258, 15493529705424787127, 1285776269413111753283, 106703936831582850735362, 8855140980751963499281763, 734869997465581387589650967, 60985354648662503206441748498
Offset: 0

Views

Author

Wolfdieter Lang, Oct 18 2004

Keywords

Comments

Used in A099372.
The proper and improper nonnegative solutions of the Pell equation x(n)^2 - 85*y(n)^2 = +4 are x(n) = a(n) and y(n) = 9*A097839(n), n >= 0. - Wolfdieter Lang, Jul 01 2013

Examples

			Pell equation: n=0: 2^2 - 85*0^2 = +4 (improper), n=1: 83^2 - 85*(9*1)^2 = +4, n=2: 6887^2 - 85*(9*83)^2 = +4. - _Wolfdieter Lang_, Jul 01 2013
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{83,-1},{2,83},20] (* Harvey P. Dale, Apr 07 2025 *)

Formula

a(n) = 83*a(n-1) - a(n-2), n >= 1; a(-1) = 83, a(0) = 2.
a(n) = S(n, 83) - S(n-2, 83) = 2*T(n, 83/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 83) = A097839(n). U-, resp. T-, are Chebyshev polynomials of the second, resp. first, case. See A049310 and A053120.
G.f.: (2 - 83*x)/(1 - 83*x + x^2).
a(n) = ap^n + am^n, with ap := (83+9*sqrt(85))/2 and am := (83-9*sqrt(85))/2.
E.g.f.: 2*exp(83*x/2)*cosh(9*sqrt(85)*x/2). - Stefano Spezia, Apr 06 2023
Showing 1-8 of 8 results.