A108299 -id:A108299 - cp's OEIS frontend

A004253 a(n) = 5*a(n-1) - a(n-2), with a(1)=1, a(2)=4.

1, 4, 19, 91, 436, 2089, 10009, 47956, 229771, 1100899, 5274724, 25272721, 121088881, 580171684, 2779769539, 13318676011, 63813610516, 305749376569, 1464933272329, 7018916985076, 33629651653051, 161129341280179, 772017054747844, 3698955932459041, 17722762607547361

Offset: 1

Frans J. Faase, Per H. Lundow

Number of domino tilings in K_3 X P_2n (or in S_4 X P_2n).
Number of perfect matchings in graph C_{3} X P_{2n}.
Number of perfect matchings in S_4 X P_2n.
In general, Sum_{k=0..n} binomial(2*n-k, k)*j^(n-k) = (-1)^n * U(2*n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,5), where L is defined as in A108299; see also A030221 for L(n,-5). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4} which do not end in 0 (e.g., at n=2, we have 02, 03, 04, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44). - Tanya Khovanova, Jan 10 2007
(sqrt(21)+5)/2 = 4.7912878... = exp(arccosh(5/2)) = 4 + 3/4 + 3/(4*19) + 3/(19*91) + 3/(91*436) + ... - Gary W. Adamson, Dec 18 2007
a(n+1) is the number of compositions of n when there are 4 types of 1 and 3 types of other natural numbers. - Milan Janjic, Aug 13 2010
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(3)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Right-shifted Binomial Transform of the left-shifted A030195. - R. J. Mathar, Oct 15 2012
Values of x (or y) in the solutions to x^2 - 5xy + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
From Wolfdieter Lang, Oct 15 2020: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 5*x*y = -3 (see the preceding comment) are given by [x(n) = S(n, 5) - S(n-1, 5), y(n) = x(n-1)], for n =-oo..+oo, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +21. There is only this family representing -3 properly with x and y positive, and there are no improper solutions.
See the  formula for a(n) = x(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
From Wolfdieter Lang, Feb 08 2021: (Start)
All proper and improper solutions of the generalized Pell equation X^2 - 21*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n+1) from the preceding comment by X(n) = x(n) + x(n-1) = S(n-1, 5) - S(n-2, 5) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 5), for all integer numbers n. For positive integers X(n) = A003501(n) and Y(n) = A004254(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper families of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer n. This follows from the mentioned paper by Robert K. Moniot. (End)
Equivalent definition: a(n) = ceiling(a(n-1)^2 / a(n-2)), with a(1)=1, a(2)=4, a(3)=19. The problem for USA Olympiad (see Andreescu and Gelca reference) asks to prove that a(n)-1 is always a multiple of 3. - Bernard Schott, Apr 13 2022

Titu Andreescu and Rǎzvan Gelca, Putnam and Beyond, New York, Springer, 2007, problem 311, pp. 104 and 466-467 (proposed for the USA Mathematical Olympiad by G. Heuer).
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.
F. A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..200
F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Preliminary version of paper that appeared in Ars Combin. 49 (1998), 129-154.
F. Faase, Counting Hamiltonian cycles in product graphs
F. Faase, Results from the counting program
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Frank A. Haight, Letter to N. J. A. Sloane, Sep 06 1976
Frank A. Haight, On a generalization of Pythagoras' theorem, pp. 73-77 of J. C. Butcher, editor, A Spectrum of Mathematics. Auckland University Press, 1971. [Annotated scanned copy]
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 422
Tanya Khovanova, Recursive Sequences
Lisa Lokteva, Constructing Rational Homology 3-Spheres That Bound Rational Homology 4-Balls, arXiv:2208.14850 [math.GT], 2022.
Per Hakan Lundow, Computation of matching polynomials and the number of 1-factors in polygraphs, Research report, No 12, 1996, Department of Math., Umea University, Sweden.
Per Hakan Lundow, Enumeration of matchings in polygraphs, 1998.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962, 2014
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
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
James A. Sellers, Domino Tilings and Products of Fibonacci and Pell Numbers, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.2
F. M. van Lamoen, Square wreaths around hexagons, Forum Geometricorum, 6 (2006) 311-325.
Index entries for sequences related to dominoes
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A003501, A004254, A030221, A049310, A004254 (partial sums), A290902 (first differences).
Row 5 of array A094954.
Cf. similar sequences listed in A238379.

a:=[1,4];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

[ n eq 1 select 1 else n eq 2 select 4 else 5*Self(n-1)-Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011

a[0]:=1: a[1]:=1: for n from 2 to 26 do a[n]:=5*a[n-1]-a[n-2] od: seq(a[n], n=1..22); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{5, -1}, {1, 4}, 22] (* Jean-François Alcover, Sep 27 2017 *)

Vec((1-x)/(1-5*x+x^2)+O(x^30)) \\ Charles R Greathouse IV, Jul 01 2013

[lucas_number1(n,5,1)-lucas_number1(n-1,5,1) for n in range(1, 23)] # Zerinvary Lajos, Nov 10 2009

G.f.: x*(1 - x) / (1 - 5*x + x^2).  Simon Plouffe in his 1992 dissertation.[offset 0]
For n>1, a(n) = A005386(n) + A005386(n-1). - Floor van Lamoen, Dec 13 2006
a(n) ~ (1/2 + 1/14*sqrt(21))*(1/2*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002[offset 0]
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then q(n, 3)=a(n). - Benoit Cloitre, Nov 10 2002 [offset 0]
For n>0, a(n)*a(n+3) = 15 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*3^k. - Paul Barry, Jul 26 2004[offset 0]
a(n) = (-1)^n*U(2n, i*sqrt(3)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005[offset 0]
[a(n), A004254(n)] = the 2 X 2 matrix [1,3; 1,4]^n * [1,0]. - Gary W. Adamson, Mar 19 2008
a(n) = ((sqrt(21)-3)*((5+sqrt(21))/2)^n + (sqrt(21)+3)*((5-sqrt(21))/2)^n)/2/sqrt(21). - Seiichi Kirikami, Sep 06 2011
a(n) = S(n-1, 5) - S(n-2, 5) = (-1)^n*S(2*n, i*sqrt(3)), n >= 1, with the Chebyshev S polynomials (A049310), and S(n-1, 5) = A004254(n), for n >= 0. See a Paul Barry formula (offset corrected). - Wolfdieter Lang, Oct 15 2020
From Peter Bala, Feb 10 2024: (Start)
a(n) = a(1-n).
a(n) = A004254(n) + A004254(1-n).
For n, j, k in Z, a(n)*a(n+j+k) - a(n+j)*a(n+k) = 3*A004254(j)*A004254(k). The case j = 1, k = 2 is given above.
a(n)^2 + a(n+1)^2 - 5*a(n)*a(n+1) = - 3.
More generally, a(n)^2 + a(n+k)^2 - (A004254(k+1) - A004254(k-1))*a(n)*a(n+k) = -3*A004254(k)^2. (End)
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/3 (telescoping series: for n >= 2,  3/(a(n) - 1/a(n)) = 1/A004254(n-1) - 1/A004254(n)). - Peter Bala, May 21 2025
E.g.f.: exp(5*x/2)*(7*cosh(sqrt(21)*x/2) - sqrt(21)*sinh(sqrt(21)*x/2))/7 - 1. - Stefano Spezia, Jul 02 2025

Additional comments from James Sellers and N. J. A. Sloane, May 03 2002

More terms from Ray Chandler, Nov 17 2003

A072256 a(n) = 10*a(n-1) - a(n-2) for n > 1, a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409, 7488979632860260401, 74133255753507979601, 733843577902219535609

Offset: 0

Lekraj Beedassy, Jul 08 2002

Any k in the sequence is followed by 5*k + 2*sqrt(2*(3*k^2 - 1)).
Gives solutions for x in 3*x^2 - 2*y^2 = 1. Corresponding y is given by A054320(n-1). [corrected by Jon E. Schoenfield, Jun 08 2018]
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7,8,9} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For n >= 2, a(n) equals the permanent of the (2n-2) X (2n-2) tridiagonal matrix with sqrt(8)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 8 = 0. - Colin Barker, Feb 09 2014
The aerated sequence [b(n)]n>=1 = [1, 0, 9, 0, 89, 0, 881, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -12, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. - Peter Bala, May 12 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Christian Aebi and Grant Cairns, Equable Parallelograms on the Eisenstein Lattice, arXiv:2401.08827 [math.CO], 2024. See p. 14.
Jean-Paul Allouche, Jeffrey Shallit, and Manon Stipulanti, Combinatorics on words and generating Dirichlet series of automatic sequences, arXiv:2401.13524 [math.CO], 2024.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 283).
Bruno Deschamps, Sur les bonnes valeurs initiales de la suite de Lucas-Lehmer, Journal of Number Theory, Volume 130, Issue 12, December 2010, Pages 2658-2670.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A031138, A046172, A054320, A108299.
Row 10 of array A094954.
First differences of A004189.
Essentially the same as A138288.

a:=[1,1];; for n in [3..20] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

[n le 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..25]]; // Vincenzo Librandi, Feb 10 2014

seq( simplify(ChebyshevU(n,5) -9*ChebyshevU(n-1,5)), n=0..20); # G. C. Greubel, Jan 14 2020

a[n_]:= a[n]= 10a[n-1] -a[n-2]; a[0]=a[1]=1; Table[ a[n], {n, 0, 20}]
CoefficientList[Series[(1-9x)/(1-10x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
Table[ChebyshevU[n, 5] -9*ChebyshevU[n-1, 5], {n,0,20}] (* G. C. Greubel, Jan 14 2020 *)
LinearRecurrence[{10,-1},{1,1},20] (* Harvey P. Dale, Jun 17 2022 *)

a(n)=([0,1; -1,10]^n*[1;1])[1,1] \\ Charles R Greathouse IV, May 10 2016

vector(21, n, polchebyshev(n-1,2,5) -9*polchebyshev(n-2,2,5) ) \\ G. C. Greubel, Jan 14 2020

a(n) = (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n.
a(n) = (2*A031138(n) + 1)/3 = sqrt((2*A054320(n-1)^2 + 1)/3), n >= 1.
a(n) = U(n-1, 5)-U(n-2, 5) = T(2*n-1, sqrt(3))/sqrt(3) with Chebyshev's U- and T- polynomials and U(-1, x) := 0, U(-2, x) := -1, T(-1, x) := x.
G.f.: (1-9*x)/(1-10*x+x^2).
6*a(n)^2 - 2 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 10 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 8) = a(n+1). - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 80 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = L(n-1,10), where L is defined as in A108299; see also A054320 for L(n,-10). - Reinhard Zumkeller, Jun 01 2005
a(n) = A138288(n-1) for n > 0. - Reinhard Zumkeller, Mar 12 2008
a(n) = sqrt(A046172(n)). - Paul Weisenhorn, May 15 2009
a(n) = ceiling(((3-sqrt(6))*(5+2*sqrt(6))^n)/6). - Paul Weisenhorn, May 23 2020
E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) - sqrt(6)*sinh(2*sqrt(6)*x))/3. - Stefano Spezia, Oct 25 2023
From Peter Bala, May 08 2025: (Start)
a(n) = (-1)^n * Dir(n-1, -5), where Dir(n, x) denotes the n-th row polynomial of A244419.
For arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = -8 with a(n) := (3-sqrt(6))/6 * (5+2*sqrt(6))^n + (3+sqrt(6))/6 * (5-2*sqrt(6))^n as given above (the particular case x = 0 is noted in the Comments section).
a(n+1/2) = 1/sqrt(3) * A001079(n).
a(n+3/4) + a(n+1/4) = sqrt(2/3) * sqrt(1 + sqrt(3)) * A001079(n).
a(n+3/4) - a(n+1/4) = 4 * sqrt(sqrt(3) - 1) * A004189(n).
a(n) divides a(3*n-1); a(n) divides a(5*n-2); in general, for k >= 0, a(n) divides a((2*k+1)*n - k).
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/8 (telescoping series: for n >= 2, 1/(a(n) - 1/a(n)) = 1/A291181(n-2) - 1/A291181(n-1).)
Product_{n >= 2} ((a(n) + 1)/(a(n) - 1))^(-1)^n = sqrt(3/2) (telescoping product: Product_{n = 2..k} (((a(n) + 1)/(a(n) - 1))^(-1)^n)^2 = 3/2 * (1 - (-1)^(k+1)/(3*A098308(k))).) (End)

Edited by Robert G. Wilson v, Jul 17 2002

A030221 Chebyshev even-indexed U-polynomials evaluated at sqrt(7)/2.

Original entry on oeis.org

1, 6, 29, 139, 666, 3191, 15289, 73254, 350981, 1681651, 8057274, 38604719, 184966321, 886226886, 4246168109, 20344613659, 97476900186, 467039887271, 2237722536169, 10721572793574, 51370141431701, 246129134364931, 1179275530392954, 5650248517599839

Offset: 0

Wolfdieter Lang

a(n) = L(n,-5)*(-1)^n, where L is defined as in A108299; see also A004253 for L(n,+5). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n) = (a(1)-1)*a(n-1) - a(n-2), a(1) >= 4; lim_{n->oo} a(n) = x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives the present sequence. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 29, 139, 3191, 15289, 350981, 1681651, ... - Ctibor O. Zizka, Sep 02 2008
Inverse binomial transform of A030240. - Philippe Deléham, Nov 19 2009
For positive n, a(n) equals the permanent of the (2n)X(2n) matrix with sqrt(7)'s along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
The aerated sequence (b(n))n>=1 = [1, 0, 6, 0, 29, 0, 139, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -3, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for a connection with Chebyshev polynomials. - Peter Bala, Mar 22 2015
From Wolfdieter Lang, Oct 26 2020: (Start)
((-1)^n)*a(n) = X(n) = ((-1)^n)*(S(n, 5) + S(n-1, 5)) and Y(n) = X(n-1) gives all integer solutions (modulo sign flip between X and Y) of X^2 + Y^2 + 5*X*Y = +7, for n = -oo..+oo, with Chebyshev S polynomials (see A049310), with S(-1, x) = 0, and S(-n, x) = - S(n-2, x), for n >= 2.
This binary indefinite quadratic form of discriminant 21, representing 7, has only this family of proper solutions (modulo sign flip), and no improper ones.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

			G.f. = 1 + 6*x + 29*x^2 + 139*x^3 + 666*x^4 + 3191*x^5 + 15289*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681. (see page 678)
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Taras Goy and Mark Shattuck, Determinants of Toeplitz-Hessenberg Matrices with Generalized Leonardo Number Entries, Ann. Math. Silesianae (2023). See p. 18.
Christian Kassel and Christophe Reutenauer, Pairs of intertwined integer sequences, arXiv:2507.15780 [math.NT], 2025. See p. 13.
Tanya Khovanova, Recursive Sequences.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=6.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-1).

Cf. A004253, A004254, A100047, A054493 (partial sums), A049310, A003501 (first differences), A299109 (subsequence of primes).

I:=[1,6]; [n le 2 select I[n] else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A030221 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,6]);
    else
        5*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

t[n_, k_?EvenQ] := I^k*Binomial[n-k/2, k/2]; t[n_, k_?OddQ] := -I^(k-1)*Binomial[n+(1-k)/2-1, (k-1)/2]; l[n_, x_] := Sum[t[n, k]*x^(n-k), {k, 0, n}]; a[n_] := (-1)^n*l[n, -5]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jul 05 2013, after Reinhard Zumkeller *)
a[ n_] := ChebyshevU[2 n, Sqrt[7]/2]; (* Michael Somos, Jan 22 2017 *)

{a(n) = simplify(polchebyshev(2*n, 2, quadgen(28)/2))}; /* Michael Somos, Jan 22 2017 */

[(lucas_number2(n,5,1)-lucas_number2(n-1,5,1))/3 for n in range(1,22)] # Zerinvary Lajos, Nov 10 2009

a(n) = 5*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = U(2*n, sqrt(7)/2).
G.f.: (1+x)/(x^2-5*x+1).
a(n) = A004254(n) + A004254(n+1).
a(n) ~ (1/2 + (1/6)*sqrt(21))*((1/2)*(5 + sqrt(21)))^n. - Joe Keane (jgk(AT)jgk.org), May 16 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then a(n) = (-1)^n*q(n, -7). - Benoit Cloitre, Nov 10 2002
A054493(2*n) = a(n)^2 for all n in Z. - Michael Somos, Jan 22 2017
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jan 22 2017
0 = -7 + a(n)*(+a(n) - 5*a(n+1)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Jan 22 2017
a(n) = S(n, 5) + S(n-1, 5) = S(2*n, sqrt(7)) (see above in terms of U), for n >= 0 with S(-1, 5) = 0, where the coefficients of the Chebyshev S polynomials are given in A049310. - Wolfdieter Lang, Oct 26 2020
From Peter Bala, May 16 2025: (Start)
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/7 (telescoping series: 7/(a(n) - 1/a(n)) = 1/A004254(n+1) + 1/A004254(n)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(7/3) (telescoping product: Product_{k = 1..n} ((a(k) + 1)/(a(k) - 1))^2 = 7/3 * (1 - 8/A231087(n+1))). (End)

A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

Original entry on oeis.org

1, 6, 41, 281, 1926, 13201, 90481, 620166, 4250681, 29134601, 199691526, 1368706081, 9381251041, 64300051206, 440719107401, 3020733700601, 20704416796806, 141910183877041, 972666870342481, 6666757908520326, 45694638489299801, 313195711516578281

Offset: 0

Clark Kimberling

In general, Sum_{k=0..n} binomial(2*n-k,k)j^(n-k) = (-1)^n*U(2n, I*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,7), where L is defined as in A108299; see also A033890 for L(n,-7). - Reinhard Zumkeller, Jun 01 2005
Take 7 numbers consisting of 5 ones together with any two successive terms from this sequence. This set has the property that the sum of their squares is 7 times their product. (R. K. Guy, Oct 12 2005.) See also A111216.
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6} which do not end in 0. - Tanya Khovanova, Jan 10 2007
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
From Wolfdieter Lang, Feb 09 2021: (Start)
All positive solutions of the Diophantine equation x^2 + y^2 - 7*x*y = -5 are given by [x(n) = S(n, 7) - S(n-1, 7), y(n) = x(n-1)], for all integer numbers n, with the Chebyshev S-polynomials (A049310), with S(-1, 0) = 0, and S(-n, x) = -S(n-2, x), for n >= 2. x(n) = a(n), for n >= 0.
This indefinite binary quadratic form has discriminant D = +45. There is only this family representing -5 properly with x and y positive, and there are no improper solutions.
All proper and improper solutions of the generalized Pell equation X^2 - 45*Y^2 = +4 are given, up to a combined sign change in X and Y, in terms of x(n) = a(n) from the preceding comment, by X(n) = x(n) + x(n-1) = S(n-1, 7) - S(n-2, 7) and Y(n) = (x(n) - x(n-1))/3 = S(n-1, 7), for all integer numbers n. For positive integers X(n) = A056854(n) and Y(n) = A004187(n). X(-n) = X(n) and Y(-n) = - Y(n), for n >= 1.
The two conjugated proper family of solutions are given by [X(3*n+1), Y(3*n+1)] and [X(3*n+2), Y(3*n+2)], and the one improper family by [X(3*n), Y(3*n)], for all integer numbers n.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)

			a(3) = L(4*3 + 2)/3 = 843/3 = 281. - _Indranil Ghosh_, Feb 06 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1193
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
John Riordan, Letter to N. J. A. Sloane, Sep 26 1980 with notes on the 1973 Handbook of Integer Sequences. Note that the sequences are identified by their N-numbers, not their A-numbers.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Row 7 of array A094954. First differences of A004187.
Cf. A002310, A002320, A049310, A049685, A056854.
Cf. similar sequences listed in A238379.

[Lucas(4*n+2)/3: n in [0..30]]; // G. C. Greubel, Dec 17 2017

Table[LucasL[4*n+2]/3, {n,0,50}] (* or *) LinearRecurrence[{7,-1}, {1,6}, 50] (* G. C. Greubel, Dec 17 2017 *)

a(n)=(fibonacci(4*n+1)+fibonacci(4*n+3))/3 \\ Charles R Greathouse IV, Jun 16 2014

[lucas_number1(n,7,1)-lucas_number1(n-1,7,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009

Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i); then q(n, 5)=a(n); a(n) = 7a(n-1) - a(n-2). - Benoit Cloitre, Nov 10 2002
From Ralf Stephan, May 29 2004: (Start)
a(n+2) = 7a(n+1) - a(n).
G.f.: (1-x)/(1-7x+x^2).
a(n)*a(n+3) = 35 + a(n+1)*a(n+2). (End)
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*5^k. - Paul Barry, Aug 30 2004
If another "1" is inserted at the beginning of the sequence, then A002310, A002320 and A049685 begin with 1, 2; 1, 3; and 1, 1; respectively and satisfy a(n+1) = (a(n)^2+5)/a(n-1). - Graeme McRae, Jan 30 2005
a(n) = (-1)^n*U(2n, i*sqrt(5)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
[a(n), A004187(n+1)] = [1,5; 1,6]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = S(n, 7) - S(n-1, 7) with Chebyshev S polynomials S(n-1, 7) = A004187(n), for n >= 0. - Wolfdieter Lang, Feb 09 2021
E.g.f.: exp(7*x/2)*(3*cosh(3*sqrt(5)*x/2) + sqrt(5)*sinh(3*sqrt(5)*x/2))/3. - Stefano Spezia, Apr 14 2025
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/9) * sqrt(1 - T(2*n+1, -7/2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 6, 0, 41, 0, 281, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -9, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/5 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A290903(n-1) - 1/A290903(n).) (End)

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

Original entry on oeis.org

1, 1, 4, 4, 10, 10, 20, 20, 35, 35, 56, 56, 84, 84, 120, 120, 165, 165, 220, 220, 286, 286, 364, 364, 455, 455, 560, 560, 680, 680, 816, 816, 969, 969, 1140, 1140, 1330, 1330, 1540, 1540, 1771, 1771, 2024, 2024, 2300, 2300, 2600, 2600, 2925, 2925, 3276, 3276

Offset: 0

Henry Bottomley, Nov 20 2000

For n >= i, i = 6,7, a(n - i) is the number of incongruent two-color bracelets of n beads, i of which are black (cf. A005513, A032280), having a diameter of symmetry. The latter means the following: if we imagine (0,1)-beads as points (with the corresponding labels) dividing a circumference of a bracelet into n identical parts, then a diameter of symmetry is a diameter (connecting two beads or not) such that a 180-degree turn of one of two sets of points around it (obtained by splitting the circumference by this diameter) leads to the coincidence of the two sets (including their labels). - Vladimir Shevelev, May 03 2011
From Johannes W. Meijer, May 20 2011: (Start)
The Kn11, Kn12, Kn13, Fi1 and Ze1 triangle sums, see A180662 for their definitions, of the Connell-Pol triangle A159797 are linear sums of shifted versions of the duplicated tetrahedral numbers, e.g., Fi1(n) = a(n-1) + 5*a(n-2) + a(n-3) + 5*a(n-4).
The Kn11, Kn12, Kn13, Kn21, Kn22, Kn23, Fi1, Fi2, Ze1 and Ze2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above. (End)
The number of quadruples of integers [x, u, v, w] that satisfy x > u > v > w >= 0, n + 5 = x + u. - Michael Somos, Feb 09 2015
Also, this sequence is the fourth column in the triangle of the coefficients of the sum of two consecutive Fibonacci polynomials F(n+1, x) and F(n, x) (n>=0) in ascending powers of x. - Mohammad K. Azarian, Jul 18 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..880
Hansraj Gupta, Enumeration of incongruent cyclic k-gons, Indian J. Pure and Appl. Math., Vol. 10, No. 8 (1979), 964-999.
Vladimir Shevelev, A problem of enumeration of two-color bracelets with several variations, arXiv:0710.1370 [math.CO], 2007-2011.
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A057884. Sum of 2 consecutive terms gives A006918, whose sum of 2 consecutive terms gives A002623, whose sum of 2 consecutive terms gives A000292, which is this sequence without the duplication. Continuing to sum 2 consecutive terms gives A000330, A005900, A001845, A008412 successively.
Cf. A096338, A005513, A032280
Cf. A000292, A190717, A190718. - Johannes W. Meijer, May 20 2011

a058187 n = a058187_list !! n
a058187_list = 1 : f 1 1 [1] where
   f x y zs = z : f (x + y) (1 - y) (z:zs) where
     z = sum $ zipWith (*) [1..x] [x,x-1..1]
-- Reinhard Zumkeller, Dec 21 2011

A058187:= proc(n) option remember; A058187(n):= binomial(floor(n/2)+3, 3) end: seq(A058187(n), n=0..51); # Johannes W. Meijer, May 20 2011

a[n_]:= Length @ FindInstance[{x>u, u>v, v>w, w>=0, x+u==n+5}, {x, u, v, w}, Integers, 10^9]; (* Michael Somos, Feb 09 2015 *)
With[{tetra=Binomial[Range[30]+2,3]},Riffle[tetra,tetra]] (* Harvey P. Dale, Mar 22 2015 *)

{a(n) = binomial(n\2+3, 3)}; /* Michael Somos, Jun 07 2005 */

[binomial((n//2)+3, 3) for n in (0..60)] # G. C. Greubel, Feb 18 2022

a(n) = A006918(n+1) - a(n-1).
a(2*n) = a(2*n+1) = A000292(n) = (n+1)*(n+2)*(n+3)/6.
a(n) = (2*n^3 + 21*n^2 + 67*n + 63)/96 + (n^2 + 7*n + 11)(-1)^n/32. - Paul Barry, Aug 19 2003
a(n) = A108299(n-3,n)*(-1)^floor(n/2) for n > 2. - Reinhard Zumkeller, Jun 01 2005
Euler transform of finite sequence [1, 3]. - Michael Somos, Jun 07 2005
G.f.: 1 / ((1 - x) * (1 - x^2)^3) = 1 / ((1 + x)^3 * (1 - x)^4). a(n) = -a(-7-n) for all n in Z.
a(n) = binomial(floor(n/2) + 3, 3). - Vladimir Shevelev, May 03 2011
a(-n) = -a(n-7); a(n) = A000292(A008619(n)). - Guenther Schrack, Sep 13 2018
Sum_{n>=0} 1/a(n) = 3. - Amiram Eldar, Aug 18 2022

A066170 Triangle read by rows: T(n,k) = (-1)^n(-1)^(floor(3k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n, n >= 0.

Original entry on oeis.org

1, -1, 1, 1, -1, -1, -1, 2, 1, -1, 1, -2, -3, 1, 1, -1, 3, 3, -4, -1, 1, 1, -3, -6, 4, 5, -1, -1, -1, 4, 6, -10, -5, 6, 1, -1, 1, -4, -10, 10, 15, -6, -7, 1, 1, -1, 5, 10, -20, -15, 21, 7, -8, -1, 1, 1, -5, -15, 20, 35, -21, -28, 8, 9, -1, -1, -1, 6, 15, -35, -35, 56, 28, -36, -9, 10, 1, -1, 1, -6, -21, 35, 70, -56, -84, 36, 45, -10, -11

Offset: 0

Floor van Lamoen, Dec 14 2001

The original name of this sequence was: Triangle giving coefficients of characteristic function of n X n matrix in which the left upper half and the antidiagonal are filled with 1's and the right lower half is filled with 0's. As was pointed out by L. Edson Jeffery this is only correct if we multiply each triangle row by (-1)^n. For the straightforward version of the coefficients of the characteristic polynomials see A187660. - Johannes W. Meijer, Aug 08 2011

			The table begins {1}; {-1, 1}; {1, -1, -1}; {-1, 2, 1, -1}; ...
The characteristic function of
( 1 1 1 )
( 1 1 0 )
( 1 0 0 )
is f(x) = x^3 - 2x^2 - x + 1, so the 3rd row is (-1)^3 times the f(x) coefficients, i.e., {-1; 2; 1; -1}.

Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001 (Chapter 14)

Indranil Ghosh, Rows 0..125, flattened
Henry W. Gould, A Variant of Pascal's Triangle, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, pp. 257-271, with corrections.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 22-31.

Cf. A007700, A059455, A065941. For another version see A030111.

A066170 := proc(n,k): (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k) end: seq(seq(A066170(n,k),k=0..n), n=0..11); // Johannes W. Meijer, Aug 08 2011

Flatten[Table[(-1)^n*(-1)^Floor[3*k/2]*Binomial[Floor[(n+k)/2],k],{n,0,12}, {k,0,n}]] (* Indranil Ghosh, Feb 19 2017 *)

From L. Edson Jeffery, Mar 23 2011: (Start)
  T(n,k) = (-1)^n*(-1)^(floor(3*k/2))*binomial(floor((n+k)/2),k);
  T(n,k) = (-1)^n*A187660(n,k). (End)
From Johannes W. Meijer, Aug 08 2011: (Start)
  abs(T(n,k)) = A046854(n,k) = abs(A108299(n,n-k))
  abs(T(n,n-k)) = A065941(n,k). (End)

More terms from Vladeta Jovovic, Jan 02 2002

Corrected and edited by Johannes W. Meijer, Aug 08 2011

A070997 a(n) = 8*a(n-1) - a(n-2), a(0)=1, a(-1)=1.

Original entry on oeis.org

1, 7, 55, 433, 3409, 26839, 211303, 1663585, 13097377, 103115431, 811826071, 6391493137, 50320119025, 396169459063, 3119035553479, 24556114968769, 193329884196673, 1522082958604615, 11983333784640247, 94344587318517361

Offset: 0

Joe Keane (jgk(AT)jgk.org), May 18 2002

A Pellian sequence.
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,8), where L is defined as in A108299; see also A057080 for L(n,-8). - Reinhard Zumkeller, Jun 01 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5,6,7} which do not end in 0. - Tanya Khovanova, Jan 10 2007
Hankel transform of A158197. - Paul Barry, Mar 13 2009
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(6)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Values of x (or y) in the solutions to x^2 - 8xy + y^2 + 6 = 0. - Colin Barker, Feb 05 2014
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*b(n)^2 - 5*a(n)^2 = -2. The corresponding b(n) are A057080(n). Note that (b(n)*b(n+2) - b(n+1)^2)/2 = -5 and (a(n)*a(n+2) - a(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n+1) + t(i))/(t(i+n+1) + t(i+n)) as long as t(i+n+1) + t(i+n) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 7*t(i+2) - 7*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i) regardless of initial values and including this sequence itself. (End)

			1 + 7*x + 55*x^2 + 433*x^3 + 3409*x^4 + 26839*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Fink, R. K. Guy and M. Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

a(n) = sqrt((3*A057080(n)^2+2)/5) (cf. Richardson comment).
Cf. A057080, A001090, A001091.
Row 8 of array A094954.
Cf. A001090.
Cf. similar sequences listed in A238379.
Cf. A041023.

I:=[1, 7]; [n le 2 select I[n] else 8*Self(n-1) - Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 26 2013

CoefficientList[Series[(1 - x)/(1 - 8*x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jan 26 2013 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
   ] (* Complement of A041023 *)
a[15, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{8,-1},{1,7},20] (* Harvey P. Dale, Dec 04 2021 *)

{a(n) = subst( 9*poltchebi(n) - poltchebi(n-1), x, 4) / 5} /* Michael Somos, Jun 07 2005 */

{a(n) = if( n<0, n=-1-n); polcoeff( (1 - x) / (1 - 8*x + x^2) + x * O(x^n), n)} /* Michael Somos, Jun 07 2005 */

[lucas_number1(n,8,1)-lucas_number1(n-1,8,1) for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all members x of the sequence, 15*x^2 - 6 is a square. Lim_{n->infinity} a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 12 2002
a(n) = (5+sqrt(15))/10 * (4+sqrt(15))^n + (5-sqrt(15))/10 * (4-sqrt(15))^n.
a(n) ~ 1/10*sqrt(10)*(1/2*(sqrt(10)+sqrt(6)))^(2*n+1)
a(n) = U(n, 4)-U(n-1, 4) = T(2*n+1, sqrt(5/2))/sqrt(5/2), with Chebyshev's U and T polynomials and U(-1, x) := 0. U(n, 4)=A001090(n+1), n>=-1.
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 6) = a(n) - Benoit Cloitre, Nov 10 2002
a(n)*a(n+3) = 48 + a(n+1)a(n+2). - Ralf Stephan, May 29 2004
a(n) = (-1)^n*U(2n, i*sqrt(6)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005
G.f.: (1-x)/(1-8*x+x^2).
a(n) = a(-1-n).
a(n) = Jacobi_P(n,-1/2,1/2,4)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006
[a(n), A001090(n+1)] = [1,6; 1,7]^(n+1) * [1,0]. - Gary W. Adamson, Mar 21 2008
For n>0, a(n) is the numerator of the continued fraction [2,3,2,3,...,2,3] with n repetitions of 2,3. For the denominators see A136325. - Greg Dresden, Sep 12 2019
From Peter Bala, Apr 30 2025: (Start)
a(n) = (1/sqrt(5)) * sqrt(1 - T(2*n+1, -4)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n+1); a(n) divides a(5*n+2); in general, for k >= 0, a(n) divides a((2*k+1)*n + k).
The aerated sequence [b(n)]n>=1 = [1, 0, 7, 0, 55, 0, 433, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -10, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1/6 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A291033(n-1) - 1/A291033(n).) (End)
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbitrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025

A057080 Even-indexed Chebyshev U-polynomials evaluated at sqrt(10)/2.

Original entry on oeis.org

1, 9, 71, 559, 4401, 34649, 272791, 2147679, 16908641, 133121449, 1048062951, 8251382159, 64962994321, 511452572409, 4026657584951, 31701808107199, 249587807272641, 1965000650073929, 15470417393318791, 121798338496476399

Offset: 0

Wolfdieter Lang, Aug 04 2000

a(n) = L(n,-8)*(-1)^n, where L is defined as in A108299; see also A070997 for L(n,+8). - Reinhard Zumkeller, Jun 01 2005
General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878. a(1)=5 gives A001834. a(1)=6 gives A030221. a(1)=7 gives A002315. a(1)=8 gives A033890. a(1)=9 gives A057080. a(1)=10 gives A057081. - Ctibor O. Zizka, Sep 02 2008
The primes in this sequence are 71, 34649, 16908641, 8251382159, 31701808107199,... - Ctibor O. Zizka, Sep 02 2008
The aerated sequence (b(n))n>=1 = [1, 0, 9, 0, 71, 0, 559, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015
From Klaus Purath, May 06 2025: (Start)
Nonnegative solutions to the Diophantine equation 3*a(n)^2 - 5*b(n)^2 = -2. The corresponding b(n) are A070997(n). Note that (a(n)*a(n+2) - a(n+1)^2)/2 = -5 and (b(n)*b(n+2) - b(n+1)^2)/2 = 3.
(a(n) + b(n))/2 = (a(n+1) - b(n+1))/2 = A001090(n+1) = Lucas U(8,1). Also a(n)*b(n+1) - a(n+1)*b(n) = -2.
a(n) = (t(i+2n+1) - t(i))/(t(i+n+1) - t(i+n)) as long as t(i+n+1) - t(i+n) != 0 for integer i and n >= 0 where (t) is a sequence satisfying t(i+3) = 9*t(i+2) - 9*t(i+1) + t(i) or t(i+2) = 8*t(i+1) - t(i), regardless of the initial values and including this sequence itself. (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Marco Abrate, Stefano Barbero, Umberto Cerruti, and Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), rhs, m=10.
Donatella Merlini and Renzo Sprugnoli, Arithmetic into geometric progressions through Riordan arrays, Discrete Mathematics 340.2 (2017): 160-174.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-1).

Cf. A033890, A100047.

a:=[1,9];; for n in [3..30] do a[n]:=8*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,9]; [n le 2 select I[n] else 8*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

A057080 := proc(n)
    option remember;
    if n <= 1 then
        op(n+1,[1,9]);
    else
        8*procname(n-1)-procname(n-2) ;
    end if;
end proc: # R. J. Mathar, Apr 30 2017

CoefficientList[Series[(1+x)/(1-8x+x^2), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Vec((1+x)/(1-8*x+x^2) + O(x^30)) \\ Michel Marcus, Mar 22 2015

[(lucas_number2(n,8,1)-lucas_number2(n-1,8,1))/6 for n in range(1, 21)] # Zerinvary Lajos, Nov 10 2009

For all elements x of the sequence, 15*x^2 + 10 is a square. Lim. n-> Inf. a(n)/a(n-1) = 4 + sqrt(15). - Gregory V. Richardson, Oct 13 2002
a(n) = 8*a(n-1) - a(n-2), a(-1)=-1, a(0)=1.
a(n) = S(n, 8) + S(n-1, 8) = S(2*n, sqrt(10)) with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 8) = A001090(n).
G.f.: (1+x)/(1-8*x+x^2).
a(n) = ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)). - Gregory V. Richardson, Oct 13 2002
a(n) = sqrt((5*A070997(n)^2 - 2)/3) (cf. Richardson comment).
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = (-1)^n*q(n,-10). - Benoit Cloitre, Nov 10 2002
a(n) = Jacobi_P(n,1/2,-1/2,4)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry, Feb 03 2006
a(n+1) = 4*a(n) + sqrt(5*(3*a(n)^2 + 2)). - Richard Choulet, Aug 30 2007
In addition to the first formula above: In general, the following applies to all recurrences (a(n)) of the form (8,-1) with a(0) = 1 and arbritrary a(1): 15*a(n)^2 + y = b^2 where y = x^2 + 8*x + 1 and x = a(1) - 8. Also y = a(k+1)^2 - a(k)*a(k+1) for any k >=0. - Klaus Purath, May 06 2025
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 4), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 8*a(n)*a(n+1) + a(n+1)^2 = 10.
More generally, for arbitrary x, a(n+x)^2 - 8*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 10 with a(n) := ( ((4+sqrt(15))^(n+1) - (4-sqrt(15))^(n+1)) + ((4+sqrt(15))^n - (4-sqrt(15))^n) )/(2*sqrt(15)) as given above.
a(n+1/2) = sqrt(10) * A001090(n+1).
a(n+3/4) + a(n+1/4) = sqrt(10)*sqrt(sqrt(10) + 2) * A001090(n+1).
a(n+3/4) - a(n+1/4) = sqrt((sqrt(40) - 4)/3) * A001091(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/10 (telescoping series: for n >= 1, 10/(a(n) - 1/a(n)) = 1/A001090(n) + 1/A001090(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(5/3) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 5/3 * (1 - 2/(1 + A001091(k+1)))). (End)

A087960 a(n) = (-1)^binomial(n+1,2).

Original entry on oeis.org

1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1

Offset: 0

W. Edwin Clark, Sep 17 2003

Period 4: repeat [1, -1, -1, 1]. - Joerg Arndt, Feb 14 2016

Also equal to the sign of product(j-i, 1<=j

Hankel transform of A097331, A097332. [Paul Barry, Aug 10 2009]

The Kn22 sums, see A180662, of triangle A108299 equal the terms of this sequence. [Johannes W. Meijer, Aug 14 2011]

			a(1) = -1 since (-1)^binomial(2,2) = (-1)^1 = -1.
G.f. = 1 - x - x^2 + x^3 + x^4 - x^5 - x^6 + x^7 + x^8 - x^9 - x^10 + ...

I. S. Gradstein and I. M. Ryshik, Tables of series, products, and integrals, Volume 1, Verlag Harri Deutsch, 1981.

Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Index entries for linear recurrences with constant coefficients, signature (0,-1).

Cf. A000217, A021913, A057077, A097331, A097332, A108299, A111125, A180662.

a087960 n = (-1) ^ (n * (n + 1) `div` 2)
a087960_list = cycle [1,-1,-1,1]  -- Reinhard Zumkeller, Nov 15 2015

[(-1)^Binomial(n+1,2) : n in [0..100]]; // Wesley Ivan Hurt, Jul 07 2016

A087960:=n->(-1)^binomial(n+1,2): seq(A087960(n), n=0..100); # Wesley Ivan Hurt, Jul 07 2016

(-1)^Binomial[Range[0,110],2] (* or *) LinearRecurrence[{0,-1},{1,1},110] (* Harvey P. Dale, Jul 07 2014 *)
a[ n_] := (-1)^(n (n + 1) / 2); (* Michael Somos, Jul 20 2015 *)
a[ n_] := (-1)^Quotient[ n + 1, 2]; (* Michael Somos, Jul 20 2015 *)

{a(n) = (-1)^((n + 1)\2)}; /* Michael Somos, Jul 20 2015 */

def A087960(n): return -1 if n+1&2 else 1 # Chai Wah Wu, Jan 31 2023

a(n) = (-1)^A000217(n).
a(n) = (-1)^floor((n+1)/2). - Benoit Cloitre and Ray Chandler, Sep 19 2003
G.f.: (1-x)/(1+x^2). - Paul Barry, Aug 10 2009
a(n) = I^(n*(n+1)). - Bruno Berselli, Oct 17 2011
a(n) = Product_{k=1..n} 2*cos(2*k*Pi/(2*n+1)) for n>=0 (for n=0 the empty product is put to 1). See the Gradstein-Ryshik reference, p. 63, 1.396 2. with x = sqrt(-1). - Wolfdieter Lang, Oct 22 2013
a(n) + a(n-2) = 0 for n>1, a(n) = a(n-4) for n>3. - Wesley Ivan Hurt, Jul 07 2016
E.g.f.: cos(x) - sin(x). - Ilya Gutkovskiy, Jul 07 2016
a(n) = Sum_{s=0..n} (-1)^(n-s)*A111125(n, s)*2^s (row polynomials of signed A111125 evaluated at 2). - Wolfdieter Lang, May 02 2021

More terms from Benoit Cloitre and Ray Chandler, Sep 19 2003

Offset and Vandermonde formula corrected by R. J. Mathar, Sep 25 2009

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cp's OEIS Frontend

A004253 a(n) = 5*a(n-1) - a(n-2), with a(1)=1, a(2)=4.

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A072256 a(n) = 10*a(n-1) - a(n-2) for n > 1, a(0) = a(1) = 1.

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A030221 Chebyshev even-indexed U-polynomials evaluated at sqrt(7)/2.

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A049685 a(n) = L(4*n+2)/3, where L=A000032 (the Lucas sequence).

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A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A058187 Expansion of (1+x)/(1-x^2)^4: duplicated tetrahedral numbers.

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A066170 Triangle read by rows: T(n,k) = (-1)^n(-1)^(floor(3k/2))*binomial(floor((n+k)/2),k), 0 <= k <= n, n >= 0.