A100237 -id:A100237 - cp's OEIS frontend

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

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

Offset: 0

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

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)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
Milan Janjić, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,1).

Row 5 of A073133, A172236, and A352361.

Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).

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

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

R:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // Marius A. Burtea, Oct 16 2019

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

LinearRecurrence[{5, 1}, {1, 5}, 30] (* Vincenzo Librandi, Feb 23 2013 *)
Table[Fibonacci[n+1, 5], {n,0,30}] (* Vladimir Reshetnikov, May 08 2016 *)

Vec(1/(1-5*x-x^2)+O(x^30)) \\ Charles R Greathouse IV, Nov 20 2011

[lucas_number1(n,5,-1) for n in range(1, 22)] # Zerinvary Lajos, Apr 24 2009

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

A100235 Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n-1 for n>0, where G100234(x) is the g.f. of A100234.

Original entry on oeis.org

1, 1, 4, 1, 8, 26, 1, 12, 63, 139, 1, 16, 116, 436, 726, 1, 20, 185, 965, 2830, 3774, 1, 24, 270, 1790, 7335, 17634, 19601, 1, 28, 371, 2975, 15505, 52444, 106827, 101784, 1, 32, 488, 4584, 28860, 124424, 358748, 633952, 528526, 1, 36, 621, 6681, 49176, 256194

Offset: 0

Paul D. Hanna, Nov 29 2004

The main diagonal forms A100236. Secondary diagonal is: T(n+1,n) = (n+1)*A100237(n). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))).

			Rows begin:
[1],
[1,4],
[1,8,26],
[1,12,63,139],
[1,16,116,436,726],
[1,20,185,965,2830,3774],
[1,24,270,1790,7335,17634,19601],
[1,28,371,2975,15505,52444,106827,101784],
[1,32,488,4584,28860,124424,358748,633952,528526],...
where row sums form 6^n-1 for n>0:
6^1-1 = 1+4 = 5
6^2-1 = 1+8+26 = 35
6^3-1 = 1+12+63+139 = 215
6^4-1 = 1+16+116+436+726 = 1295
6^5-1 = 1+20+185+965+2830+3774 = 7775.
The main diagonal forms A100236 = [1,4,26,139,726,3774,...],
where Sum_{n>=1} A100236(n)/n*x^n = log((1-x)/(1-5*x-x^2)).

Cf. A100234, A100236, A100237, A100232.

row[n_] := CoefficientList[ Series[ (1 + 5*x + Sqrt[1 + 6*x + 29*x^2])^n/2^n, {x, 0, n}], x]; Flatten[ Table[ row[n], {n, 0, 9}]](* Jean-François Alcover, May 11 2012, after PARI *)

PARI
```
T(n,k,m=6)=if(n
				
```

G.f.: A(x, y)=(1-2*x*y+6*x^2*y^2)/((1-x*y)*(1-5*x*y-x^2*y^2-x*(1-x*y))).

cp's OEIS Frontend

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

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