A110679 -id:A110679 - cp's OEIS frontend

A001076 Denominators of continued fraction convergents to sqrt(5).

0, 1, 4, 17, 72, 305, 1292, 5473, 23184, 98209, 416020, 1762289, 7465176, 31622993, 133957148, 567451585, 2403763488, 10182505537, 43133785636, 182717648081, 774004377960, 3278735159921, 13888945017644, 58834515230497, 249227005939632, 1055742538989025

Offset: 0

N. J. A. Sloane

a(2*n+1) with b(2*n+1) := A001077(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = -1, a(2*n) with b(2*n) := A001077(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 5*a^2 = +1 (cf. Emerson reference).
Bisection: a(2*n+1) = T(2*n+1, sqrt(5))/sqrt(5) = A007805(n), n >= 0 and a(2*n) = 4*S(n-1,18), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. S(n,18)=A049660(n+1). - Wolfdieter Lang, Jan 10 2003
Apart from initial terms, this is the Pisot sequence E(4,17), a(n) = floor(a(n-1)^2/a(n-2) + 1/2).
This is also the Horadam sequence (0,1,1,4), having the recurrence relation a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 4, r = 1. a(n) / a(n-1) converges to 5^1/2 + 2 as n approaches infinity. 5^(1/2) + 2 can also be written as (2 * Phi) + 1 and Phi^2 + Phi. - Ross La Haye, Aug 18 2003
Numerators of continued fraction [4, 4, 4, ...], where the convergents to [4, 4, 4, ...] = (4/1, 17/4, 72/17, ...). Let X = the 2 X 2 matrix [0, 1; 1, 4]; then X^n = [a(n-1), a(n); a(n), a(n+1)]; e.g., X^3 = [4, 17; 17, 72]. Let C = the limit of a(n)/a(n-1) = 2 + sqrt(5) = 4.236067977...; then C^n = a(n+1) + (1/C)*a(n), where (1/C) = 0.236067977... . Example: C^3 = 76.01315556..., = 72 + 17*(0.2360679...). - Gary W. Adamson, Dec 15 2007, corrected by Greg Dresden, Sep 16 2019, corrected by Alex Mark, Jul 21 2020
Sqrt(5) = 4/2 + 4/17 + 4/(17*305) + 4/(305*5473) + 4/(5473*98209) + ... . - Gary W. Adamson, Dec 15 2007
a(p) == 20^((p-1)/2) (mod p) for odd primes p. - Gary W. Adamson, Feb 22 2009
a(n) = A167808(3*n). - Reinhard Zumkeller, Nov 12 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 4's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Moreover, a(n) is the second binomial transform of (0,1,0,5,0,25,...) (see also A033887). This fact can be proved similarly like the proof of Paul Barry's remark in A033887 by using the following scaling identity for delta-Fibonacci numbers: y^n b(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) b(k;x) and the fact that b(n;2) = (1-(-1)^n) 5^floor(n/2). - Roman Witula, Jul 12 2012
Binomial transform of 0, 1, 2, 8, 24, 80, 256, ... (A063727 with offset 1). - R. J. Mathar, Feb 05 2014
For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,...,4} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
With offset 1 is the INVERT transform of A006190: (1, 3, 10, 33, 109, 360, ...). - Gary W. Adamson, Jul 24 2015
From Rogério Serôdio, Mar 30 2018: (Start)
This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
The initial 0 of this sequence is in contradiction with the fact that 0 is no valid denominator and according to all standard references, the first convergent of a continued fraction is p(0)/q(0) = b(0)/1 where b(0) is the first term of the continued fraction, given by the integer part of the number. One may artificially define q(-1) = 0 to have a recurrent relation q(n) = b(n)*q(n-1) + q(n-2), n >= 1, but then its index should be -1. - M. F. Hasler, Nov 01 2019
Number of 4-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
From Michael A. Allen, Feb 15 2023: (Start)
Also called the 4-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 4 kinds of squares available. (End)
a(n) is the smallest nonnegative integer that is the sum of n, but no fewer, Fibonacci numbers including negative-index Fibonacci numbers (A039834), with that sum being a(n) = Sum_{i=0..n-1} A000045(3*i+1). a(n) is also the smallest nonnegative integer that is the sum of n, but no fewer, terms each of which is either a Fibonacci number or the negative of a Fibonacci number. (See A027941 for negatives disallowed.) - Mike Speciner, Oct 08 2023
From Enrique Navarrete, Dec 16 2023: (Start)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n > = 1, where P(k) = A006190(k) is the k-th 3-metallonacci number (see example below).
In general, the number of compositions with k-metallonacci number of parts is counted by the (k+1)-st metallonacci sequence (note k=1 and k=2 are the Fibonacci and the Pell numbers, respectively). (End).
a(n) is the number of tilings of a 2 X n rectangle missing the top right 1 X 1 cell, using 1 X 1 squares, dominoes and right trominoes. Compare to A110679 which is the same problem but without the missing top right cell. - Greg Dresden and Yilin Zhu, Jul 10 2025

			1 2 9 38 161 (A001077)
-,-,-,--,---, ...
0 1 4 17 72 (A001076)
G.f. = x + 4*x^2 + 17*x^3 + 72*x^4 + 305*x^5 + 1292*x^6 + 5473*x^7 + 23184*x^8 + ...
From _Enrique Navarrete_, Dec 16 2023: (Start)
From the comment on compositions with 3-metallonacci sorts of parts, A006190(k), there are A006190(1)=1 type of 1, A006190(2)=3 types of 2, A006190(3)=10 types of 3, A006190(4)=33 types of 4, A006190(5)=109 types of 5 and A006190(6)=360 types of 6. The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
 6,              1,      360;
 5+1,            2,      218;
 4+2,            2,      198;
 3+3,            1,      100;
 4+1+1,          3,       99;
 3+2+1,          6,      180;
 2+2+2,          1,       27;
 3+1+1+1,        4,       40;
 2+2+1+1,        6,       54;
 2+1+1+1+1,      5,       15;
 1+1+1+1+1+1,    1,        1;
for a total of a(6)=1292 compositions of n=6. (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 23.
S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80. See Table 1.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 282.

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Paraskevas K. Alvanos and Konstantinos A. Draziotis, Integer Solutions of the Equation y^2 = Ax^4 + B, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.4.
Elwyn Berlekamp and Richard K. Guy, Fibonacci Plays Billiards (2003), arXiv:2002.03705 [math.HO], 2020. See p. 5.
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.
D. W. Boyd, Some integer sequences related to the Pisot sequences, Acta Arithmetica, 34 (1979), 295-305.
D. W. Boyd, Linear recurrence relations for some generalized Pisot sequences, Advances in Number Theory ( Kingston ON, 1991) 333-340, Oxford Sci. Publ., Oxford Univ. Press, New York, 1993.
Latham Boyle and Paul J. Steinhardt, Self-Similar One-Dimensional Quasilattices, arXiv preprint arXiv:1608.08220 [math-ph], 2016.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
M. C. Firengiz and A. Dil, Generalized Euler-Seidel method for second order recurrence relations, Notes on Number Theory and Discrete Mathematics, Vol. 20, 2014, No. 4, 21-32.
Yixing Fu, E. J. König, J. H. Wilson, Yang-Zhi Chou, and J. H. Pixley, Magic-angle semimetals, arXiv:1809.04604 [cond-mat.str-el], 2018.
Juan B. Gil and Aaron Worley, Generalized metallic means, arXiv:1901.02619 [math.NT], 2019.
Brian Hopkins and Stéphane Ouvry, Combinatorics of Multicompositions, arXiv:2008.04937 [math.CO], 2020.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 398
Milan Janjic, 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
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Kai Wang, On k-Fibonacci Sequences And Infinite Series List of Results and Examples, 2020.
Shaoxiong (Steven) Yuan, Generalized Identities of Certain Continued Fractions, arXiv:1907.12459 [math.NT], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Row n=4 of A073133, A172236 and A352361.
Cf. A000045, A001077, A015448, A175183 (Pisano periods).
Cf. A049660, A007805, A110679.
Partial sums of A033887. First differences of A049652. Bisection of A059973.
Third column of array A028412.

a:=[0,1];; for n in [3..30] do a[n]:=4*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018

I:=[0,1]; [n le 2 select I[n] else 4*Self(n-1) + Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A001076:=-1/(-1+4*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{0}, Denominator[Convergents[Sqrt[5], 30]]] (* Harvey P. Dale, Dec 10 2011 *)
a[ n_] := Fibonacci[3*n] / 2; (* Michael Somos, Feb 23 2014 *)
a[ n_] := ((2 + Sqrt[5])^n - (2 - Sqrt[5])^n) /(2 Sqrt[5]) // Simplify; (* Michael Somos, Feb 23 2014 *)
LinearRecurrence[{4, 1}, {0, 1}, 26] (* Jean-François Alcover, Sep 23 2017 *)
a[ n_] := Fibonacci[n, 4]; (* Michael Somos, Nov 02 2021 *)

a(n):=sum(4^(n-1-2*k)*binomial(n-k-1,n-2*k-1),k,0,floor((n)/2));/* Vladimir Kruchinin, Oct 02 2022 */

numlib::fibonacci(3*n)/2 $ n = 0..30; // Zerinvary Lajos, May 09 2008

{a(n) = fibonacci(3*n) / 2}; /* Michael Somos, Aug 11 2009 */

{a(n) = imag( (2 + quadgen(20))^n )}; /* Michael Somos, Feb 23 2014 */

{a(n) = polchebyshev(n-1, 2, 2*I)/I^(n-1)}; /* Michael Somos, Nov 02 2021 */

[lucas_number1(n,4,-1) for n in range(23)] # Zerinvary Lajos, Apr 23 2009

[fibonacci(3*n)/2 for n in range(23)] # Zerinvary Lajos, May 15 2009

a(n) = 4*a(n-1) + a(n-2), n > 1. a(0)=0, a(1)=1.
G.f.: x/(1 - 4*x - x^2).
a(n) = ((2+sqrt(5))^n - (2-sqrt(5))^n)/(2*sqrt(5)).
a(n) = A014445(n)/2 = F(3n)/2.
a(n) = ((-i)^(n-1))*S(n-1, 4*i), with i^2 = -1 and S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0.
a(n) = Sum_{i=0..n} Sum_{j=0..n} Fibonacci(i+j)*n!/(i!j!(n-i-j)!)/2. - Paul Barry, Feb 06 2004
E.g.f.: exp(2*x)*sinh(sqrt(5)*x)/sqrt(5). - Vladeta Jovovic, Sep 01 2004
a(n) = F(1) + F(4) + F(7) + ... + F(3n-2), for n > 0.
Conjecture: 2a(n+1) = a(n+2) - A001077(n+1). - Creighton Dement, Nov 28 2004
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n, j)*C(j, k)*F(j)/2. - Paul Barry, Feb 14 2005
a(n) = A048876(n) - A048875(n). - Creighton Dement, Mar 19 2005
Let M = {{0, 1}, {1, 4}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = v[n][[1]]. - Roger L. Bagula, May 29 2005
a(n) = F(n, 4), the n-th Fibonacci polynomial evaluated at x=4. - T. D. Noe, Jan 19 2006
[A015448(n), a(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = (Sum_{k=0..n} Fibonacci(3*k-2)) + 1. - Gary Detlefs, Dec 26 2010
a(n) = (3*(-1)^n*F(n) + 5*F(n)^3)/2, n >= 0. See the general D. Jennings formula given in a comment on triangle A111125, where also the reference is given. Here the second (k=1) row [3,1] applies. - Wolfdieter Lang, Sep 01 2012
Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (Sum_{k>=1} (-1)^(k-1)/(F_k*F_(k+1)))^3 = phi^(-3), where F_n is the n-th Fibonacci numbers (A000045) and phi is golden ratio (A001622). - Vladimir Shevelev, Feb 23 2013
G.f.: Q(0)*x/(2-4*x), where Q(k) = 1 + 1/(1 - x*(5*k-4)/(x*(5*k+1) - 2/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Oct 11 2013
a(-n) = -(-1)^n * a(n). - Michael Somos, Feb 23 2014
The o.g.f. A(x) = x/(1 - 4*x - x^2) satisfies A(x) + A(-x) + 8*A(x)*A(-x) = 0 or equivalently (1 + 8*A(x))*(1 + 8*A(-x)) = 1. The o.g.f. for A049660 equals -A(sqrt(x))*A(-sqrt(x)). - Peter Bala, Apr 02 2015
From Rogério Serôdio, Mar 30 2018: (Start)
Some properties:
(1) a(n)*a(n+1) = 4*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 4*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(n-1)*a(n+1) = a(n)^2 + (-1)^n (particular case of (5)!);
(7) a(2*n) = 2*a(n)*(2*a(n) + a(n-1));
(8) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(9) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (2+sqrt(5))^(2*k-1) + (2-sqrt(5))^(2*k-1);
(10) 31|Sum_{k=n..n+9} a(k), for all positive n. (End)
O.g.f.: x*exp(Sum_{n >= 1} Lucas(3*n)*x^n/n) = x + 4*x^2 + 17*x^3 + .... - Peter Bala, Oct 11 2019
a(n) = Sum_{k=0..floor(n/2)} 4^(n-2*k-1)*C(n-k-1,n-2*k-1). - Vladimir Kruchinin, Oct 02 2022
a(n) = i^(n-1)*S(n-1, -4*i), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
G.f.: x/(1 - 4*x - x^2) = Sum_{n >= 0} x^(n+1) * ( Product_{k = 1..n} (m*k + 4 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
a(n) = 4^(n-1)*hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4) for n > 0. - Peter Luschny, Mar 30 2025
a(n) = a(n-1) + A110679(n-1) + A110679(n-2) = a(n-1) + Fibonacci(3*n-2). - Greg Dresden and Yilin Zhu, Jul 10 2025

A077259 First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.

Original entry on oeis.org

0, 2, 6, 44, 116, 798, 2090, 14328, 37512, 257114, 673134, 4613732, 12078908, 82790070, 216747218, 1485607536, 3889371024, 26658145586, 69791931222, 478361013020, 1252365390980, 8583840088782, 22472785106426, 154030760585064, 403257766524696, 2763969850442378

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 01 2002

			a(3) = (2*6) - 2 + (2*17) = 12 - 2 + 34 = 44.
G.f. = 2*x + 6*x^2 + 44*x^3 + 116*x^4 + 798*x^5 + 2090*x^6 + 14328*x^7 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A007805, A077260, A077261, A077262.

Cf. A053141.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))); // G. C. Greubel, Jul 15 2018

f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 18*a(n - 2) - a(n - 4) + 8}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020

LinearRecurrence[{1, 18, -18, -1, 1}, {0, 2, 6, 44, 116}, 30] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ 2 x (x + 1)^2 / ((1 - x) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

my(x='x+O('x^30)); concat([0], Vec(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))) \\ G. C. Greubel, Jul 15 2018

Let b(n) be A007805(n). Then with a(0)=0, a(1)=2, a(2*n+2) = 2*a(2*n+1) - a(2*n) + 2*b(n), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + 2*b(n+1).
a(n) = (A000045(A007310(n+1))-1)/2. - Vladeta Jovovic, Nov 02 2002 [corrected by R. J. Mathar, Sep 16 2009]
a(0)=0, a(1)=2, a(n+2) = 4 + 9*a(n) + 2*sqrt(1 +20*a(n) +20*a(n)^2). - Herbert Kociemba, May 12 2008
a(0)=0, a(1)=2, a(2)=6, a(3)=44, a(n) = 18*a(n-2) - a(n-4) + 8. - Robert Phillips, Sep 01 2008
G.f.: 2*x*(1+x)^2/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(2*n) = A049651(2*n); a(2*n+1) = A110679(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
From Vladimir Pletser, Feb 07 2021: (Start)
a(n) = ((5+sqrt(5))*(2+sqrt(5))^n + (5-sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for even n;
a(n) = ((5+3*sqrt(5))*(2+sqrt(5))^n + (5-3*sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for odd n. (End)

More terms from Colin Barker, Mar 23 2014

A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

Original entry on oeis.org

1, 2, 11, 28, 189, 494, 3383, 8856, 60697, 158906, 1089155, 2851444, 19544085, 51167078, 350704367, 918155952, 6293134513, 16475640050, 112925716859, 295643364940, 2026369768941, 5305104928862, 36361730124071, 95196245354568, 652484772464329

Offset: 1

Colin Barker, Feb 03 2015

Also positive integers y in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of x being A254626.

Also indices of centered pentagonal numbers (A005891) that are also hexagonal numbers (A000384). - Colin Barker, Feb 11 2015

			2 is in the sequence because the 2nd centered pentagonal number is 6, which is also the 3rd triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A000217, A005891, A254626, A254628.

Cf. A000384, A254962.

[(2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4 : n in [1..30]]; // G. C. Greubel, Apr 19 2019

CoefficientList[Series[x (x^3 + 9 x^2 - x - 1)/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 25}], x] (* Michael De Vlieger, Jun 06 2016 *)
LinearRecurrence[{1,18,-18,-1,1},{1,2,11,28,189},30] (* Harvey P. Dale, Apr 23 2017 *)

Vec(x*(x^3+9*x^2-x-1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^30))

{a(n) = (2 +(1+3*(-1)^n)*fibonacci(3*n) - 2*(-1)^n*fibonacci(3*n+1))/4}; \\ G. C. Greubel, Apr 19 2019

[(2 +(1+3*(-1)^n)*fibonacci(3*n) -2*(-1)^n*fibonacci(3*n+1))/4 for n in (1..30)] # G. C. Greubel, Apr 19 2019

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+x-9*x^2-x^3)/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)).
a(n) = (10 - sqrt(5)*(2-sqrt(5))^n - 5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + sqrt(5)*(2+sqrt(5))^n + (-2-sqrt(5))^n*(-5+2*sqrt(5)))/20. - Colin Barker, Jun 06 2016
a(2*n+2) = A232970(2*n+1); a(2*n+1) = A110679(2*n). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = (2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4. - G. C. Greubel, Apr 19 2019

A232970 Expansion of (1-3x)/(1-5x+3*x^2+x^3).

Original entry on oeis.org

1, 2, 7, 28, 117, 494, 2091, 8856, 37513, 158906, 673135, 2851444, 12078909, 51167078, 216747219, 918155952, 3889371025, 16475640050, 69791931223, 295643364940, 1252365390981, 5305104928862, 22472785106427, 95196245354568, 403257766524697, 1708227311453354, 7236167012338111, 30652895360805796

Offset: 0

N. J. A. Sloane, Dec 05 2013

For n > 2, a(n) is the number of tilings of (a 2 X (n+1) rectangle missing the top right and top left 1 X 1 cells) using 1 X 1 squares, dominoes and right trominoes. Compare with similar tiling sequences A001076 and A110679. - Greg Dresden and Yilin Zhu, Jul 10 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Dziemianczuk, Counting Lattice Paths With Four Types of Steps, Graphs and Combinatorics, September 2013, DOI 10.1007/s00373-013-1357-1.
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A001076, A110679.

I:=[1,2,7]; [n le 3 select I[n] else 5*Self(n-1)- 3*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 24 2017

LinearRecurrence[{5, -3, -1}, {1, 2, 7}, 30] (* Vincenzo Librandi, Jun 24 2017 *)
CoefficientList[Series[(1-3x)/(1-5x+3x^2+x^3),{x,0,30}],x] (* Harvey P. Dale, Oct 19 2024 *)

Vec((1-3*x)/(1-5*x+3*x^2+x^3) + O(x^30)) \\ Felix Fröhlich, Apr 15 2019

[(fibonacci(3*n+1) +1)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3). - N. J. A. Sloane, Jun 23 2017
a(n) = (Fibonacci(3*n+1) + 1)/2 = Sum_{k=0..n} Fibonacci(3*k-1). - Ehren Metcalfe, Apr 15 2019
a(2*n) = A294262(2*n); a(2*n+1) = A254627(2*n+2). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019

A353877 Triangle read by rows: T(n,k) = number of tilings of a n X k rectangle using right trominoes, dominoes and 1 X 1 tiles, n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 1, 1, 2, 11, 1, 3, 44, 369, 1, 5, 189, 3633, 83374, 1, 8, 798, 34002, 1817897, 90916452, 1, 13, 3383, 323293, 40220893, 4635661331, 546063639624, 1, 21, 14328, 3058623, 886130549, 235025597912, 63919977468729, 17259079054003609, 1, 34, 60697, 28982628, 19546906987, 11935601703140, 7495901454256347, 4669873251135795702, 2916019543694306398589

Offset: 0

Gerhard Kirchner, May 09 2022

Tiling algorithm, see A351322.

Reading the sequence {T(n,k)} for k>n, use T(k,n) instead of T(n,k).

			Triangle begins
n\k_0__1____2______3________4__________5____________6
0:  1
1:  1  1
2:  1  2   11
3:  1  3   44    369
4:  1  5  189   3633    83374
5:  1  8  798  34002  1817897   90916452
6:  1 13 3383 323293 40220893 4635661331 546063639624

Liang Kai, Rows n=0..17, flattened

Row/columns 0..4 are A000012, A000045(n+1), A110679, A353878, A353879.
Main diagonal is A353934.
Cf. A219987, A351322, A352589.

Maxima
```
See A352589.
```

A110527 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 0, a(1) = 1, a(2) = 8.

Original entry on oeis.org

0, 1, 8, 29, 128, 537, 2280, 9653, 40896, 173233, 733832, 3108557, 13168064, 55780809, 236291304, 1000946021, 4240075392, 17961247585, 76085065736, 322301510525, 1365291107840, 5783465941881, 24499154875368

Offset: 0

Creighton Dement, Jul 24 2005

A048878(n) = a(n) + a(n+1). Compare with A110526.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,5,1).

Cf. A110526, A110528, A033887, A001076, A049661, A033887.

seriestolist(series(-x*(1+5*x)/((1+x)*(x^2+4*x-1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: 1lesseq[(- 'i + 'j - i' + j' - 'kk' - 'ik' - 'jk' - 'ki' - 'kj')(+ .5'i + .5i' + .5'jj' + .5'kk')], apart from initial term.

LinearRecurrence[{3,5,1},{0,1,8},30] (* Harvey P. Dale, Feb 12 2015 *)

x='x+O('x^50); concat(0, Vec(-x*(1+5*x)/((1+x)*(x^2+4*x-1)))) \\ G. C. Greubel, Aug 30 2017

G.f.: -x*(1+5*x)/((1+x)*(x^2+4*x-1)).
a(n) = (-1)^n + 3*A001076(n) - A015448(n). - Ehren Metcalfe, Nov 18 2017
a(n) = (-1)^n + 2*A110526(n) + A110679(n-2) for n >= 2. - Yomna Bakr and Greg Dresden, May 25 2024

A110687 Expansion of -(7x^2+3x-1)(2x^2+2x+1) / ((3x^2+3x+1)(2x^3+2x^2+4*x+1)).

Original entry on oeis.org

1, -8, 28, -100, 358, -1276, 4558, -16342, 58732, -211306, 760498, -2737168, 9851098, -35452510, 127584124, -459135130, 1652275834, -5945992576, 21397667026, -77003195254, 277109379628, -997226422690, 3588693361378, -12914539595584, 46475225095450

Offset: 0

Creighton Dement, Aug 02 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-7,-17,-20,-12,-6).

Cf. A110688, A110689, A110679.

seriestolist(series(-(7*x^2+3*x-1)*(2*x^2+2*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: tessum(infty)-4basekforsumseq[ + 'i - .25'j + .25'k - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e], Sumtype is set to: default; Fortype is set to: 1A.

CoefficientList[Series[-(7*x^2 + 3*x - 1)*(2*x^2 + 2*x + 1)/((3*x^2 + 3*x + 1)*(2*x^3 + 2*x^2 + 4*x + 1)), {x,0,50}], x] (* G. C. Greubel, Sep 06 2017 *)

Vec(-(7*x^2+3*x-1)*(2*x^2+2*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = -7*a(n-1) - 17*a(n-2) - 20*a(n-3) - 12*a(n-4) - 6*a(n-5) for n>4. - Colin Barker, May 19 2019

A110688 Expansion of -(2x+1)(6x^2+4x+1)/((3x^2+3x+1)(2x^3+2x^2+4x+1)).

Original entry on oeis.org

-1, 1, -4, 19, -73, 262, -931, 3319, -11884, 42679, -153505, 552430, -1988311, 7156123, -25754188, 92683315, -333539317, 1200299014, -4319477491, 15544370887, -55939087228, 201306503071, -724436520553, 2607011250526, -9381785144287

Offset: 0

Creighton Dement, Aug 02 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-7,-17,-20,-12,-6).

Cf. A110687, A110689, A110679.

seriestolist(series(-(2*x+1)*(6*x^2+4*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: tessum(infty)-4jbaseforsumseq[ + 'i - .25'j + .25'k - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e], Sumtype is set to: default; Fortype is set to: 1A.

CoefficientList[Series[-(2*x + 1)*(6*x^2 + 4*x + 1)/((3*x^2 + 3*x + 1)*(2*x^3 + 2*x^2 + 4*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Sep 06 2017 *)

Vec(-(2*x+1)*(6*x^2+4*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

A110689 Expansion of (2x+1)(4x^2+8x+1)/((3x^2+3x+1)(2x^3+2x^2+4x+1)).

Original entry on oeis.org

1, 3, -18, 63, -207, 696, -2415, 8565, -30714, 110583, -398439, 1435152, -5167083, 18598065, -66931314, 240862563, -866772819, 3119198160, -11224913079, 40394716341, -145367356794, 523129840335, -1882574375679, 6774773362320, -24380205972915

Offset: 0

Creighton Dement, Aug 02 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-7,-17,-20,-12,-6).

Cf. A110687, A110688, A110679.

seriestolist(series((2*x+1)*(4*x^2+8*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1)), x=0,25)); -or- Floretion Algebra Multiplication Program, FAMP Code: tessum(infty)-4basekforsumseq[ + 'i - .25'j + .25'k - .25j' + .25k' - .5'ii' - .25'ij' - .25'ik' - .25'ji' - .25'ki' - .5e], Sumtype is set to: default; Fortype is set to: 1A.

CoefficientList[Series[(2*x + 1)*(4*x^2 + 8*x + 1)/((3*x^2 + 3*x + 1)*(2*x^3 + 2*x^2 + 4*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Sep 06 2017 *)

Vec((2*x+1)*(4*x^2+8*x+1)/((3*x^2+3*x+1)*(2*x^3+2*x^2+4*x+1))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

A294262 a(n) = 3a(n-1) + 5a(n-2) + a(n-3), with a(0) = a(1) = 1 and a(2) = 7, a linear recurrence which is a trisection of A005252.

Original entry on oeis.org

1, 1, 7, 27, 117, 493, 2091, 8855, 37513, 158905, 673135, 2851443, 12078909, 51167077, 216747219, 918155951, 3889371025, 16475640049, 69791931223, 295643364939, 1252365390981, 5305104928861, 22472785106427, 95196245354567, 403257766524697, 1708227311453353, 7236167012338111, 30652895360805795, 129847748455561293, 550043889183050965

Offset: 0

Jean-François Alcover and Paul Curtz, Oct 26 2017

Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3,5,1).

Cf. A001076, A005252, A110679.

[(Fibonacci(3*n+1) +(-1)^n)/2 : n in [0..30]]; // G. C. Greubel, Apr 19 2019

Mathematica
```
LinearRecurrence[{3,5,1},{1,1,7},30]
```

{a(n) = (fibonacci(3*n+1) +(-1)^n)/2}; \\ G. C. Greubel, Apr 19 2019

[(fibonacci(3*n+1) +(-1)^n)/2 for n in (0..30)] # G. C. Greubel, Apr 19 2019

a=1
b=1
c=7
print 0," ",a,"\n"
print 1," ",b,"\n"
print 2," ",c,"\n"
for(x=3;x<=1000;++x){
d=3*c+5*b+1*a
print x," ",d,"\n"
a=b
b=c
c=d
} # Hermann Stamm-Wilbrandt, Apr 18 2019

G.f.: (1 - 2*x - x^2)/(1 - 3*x - 5*x^2 - x^3).
a(n) = (1/20)*(10*(-1)^n + (2-sqrt(5))^n*(5-sqrt(5)) + (2+sqrt(5))^n*(5+sqrt(5))).
a(n) = A005252(3*n).
a(n) = 4*a(n-1) + a(n-2) + 2*(-1)^n for n >= 2.
a(n) = Sum_{k=0..floor(3*n/4)} binomial(3*n-2*k, 2*k).
a(n) = A110679(n) - A001076(n).
a(n) = (Fibonacci(3*n + 1) + (-1)^n)/2.
a(2*n) = A232970(2*n); a(2*n+1) = A049651(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019

cp's OEIS Frontend

A001076 Denominators of continued fraction convergents to sqrt(5).

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A077259 First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.

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A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

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A232970 Expansion of (1-3*x)/(1-5*x+3*x^2+x^3).

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A353877 Triangle read by rows: T(n,k) = number of tilings of a n X k rectangle using right trominoes, dominoes and 1 X 1 tiles, n >= 0, k = 0..n.

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A110527 a(n+3) = 3*a(n+2) + 5*a(n+1) + a(n), a(0) = 0, a(1) = 1, a(2) = 8.

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A232970 Expansion of (1-3x)/(1-5x+3*x^2+x^3).

A110527 a(n+3) = 3a(n+2) + 5a(n+1) + a(n), a(0) = 0, a(1) = 1, a(2) = 8.

A110687 Expansion of -(7x^2+3x-1)(2x^2+2x+1) / ((3x^2+3x+1)(2x^3+2x^2+4*x+1)).

A110688 Expansion of -(2x+1)(6x^2+4x+1)/((3x^2+3x+1)(2x^3+2x^2+4x+1)).

A110689 Expansion of (2x+1)(4x^2+8x+1)/((3x^2+3x+1)(2x^3+2x^2+4x+1)).

A294262 a(n) = 3a(n-1) + 5a(n-2) + a(n-3), with a(0) = a(1) = 1 and a(2) = 7, a linear recurrence which is a trisection of A005252.