A212336 -id:A212336 - cp's OEIS frontend

A027941 a(n) = Fibonacci(2*n + 1) - 1.

0, 1, 4, 12, 33, 88, 232, 609, 1596, 4180, 10945, 28656, 75024, 196417, 514228, 1346268, 3524577, 9227464, 24157816, 63245985, 165580140, 433494436, 1134903169, 2971215072, 7778742048, 20365011073, 53316291172, 139583862444, 365435296161, 956722026040

Offset: 0

Clark Kimberling

Also T(2n+1,n+1), T given by A027935. Also first row of Inverse Stolarsky array.
Third diagonal of array defined by T(i, 1)=T(1, j)=1, T(i, j)=Max(T(i-1, j)+T(i-1, j-1); T(i-1, j-1)+T(i, j-1)). - Benoit Cloitre, Aug 05 2003
Number of Schroeder paths of length 2(n+1) having exactly one up step starting at an even height (a Schroeder path is a lattice path starting from (0,0), ending at a point on the x-axis, consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis). Schroeder paths are counted by the large Schroeder numbers (A006318). Example: a(1)=4 because among the six Schroeder paths of length 4 only the paths (U)HD, (U)UDD, H(U)D, (U)DH have exactly one U step that starts at an even height (shown between parentheses). - Emeric Deutsch, Dec 19 2004
Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023
Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005
Successive extremal petal bends beta(n) = a(n-2). See the Ring Lemma of Rodin and Sullivan in K. Stephenson, Introduction to Circle Packing (Cambridge U. P., 2005), pp. 73-74 and 318-321. - David W. Cantrell (DWCantrell(AT)sigmaxi.net)
a(n+1)= AAB^(n)(1), n>=1, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 4=`110`, 12=`1100`, 33=`11000`, 88=`110000`, ..., in Wythoff code. AA(1)=1=a(1) but for uniqueness reason 1=A(1) in Wythoff code. - N. J. A. Sloane, Jun 29 2008
Start with n. Each n generates a sublist {n-1,n-1,n-2,..,1}. Each element of each sublist also generates a sublist. Add numbers in all terms. For example, 3->{2,2,1} and both 2->{1,1}, so a(3) = 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12. - Jon Perry, Sep 01 2012
For n>0: smallest number such that the inner product of Zeckendorf binary representation and its reverse equals n: A216176(a(n)) = n, see also A189920. - Reinhard Zumkeller, Mar 10 2013
Also, numbers m such that 5*m*(m+2)+1 is a square. - Bruno Berselli, May 19 2014
Also, number of nonempty submultisets of multisets of weight n that span an initial interval of integers (see 2nd example). - Gus Wiseman, Feb 10 2015
From Robert K. Moniot, Oct 04 2020: (Start)
Including a(-1):=0, consecutive terms (a(n-1),a(n))=(u,v) or (v,u) give all points on the hyperbola u^2-u+v^2-v-4*u*v=0 with both coordinates nonnegative integers.  Note that this follows from identifying (1,u+1,v+1) with the Markov triple (1,Fibonacci(2n-1),Fibonacci(2n+1)).  See A001519 (comments by Robert G. Wilson, Oct 05 2005, and Wolfdieter Lang, Jan 30 2015).
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/5. (End)

			a(5) = 88 = 2*33 + 12 + 4 + 1 + 5. a(6) = 232 = 2*88 + 33 + 12 + 4 + 1 + 6. - _Jon Perry_, Sep 01 2012
a(4) = 33 counts all nonempty submultisets of the last row: [1][2][3][4], [11][12][13][14][22][23][24][33][34], [111][112][113][122][123][124][133][134][222][223][233][234], [1111][1112][1122][1123][1222][1223][1233][1234]. - _Gus Wiseman_, Feb 10 2015

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 12.

T. D. Noe, Table of n, a(n) for n = 0..200
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Russ Euler and Jawad Sadek, Problem B-912, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 39, No. 1 (2001), p. 85; From a Product to a Sum, Solution to Problem B-912 by Charles K. Cook, ibid., Vol. 39, No. 5 (2001), pp. 468-469.
Sela Fried, Even-up words and their variants, arXiv:2505.14196 [math.CO], 2025. See p. 7.
Clark Kimberling, Interspersions.
Clark Kimberling, Interspersions and dispersions, Proceedings of the American Mathematical Society, 117 (1993) 313-321.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
R. J. Mathar, Paving rectangular regions with rectangular tiles: tatami and non-tatami tilings, arXiv:1311.6135 [math.CO], 2013, Table 60 (doubled).
Luis A. Medina and Armin Straub, On multiple and infinite log-concavity, Annals of Combinatorics, Vol. 20, No. 1 (2016), pp. 125-138; arXiv preprint, arXiv:1405.1765 [math.CO], 2014; preprint, 2014.
László Németh, Hyperbolic Pascal pyramid, arXiv:1511.02067 [math.CO], 2015 (2nd line of Table 1 is 3*a(n-2)).
László Németh, Pascal pyramid in the space H^2×R, arXiv:1701.06022 [math.CO], 2016. See bn in Table 1 p.10.
N. J. A. Sloane, Classic Sequences.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A035507, A001906, A006318, A000032.
Related to partial sums of Fibonacci(k*n) over n: A000071, A099919, A058038, A138134, A053606; this sequence is the case k=2.
Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).
Cf. A000225 (sublist connection).
Cf. A258993 (row sums, n > 0), A000967.

a027941 = (subtract 1) . a000045 . (+ 1) . (* 2)
-- Reinhard Zumkeller, Mar 10 2013

[Fibonacci(2*n+1)-1: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

with(combinat): seq(fibonacci(2*n+1)-1,n=1..27); # Emeric Deutsch, Dec 19 2004
a:=n->sum(binomial(n+k+1,2*k), k=0..n): seq(a(n), n=-1..26); # Zerinvary Lajos, Oct 02 2007

Table[Fibonacci[2*n+1]-1,{n,0,17}] (* Vladimir Joseph Stephan Orlovsky, Jul 21 2008 *)
LinearRecurrence[{4,-4,1},{0,1,4},40] (* Harvey P. Dale, Aug 17 2021 *)

a(n):=sum(binomial(n+1,k+1)*fib(k),k,0,n); /* Vladimir Kruchinin, Oct 14 2016 */

a(n)=fibonacci(2*n+1)-1 \\ Charles R Greathouse IV, Nov 20 2012

concat(0, Vec(x/((1-x)*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Jun 03 2016

a(n) = Sum_{i=1..n} binomial(n+i, n-i). - Benoit Cloitre, Oct 15 2002
G.f.: Sum_{k>=1} x^k/(1-x)^(2*k+1). - Benoit Cloitre, Apr 21 2003
a(n) = Sum_{k=1..n} F(2*k), i.e., partial sums of A001906. - Benoit Cloitre, Oct 27 2003
a(n) = Sum_{k=0..n-1} U(k, 3/2) = Sum_{k=0..n-1} S(k, 3), with S(k, 3) = A001906(k+1). - Paul Barry, Nov 14 2003
G.f.: x/((1-x)*(1-3*x+x^2)) = x/(1-4*x+4*x^2-x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3) with n>=2, a(-1)=0, a(0)=0, a(1)=1.
a(n) = 3*a(n-1) - a(n-2) + 1 with n>=1, a(-1)=0, a(0)=0.
a(n) = Sum_{k=1..n} F(k)*L(k), where L(k) = Lucas(k) = A000032(k) = F(k-1) + F(k+1). - Alexander Adamchuk, May 18 2007
a(n) = 2*a(n-1) + (Sum_{k=1..n-2} a(k)) + n. - Jon Perry, Sep 01 2012
Sum {n >= 1} 1/a(n) = 3 - phi, where phi = 1/2*(1 + sqrt(5)) is the golden ratio. The ratio of adjacent terms r(n) := a(n)/a(n-1) satisfies the recurrence r(n+1) = (4*r(n) - 1)/(r(n) + 1) for n >= 2. - Peter Bala, Dec 05 2013
a(n) = S(n, 3) - S(n-1, 3) - 1, n >= 0, with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0. - Wolfdieter Lang, Aug 28 2014
a(n) = -1 + (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Jun 03 2016
E.g.f.: (sqrt(5)*sinh(sqrt(5)*x/2) + 5*cosh(sqrt(5)*x/2))*exp(3*x/2)/5 - exp(x). - Ilya Gutkovskiy, Jun 03 2016
a(n) = Sum_{k=0..n} binomial(n+1,k+1)*Fibonacci(k). - Vladimir Kruchinin, Oct 14 2016
a(n) = Sum_{k=0..n-1} Sum_{i=0..n-1} C(k+i+1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Robert K. Moniot, Aug 23 2020
a(n) = Sum_{k=1..n} C(2*n-k,k). - Wesley Ivan Hurt, Dec 22 2020
a(n) = Sum_{k = 1..2*n+2} (-1)^k*Fibonacci(k). - Peter Bala, Nov 14 2021
a(n) = (2*cosh((1 + 2*n)*arccsch(2)))/sqrt(5) - 1. - Peter Luschny, Nov 21 2021
a(n) = F(n + (n mod 2)) * L(n+1 - (n mod 2)), where L(n) = A000032(n) and F(n) = A000045(n) (Euler and Sadek, 2001). - Amiram Eldar, Jan 13 2022

More terms from James Sellers, Sep 08 2000

Paul Barry's Nov 14 2003 formula, recurrences and g.f. corrected for offset 0 and index link for Chebyshev polynomials added by Wolfdieter Lang, Aug 28 2014

A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

Original entry on oeis.org

0, 1, 5, 20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset: 0

Henry Bottomley, Jun 04 2001

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002
On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011
Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007
Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)
Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008
From John P. McSorley, May 26 2020: (Start)
Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.
Also related to the block sizes of small multi-set designs. (End)
If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

			a(2)=5 and T(5)=15 which is 1/3 of 45=T(9).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
Brian Lawrence and Will Sawin, The Shafarevich conjecture for hypersurfaces in abelian varieties, arXiv:2004.09046 [math.NT], 2020.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
L. A. Medina and A. Straub, On multiple and infinite log-concavity, 2013.
S. Morier-Genoud, V. Ovsienko and S. Tabachnikov, 2-frieze patterns and the cluster structure of the space of polygons, Annales de l'institut Fourier, 62 no. 3 (2012), 937-987. - From _N. J. A. Sloane_, Dec 26 2012
Robert Phillips, A triangular number result, 2009.
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, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Eric Weisstein, Centered Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001571, A001834, A001835, A079935, A101265. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2) + 1: n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

f:= gfun:-rectoproc({a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3),a(1)=1,a(0)=0,a(-1)=0},a(n),remember):
map(f, [$0..50]); # Robert Israel, Jun 05 2015

CoefficientList[Series[x/(1 - 5*x + 5*x^2 - x^3), {x, 0, nn}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{5,-5,1},{0,1,5},30] (* Harvey P. Dale, Dec 23 2012 *)

M = [1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1, 30, print1(([1, 0, 0]*M^i)[3], ",")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

a(n) = 4*a(n-1) - a(n-2) + 1.
a(n) = A001075(n) - a(n-1) - 1.
a(n) = (A001835(n+1) - 1)/2 = (A001353(n+1) - A001353(n) - 1)/2.
a(n) = a(n-1) + A001353(n), i.e., partial sum of A001353.
From Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002: (Start)
a(n+2) = 4*a(n+1) - a(n) + 1 for a(0)=0, a(1)=1.
G.f.: x/((1 - x)*(1 - 4*x + x^2)).
a(n) = (1/12)*((3 - sqrt(3))*(2 - sqrt(3))^n + (3 + sqrt(3))*(2 + sqrt(3))^n-6). (End)
a(n) = (1/12)*(A003500(n) + A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003
a(n+1) = Sum_{k=0..n} U(k, 2) = Sum_{k=0..n} S(k, 4), where U(n,x) and S(n,x) are Chebyshev polynomials. - Paul Barry, Nov 14 2003
G.f.: x/(1 - 5*x + 5*x^2 - x^3).
a(n) = A079935(n+1) + A001571(n) for n>0, a(0)=0. - Gerry Martens, Jun 05 2015
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Bruno Berselli, Nov 29 2016
From John P. McSorley, May 25 2020: (Start)
a(n)^2 - a(n) + a(n-1)^2 - a(n-1) - 4*a(n)*a(n-1) = 0.
a(n) = A001834(n-1) + a(n-2). (End)
(T(a(n)-1) + T(a(n+1)-1))/T(a(n) + a(n+1) - 1) = 2/3 where T(i) is the i-th triangular number. - Robert K. Moniot, Oct 11 2020
E.g.f.: exp(x)*(exp(x)*(3*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - 3)/6. - Stefano Spezia, Feb 05 2021
a(n) = A101265(n) - 1. - Jon E. Schoenfield, Jan 01 2022

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

A076139 Triangular numbers that are one-third of another triangular number: T(m) such that 3*T(m) = T(k) for some k.

Original entry on oeis.org

0, 1, 15, 210, 2926, 40755, 567645, 7906276, 110120220, 1533776805, 21362755051, 297544793910, 4144264359690, 57722156241751, 803965923024825, 11197800766105800, 155965244802456376, 2172315626468283465, 30256453525753512135, 421418033734080886426

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002

Both triangular and generalized pentagonal numbers: intersection of A000217 and A001318. - Vladeta Jovovic, Aug 29 2004

Partial sums of Chebyshev polynomials S(n,14).

			G.f. = x + 15*x^2 + 210*x^3 + 2926*x^4 + 40755*x^5 + 567645*x^6 + ...
a(3)=210=T(20) and 3*210=630=T(35).

Colin Barker, Table of n, a(n) for n = 0..874
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
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.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

The m values are in A061278, the k values are in A001571.
Cf. A014979, A076140, A108281.
Cf. A045899, A123480, A217855.
Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

[(Evaluate(ChebyshevU(n+1), 7) - Evaluate(ChebyshevU(n), 7) - 1)/12 : n in [0..30]]; // G. C. Greubel, Feb 03 2022

a[n_] := a[n] = 14*a[n-1] - a[n-2] + 1; a[0] = 0; a[1] = 1; Table[ a[n], {n, 0, 17}] (* Jean-François Alcover, Dec 15 2011, after given formula *)

{a(n) = polchebyshev( n, 2, 7) / 14 + polchebyshev( n, 1, 7)/ 84 - 1 / 12}; /* Michael Somos, Jun 16 2011 */

concat(0, Vec(-x/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

[(chebyshev_U(n,7) - chebyshev_U(n-1,7) - 1)/12 for n in (0..30)] # G. C. Greubel, Feb 03 2022

G.f.: x / ((1 - x) * (1 - 14*x +x^2)).
a(n+1) = Sum_{k=0..n} S(k, 14), n >= 0, where S(k, 14) = U(k, 7) = A007655(k+2).
a(n+1) = (S(n+1, 14) - S(n, 14) - 1)/12, n >= 0.
a(n) = 14 * a(n-1) - a(n-2) + 1. a(0)=0, a(1)=1.
a(-n) = a(n-1).
a(n) = A061278(n)*(A061278(n)+1)/2.
a(n) = (1/288)*(-24 + (12-6*sqrt(3))*(7-4*sqrt(3))^n + (12+6*sqrt(3))*(7+4*sqrt(3))^n).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3) with a(0)=0, a(1)=1, a(2)=15. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002
a(2*n) = A108281(n + 1). a(2*n + 1) = A014979(n + 2). - Michael Somos, Jun 16 2011
a(n) = (1/2)*A217855(n) = (1/3)*A076140(n) = (1/4)*A123480(n) = (1/8)*A045899(n). - Peter Bala, Dec 31 2012
a(n) = A001353(n) * A001353(n-1) / 4. - Richard R. Forberg, Aug 26 2013
a(n) = ((2+sqrt(3))^(2*n+1) + (2-sqrt(3))^(2*n+1))/48 - 1/12. - Vladimir Pletser, Jan 15 2021

More terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 01 2002

Chebyshev comments from Wolfdieter Lang, Aug 31 2004

A053142 a(n) = A053141(n)/2.

Original entry on oeis.org

0, 1, 7, 42, 246, 1435, 8365, 48756, 284172, 1656277, 9653491, 56264670, 327934530, 1911342511, 11140120537, 64929380712, 378436163736, 2205687601705, 12855689446495, 74928449077266, 436715005017102, 2545361581025347, 14835454481134981, 86467365305784540

Offset: 0

Wolfdieter Lang

Partial sums of A001109. - Barry Williams, May 03 2000.
Number m such that 16*m*(2*m+1)+1 is a square. - Bruno Berselli, Oct 19 2012
From Robert K. Moniot, Sep 21 2020: (Start)
Consecutive terms (a(n-1),a(n))=(u,v) give all points on the hyperbola u^2-u+v^2-v-6*u*v=0 in quadrant 1 with both coordinates an integer.
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/4.
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001109, A001653, A053141, A001652, A046090, A049310.

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018

Join[{a=0,b=1}, Table[c=6*b-a+1; a=b; b=c, {n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Table[(Fibonacci[2n + 1, 2] - 1)/4, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
LinearRecurrence[{7, -7, 1}, {0, 1, 7}, 30] (* G. C. Greubel, Jul 15 2018 *)

{a=1+sqrt(2); b=1-sqrt(2); P(n) = (a^n - b^n)/(a-b)};
for(n=0, 30, print1(round((P(2*n+1) - 1)/4), ", ")) \\ G. C. Greubel, Jul 15 2018

x='x+O('x^30); Vec(x/((1-x)*(1-6*x+x^2))) \\ G. C. Greubel, Jul 15 2018

a(n) = (A001653(n)-1)/4.
a(n) = 6*a(n-1)-a(n-2)+1, a(0)=0, a(1)=1.
G.f.: x/((1-x)*(1-6*x+x^2)).
From Paul Barry, Nov 14 2003: (Start)
a(n+1) = Sum_{k=0..n} S(k, 6) = Sum_{k=0..n} U(n, 3), Chebyshev polynomials of 2nd kind, A049310.
a(n+1) = (sqrt(2)-1)^(2*n)(5/8-7*sqrt(2)/16)+(sqrt(2)+1)^(2*n)*(7*sqrt(2)/16 + 5/8)-1/4. (End)
From Antonio Alberto Olivares, Jan 13 2004: (Start)
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
a(n) = -(1/4) + (1-sqrt(2))/(-8*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(8*sqrt(2))*(3+2*sqrt(2))^n. (End)
a(n) = Sum_{k=0..n} Sum_{j=0..2*k} (-1)^(j+1)*A000129(j)*A000129(2*k-j). Paul Barry, Oct 23 2009
a(2*k) = A001109(k)*(A001109(k) + A001109(k-1)) and a(2*k-1) = A001109(k)*(A001109(k) + A001109(k+1)). Kenneth J Ramsey, Sep 10 2010
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Robert K. Moniot, Sep 21 2020
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/8. - Stefano Spezia, Mar 16 2024

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

Original entry on oeis.org

1, 8, 56, 385, 2640, 18096, 124033, 850136, 5826920, 39938305, 273741216, 1876250208, 12860010241, 88143821480, 604146740120, 4140883359361, 28382036775408, 194533374068496, 1333351581704065, 9138927697859960

Offset: 1

K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, Apr 06 2004

a(n) such that 9*(T(a(n)-1) + T(a(n+1)-1)) = 7*(T(a(n) + a(n+1) - 1)), where T(i) denotes the i-th triangular number.
Partial sums of Chebyshev sequence S(n,7) = U(n,7/2) = A004187(n+1). - Wolfdieter Lang, Aug 31 2004
From Klaus Purath, Aug 06 2025: (Start)
Numbers k such that both 3*k + 1 and 15*k + 1 are perfect squares. Also the sum of two consecutive terms is a square.
Take any recurrence (r) of the form (3,-1) with initial value 0 followed by an arbitrary positive integer i. Then the product of two consecutive terms of r divided by 3*i^2 gives the current sequence. (End)

			G.f. = x + 8*x^2 + 56*x^3 + 385*x^4 + 2640*x^5 + 18096*x^6 + ... - _Michael Somos_, Jan 23 2025

Michael De Vlieger, Table of n, a(n) for n = 1..1197
Francesca Arici and Jens Kaad, Gysin sequences and SU(2)-symmetries of C*-algebras, arXiv:2012.11186 [math.OA], 2020.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A212336 for more sequences with g.f. of the type 1/(1 - k*x + k*x^2 - x^3).

Cf. A000032, A004187, A058038, A081079.

A092521:= func< n | (Lucas(4*n+2) -3)/15 >; // G. C. Greubel, Jun 12 2025

a[1] = 1; a[2] = 8; a[3] = 56; a[n_] := a[n] = 8 a[n - 1] - 8 a[n - 2] + a[n - 3]; Table[ a[n], {n, 20}] (* Robert G. Wilson v, Apr 08 2004 *)
Table[(LucasL[4n+2]-3)/15, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 28 2015 *)
LinearRecurrence[{8,-8,1},{1,8,56},30] (* Harvey P. Dale, Dec 27 2015 *)

Vec(x/((1-x)*(1-7*x+x^2)) + O(x^100)) \\ Altug Alkan, Oct 29 2015

def A092521(n): return (lucas_number2(4*n+2,1,-1) -3)//15 # G. C. Greubel, Jun 12 2025

G.f.: x/(1 - 8*x + 8*x^2 - x^3) = x/((1 - x)*(1 - 7*x + x^2)).
a(n) = 7*a(n-1) - a(n-2) + 1, n>=2, a(0):=0, a(1)=1.
a(n) = (S(n, 7)-S(n-1, 7) -1)/5, n>=1, with S(n, 7) = U(n, 7/2) = A004187(n+1).
a(n) = A058038(n)/3.
a(n) = (1/3)*Sum_{k=0..n} Fibonacci(4*k). - Gary Detlefs, Dec 07 2010
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 23 2025
From G. C. Greubel, Jun 12 2025: (Start)
a(n) = A081079(n)/15.
E.g.f.: (1/15)*( exp(7*x/2)*( 3*cosh(p*x) + sqrt(5)*sinh(p*x) ) - 3*exp(x) ), where p = 3*sqrt(5)/2. (End)

Edited and extended by Robert G. Wilson v, Apr 08 2004

A076765 Partial sums of Chebyshev sequence S(n,8) = U(n,4) = A001090(n+1).

Original entry on oeis.org

1, 9, 72, 568, 4473, 35217, 277264, 2182896, 17185905, 135304345, 1065248856, 8386686504, 66028243177, 519839258913, 4092685828128, 32221647366112, 253680493100769, 1997222297440041, 15724097886419560, 123795560793916440

Offset: 0

Maurice MARGENSTERN (margens(AT)lita.univ-metz.fr), Nov 14 2002

In the tiling {5,3,4} of 3-dimensional hyperbolic space, the number of regular dodecahedra with right angles of the n generation which are contained in an eighth of space (intersection of three pairwise perpendicular hyperplanes which are supported by the faces of a dodecahedron at a vertex).

Let beta be the greatest real root of the polynomial which is defined by the above recurrent equation. Consider the representation of positive numbers in the basis beta. Then the language which consists of the maximal representations of positive numbers is neither regular nor context-free (M. Margenstern's theorem, see second reference, above).

M. Margenstern and G. Skordev, Tools for devising cellular automata in the 3-dimensional hyperbolic space, I - the geometrical part, proceedings of SCI'2002, Orlando, Florida, Jul 14-18, (2002), vol. XI, 542-547 Vol. 100 (1993), pp. 1-25.
M. Margenstern and G. Skordev, Tools for devising cellular automata in the 3-dimensional hyperbolic space, II - the numeric algorithms, proceedings of SCI'2002, Orlando, Florida, Jul 14-18, (2002), vol. XI, 548-552

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Margenstern, Number of polyhedra at distance n in {5,3,4}
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-9,1).

Cf. A092521 (partial sums of S(n, 7)).

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

Join[{a=1,b=9},Table[c=8*b-a+1; a=b; b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 19 2011 *)
LinearRecurrence[{9,-9,1},{1,9,72},30] (* Harvey P. Dale, Mar 13 2014 *)
CoefficientList[Series[1/((1 - x) (1 - 8 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 15 2014 *)

a(n+3) = 9*a(n+2) - 9*a(n+1) + a(n); initial values: a(0) = 1, a(1) = 9, a(2) = 72
a(n) = Sum_{k=0..n} S(k, 8) with S(k, x) = U(k, x/2) Chebyshev's polynomials of the second kind.
G.f.: 1/((1-x)*(1 - 8*x + x^2)) = 1/(1 - 9*x + 9*x^2 - x^3).
a(n) = 8*a(n-1) - a(n-2) + 1; a(-1)=0, a(0)=1.
a(n) = (S(n+1, 8) - S(n, 8) - 1)/6, n >= 0.

Extension and Chebyshev comments from Wolfdieter Lang, Aug 31 2004

A089817 a(n) = 5*a(n-1) - a(n-2) + 1 with a(0)=1, a(1)=6.

Original entry on oeis.org

1, 6, 30, 145, 696, 3336, 15985, 76590, 366966, 1758241, 8424240, 40362960, 193390561, 926589846, 4439558670, 21271203505, 101916458856, 488311090776, 2339638995025, 11209883884350, 53709780426726, 257339018249281

Offset: 0

Paul Barry, Nov 14 2003

Partial sums of Chebyshev sequence S(n,5) = U(n,5/2) = A004254(n) (Chebyshev's polynomials of the second kind, see A049310). - Wolfdieter Lang, Aug 31 2004

In this sequence 4*a(n)*a(n+2)+1 is a square. - Bruno Berselli, Jun 19 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
F. M. van Lamoen, Square wreaths around hexagons, Forum Geometricorum, 6 (2006) 311-325.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-6,1).

Cf. A061278, A053142, A101368.

See. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

[Round((2/3 - Sqrt(21)/7)*(5/2 - Sqrt(21)/2)^n + (2/3 + Sqrt(21)/7)*(5/2 + Sqrt(21)/2)^n - 1/3): n in [0..30]]; // G. C. Greubel, Nov 20 2017

Join[{a=1,b=6},Table[c=5*b-a+1;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2011*)
CoefficientList[Series[1/(1 - 6*x + 6*x^2 - x^3), {x, 0, 50}], x] (* G. C. Greubel, Nov 20 2017 *)

a(n)=([0,1,0; 0,0,1; 1,-6,6]^n*[1;6;30])[1,1] \\ Charles R Greathouse IV, Nov 29 2016

x='x+O('x^50); Vec(1/(1-6*x+6*x^2-x^3)) \\ G. C. Greubel, Nov 20 2017

For n > 0, a(n-1) = Sum_{i=1..n} Sum_{j=1..i} b(n) with b(n) as in A004253.
a(n) = (2/3 - sqrt(21)/7)*(5/2 - sqrt(21)/2)^n + (2/3 + sqrt(21)/7)*(5/2 + sqrt(21)/2)^n - 1/3.
G.f.: 1/((1-x)*(1 - 5*x + x^2)) = 1/(1 - 6*x + 6*x^2 - x^3).
a(n) = 6*a(n-1) - 6*a(n-2) + a(n-3) for n >= 2, a(-1):=0, a(0)=1, a(1)=6.
a(n) = (S(n+1, 5) - S(n, 5) - 1)/3 for n >= 0.
a(n)*a(n-2) = a(n-1)*(a(n-1)-1) for n > 1. - Bruno Berselli, Nov 29 2016

A097784 Partial sums of Chebyshev sequence S(n,10) = U(n,5) = A004189(n+1).

Original entry on oeis.org

1, 11, 110, 1090, 10791, 106821, 1057420, 10467380, 103616381, 1025696431, 10153347930, 100507782870, 994924480771, 9848737024841, 97492445767640, 965075720651560, 9553264760747961, 94567571886828051, 936122454107532550, 9266656969188497450

Offset: 0

Wolfdieter Lang, Aug 31 2004

Colin Barker, Table of n, a(n) for n = 0..1000
D. Fortin, B-spline Toeplitz Inverse Under Corner Perturbations, International Journal of Pure and Applied Mathematics, Volume 77, No. 1, 2012, 107-118. - From _N. J. A. Sloane_, Oct 22 2012
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A098296.

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

a:=[1,11,110];; for n in [4..30] do a[n]:=11*a[n-1]-11*a[n-2]+ a[n-3]; od; a; # G. C. Greubel, May 24 2019

I:=[1,11,110]; [n le 3 select I[n] else 11*Self(n-1)-11*Self(n-2) +Self(n-3): n in [1..30]]; // G. C. Greubel, May 24 2019

LinearRecurrence[{11,-11,1}, {1,11,110}, 30] (* G. C. Greubel, May 24 2019 *)
CoefficientList[Series[1/(1-11x+11x^2-x^3),{x,0,30}],x] (* Harvey P. Dale, Aug 24 2021 *)

Vec(1/((1-x)*(1-10*x+x^2)) + O(x^30)) \\ Colin Barker, Jun 14 2015

(1/((1-x)*(1 - 10*x + x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019

a(n) = Sum_{k=0..n} S(k, 10) with S(k, 10) = U(k, 5) = A004189(k+1) Chebyshev's polynomials of the second kind.
G.f.: 1/((1-x)*(1 - 10*x + x^2)) = 1/(1 - 11*x + 11*x^2 - x^3).
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) with n >= 2, a(-1)=0, a(0)=1, a(1)=11.
a(n) = 10*a(n-1) - a(n-2) + 1 with n >= 1, a(-1)=0, a(0)=1.
a(n) = (S(n+1, 10) - S(n, 10) - 1)/8.
a(n) = (-6 + (27-11*sqrt(6))*(5 - 2*sqrt(6))^n + (5 + 2*sqrt(6))^n*(27 + 11*sqrt(6)))/48. - Colin Barker, Mar 05 2016

A049664 a(n) = (F(6*n+3) - 2)/32, where F = A000045 (the Fibonacci sequence).

Original entry on oeis.org

0, 1, 19, 342, 6138, 110143, 1976437, 35465724, 636406596, 11419853005, 204920947495, 3677157201906, 65983908686814, 1184033199160747, 21246613676206633, 381255012972558648, 6841343619829849032, 122762930143964723929, 2202891398971535181691

Offset: 0

Clark Kimberling

Partial sums of Chebyshev polynomials S(n,18).

G. C. Greubel, Table of n, a(n) for n = 0..750
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000045, A001622, A049660, A053606.

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

[(Fibonacc9(6*n+3)-2)/32: n in [0..30]]; // G. C. Greubel, Dec 02 2017

LinearRecurrence[{19, -19, 1}, {0, 1, 19}, 50] (* or *) Table[(Fibonacci[ 6*n +3] - 2)/32, {n,0,30}] (* G. C. Greubel, Dec 02 2017 *)

a(n)=fibonacci(6*n+3)\32 \\ Charles R Greathouse IV, Oct 07 2016

G.f.: x/(1-19*x+19*x^2-x^3) = x/((1-x)*(1-18*x+x^2)).
a(n+1) = Sum_{k=0..n} S(k, 18), with n>=0, S(k, 18) = U(k, 9) = A049660(k+1).
a(n) = 19*a(n-1) - 19*a(n-2) + a(n-3), n>=3, a(0)=0, a(1)=1, a(2)=19.
a(n) = 18*a(n-1) - a(n-2) + 1, n>=2, a(0)=0, a(1)=1.
a(n+1) = (S(n+1, 18) - S(n, 18) - 1)/16, n>=0.
a(n) = (1/8)*Sum_{k=0..n} Fibonacci(6*k). - Gary Detlefs, Dec 07 2010
Product_{n>=2} (1 - 1/a(n)) = phi^6/19 = (4*sqrt(5)+9)/19, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

Chebyshev comments from Wolfdieter Lang, Aug 31 2004

cp's OEIS Frontend

A027941 a(n) = Fibonacci(2*n + 1) - 1.

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A061278 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

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A076139 Triangular numbers that are one-third of another triangular number: T(m) such that 3*T(m) = T(k) for some k.

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A053142 a(n) = A053141(n)/2.

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A092521 a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.

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A076765 Partial sums of Chebyshev sequence S(n,8) = U(n,4) = A001090(n+1).

A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

A092521 a(n) = 8a(n-1) - 8a(n-2) + a(n-3), with a(1) = 1, a(2) = 8, a(3) = 56.