A072256 -id:A072256 - cp's OEIS frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

0, 1, 10, 99, 980, 9701, 96030, 950599, 9409960, 93149001, 922080050, 9127651499, 90354434940, 894416697901, 8853812544070, 87643708742799, 867583274883920, 8588189040096401, 85014307126080090, 841554882220704499, 8330534515080964900, 82463790268588944501, 816307368170808480110

Offset: 0

N. J. A. Sloane

Indices of square numbers which are also generalized pentagonal numbers.
If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008
The n-th term is a(n) = ((5+sqrt(24))^n - (5-sqrt(24))^n)/(2*sqrt(24)). - Sture Sjöstedt, May 31 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 10'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
a(n) and b(n) (A001079) are the nonnegative proper solutions of the Pell equation b(n)^2 - 6*(2*a(n))^2 = +1. See the cross reference to A001079 below. - Wolfdieter Lang, Jun 26 2013
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,9}. - Milan Janjic, Jan 25 2015
For n > 1, this also gives the number of (n-1)-decimal-digit numbers which avoid a particular two-digit number with distinct digits. For example, there are a(5) = 9701 4-digit numbers which do not include "39" as a substring; see Wikipedia link. - Charles R Greathouse IV, Jan 14 2016
All possible solutions for y in Pell equation x^2 - 24*y^2 = 1. The values for x are given in A001079. - Herbert Kociemba, Jun 05 2022
Dickson on page 384 gives the Diophantine equation "(20)  24x^2 + 1 = y^2" and later states "... three consecutive sets (x_i, y_i) of solutions of (20) or 2x^2 + 1 = 3y^2 satisfy x_{n+1} = 10x_n - x_{n-1}, y_{n+1} = 10y_n - y_{n-1} with (x_1, y_1) = (0, 1) or (1, 1), (x_2, y_2) = (1, 5) or (11, 9), respectively." The first set of values (x_n, y_n) = (A001079(n-1), a(n-1)). - Michael Somos, Jun 19 2023

			a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - _Michael Somos_, Sep 05 2006
G.f. = x + 10*x^2 + 99*x^3 + 980*x^4 + 9701*x^5 + 96030*x^6 + ...

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.

Vincenzo Librandi, 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.
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 12.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=10, q=-1.
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
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=12.
Roger B. Nelson, Multi-Polygonal Numbers, Mathematics Magazine, Vol. 89, No. 3 (June 2016), pp. 159-164.
Robert Phillips, Polynomials of the form 1+4ke+4ke^2, 2008.
Wikipedia, Curse of 39
Jianqiang Zhao, Finite Multiple zeta Values and Finite Euler Sums, arXiv:1507.04917 [math.NT], 2015.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (10,-1).

Cf. A001079, A001109, A001353, A001906, A004187, A004254, A018913, A046173, A049310, A054320, A072256, A101950, A105138, A108741 (squares).
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), this sequence (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=5;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[ n eq 1 select 0 else n eq 2 select 1 else 10*Self(n-1)-Self(n-2): n in [1..20] ]; // Vincenzo Librandi, Aug 19 2011

A004189 := proc(n)
    option remember;
    if n <= 1 then
        n ;
    else
        10*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A004189(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
seq( simplify(ChebyshevU(n-1, 5)), n=0..20); # G. C. Greubel, Dec 23 2019

Table[GegenbauerC[n-1,1,5], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008; modified by G. C. Greubel, Jun 06 2019 *)
LinearRecurrence[{10, -1}, {0, 1}, 20] (* Jean-François Alcover, Nov 15 2017 *)
ChebyshevU[Range[21] -2, 5] (* G. C. Greubel, Dec 23 2019 *)

{a(n) = subst(poltchebi(n+1) - 5*poltchebi(n), 'x, 5) / 24}; /* Michael Somos, Sep 05 2006 */

a(n)=([9,1;8,1]^(n-1)*[1;1])[1,1] \\ Charles R Greathouse IV, Jan 14 2016

vector(21, n, n--; polchebyshev(n-1, 2, 5) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,10,1) for n in range(22)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n-1,5) for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.
A001079(n) = sqrt(24*(a(n)^2)+1), that is a(n) = sqrt((A001079(n)^2-1)/24).
From Barry E. Williams, Aug 18 2000: (Start)
a(n) = ( (5+2*sqrt(6))^n - (5-2*sqrt(6))^n )/(4*sqrt(6)).
G.f.: x/(1-10*x+x^2). (End)
a(-n) = -a(n). - Michael Somos, Sep 05 2006
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = 10*a(n-1) - a(n-2). (End)
a(n+1) = Sum_{k=0..n} A101950(n,k)*9^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = 1/2*(2 + sqrt(6)).
Product {n >= 2} (1 - 1/a(n)) = 1/5*(2 + sqrt(6)). (End)
a(n) = (A054320(n-1) + A072256(n))/2. - Richard R. Forberg, Nov 21 2013
a(2*n - 1) = A046173(n).
E.g.f.: exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, Dec 12 2022
a(n) = Sum_{k = 0..n-1} binomial(n+k, 2*k+1)*8^k = Sum_{k = 0..n-1} (-1)^(n+k+1)* binomial(n+k, 2*k+1)*12^k. - Peter Bala, Jul 18 2023

A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, Jun 01 2005

Matrix inverse of A124645.
Let L(n,x) = Sum_{k=0..n} T(n,k)*x^(n-k) and Pi=3.14...:
L(n,x) = Product_{k=1..n} (x - 2*cos((2*k-1)*Pi/(2*n+1)));
Sum_{k=0..n} T(n,k) = L(n,1) = A010892(n+1);
Sum_{k=0..n} abs(T(n,k)) = A000045(n+2);
abs(T(n,k)) = A065941(n,k), T(n,k) = A065941(n,k)*A087960(k);
T(2*n,k) + T(2*n+1,k+1) = 0 for 0 <= k <= 2*n;
T(n,0) = A000012(n) = 1; T(n,1) = -1 for n > 0;
T(n,2) = -(n-1) for n > 1; T(n,3) = A000027(n)=n for n > 2;
T(n,4) = A000217(n-3) for n > 3; T(n,5) = -A000217(n-4) for n > 4;
T(n,6) = -A000292(n-5) for n > 5; T(n,7) = A000292(n-6) for n > 6;
T(n,n-3) = A058187(n-3)*(-1)^floor(n/2) for n > 2;
T(n,n-2) = A008805(n-2)*(-1)^floor((n+1)/2) for n > 1;
T(n,n-1) = A008619(n-1)*(-1)^floor(n/2) for n > 0;
T(n,n) = L(n,0) = (-1)^floor((n+1)/2);
L(n,1) = A010892(n+1); L(n,-1) = A061347(n+2);
L(n,2) = 1; L(n,-2) = A005408(n)*(-1)^n;
L(n,3) = A001519(n); L(n,-3) = A002878(n)*(-1)^n;
L(n,4) = A001835(n+1); L(n,-4) = A001834(n)*(-1)^n;
L(n,5) = A004253(n); L(n,-5) = A030221(n)*(-1)^n;
L(n,6) = A001653(n); L(n,-6) = A002315(n)*(-1)^n;
L(n,7) = A049685(n); L(n,-7) = A033890(n)*(-1)^n;
L(n,8) = A070997(n); L(n,-8) = A057080(n)*(-1)^n;
L(n,9) = A070998(n); L(n,-9) = A057081(n)*(-1)^n;
L(n,10) = A072256(n+1); L(n,-10) = A054320(n)*(-1)^n;
L(n,11) = A078922(n+1); L(n,-11) = A097783(n)*(-1)^n;
L(n,12) = A077417(n); L(n,-12) = A077416(n)*(-1)^n;
L(n,13) = A085260(n);
L(n,14) = A001570(n); L(n,-14) = A028230(n)*(-1)^n;
L(n,n) = A108366(n); L(n,-n) = A108367(n).
Row n of the matrix inverse (A124645) has g.f.: x^floor(n/2)*(1-x)^(n-floor(n/2)). - Paul D. Hanna, Jun 12 2005
From L. Edson Jeffery, Mar 12 2011: (Start)
Conjecture: Let N=2*n+1, with n > 2. Then T(n,k) (0 <= k <= n) gives the k-th coefficient in the characteristic function p_N(x)=0, of degree n in x, for the n X n tridiagonal unit-primitive matrix G_N (see [Jeffery]) of the form
  G_N=A_{N,1}=
  (0 1 0 ... 0)
  (1 0 1 0 ... 0)
  (0 1 0 1 0 ... 0)
  ...
  (0 ... 0 1 0 1)
  (0 ... 0 1 1),
with solutions phi_j = 2*cos((2*j-1)*Pi/N), j=1,2,...,n. For example, for n=3,
  G_7=A_{7,1}=
  (0 1 0)
  (1 0 1)
  (0 1 1).
We have {T(3,k)}=(1,-1,-2,1), while the characteristic function of G_7 is p(x) = x^3-x^2-2*x+1 = 0, with solutions phi_j = 2*cos((2*j-1)*Pi/7), j=1,2,3. (End)
The triangle sums, see A180662 for their definitions, link A108299 with several sequences, see the crossrefs. - Johannes W. Meijer, Aug 08 2011
The roots to the polynomials are chaotic using iterates of the operation (x^2 - 2), with cycle lengths L and initial seeds returning to the same term or (-1)* the seed. Periodic cycle lengths L are shown in A003558 such that for the polynomial represented by row r, the cycle length L is A003558(r-1). The matrices corresponding to the rows as characteristic polynomials are likewise chaotic [cf. Kappraff et al., 2005] with the same cycle lengths but substituting 2*I for the "2" in (x^2 - 2), where I = the Identity matrix. For example, the roots to x^3 - x^2 - 2x + 1 = 0 are 1.801937..., -1.246979..., and 0.445041... With 1.801937... as the initial seed and using (x^2 - 2), we obtain the 3-period trajectory of 8.801937... -> 1.246979... -> -0.445041... (returning to -1.801937...). We note that A003558(2) = 3. The corresponding matrix M is: [0,1,0; 1,0,1; 0,1,1,]. Using seed M with (x^2 - 2*I), we obtain the 3-period with the cycle completed at (-1)*M. - Gary W. Adamson, Feb 07 2012

			Triangle begins:
  1;
  1,  -1;
  1,  -1,  -1;
  1,  -1,  -2,   1;
  1,  -1,  -3,   2,   1;
  1,  -1,  -4,   3,   3,  -1;
  1,  -1,  -5,   4,   6,  -3,  -1;
  1,  -1,  -6,   5,  10,  -6,  -4,   1;
  1,  -1,  -7,   6,  15, -10, -10,   4,   1;
  1,  -1,  -8,   7,  21, -15, -20,  10,   5,  -1;
  1,  -1,  -9,   8,  28, -21, -35,  20,  15,  -5,  -1;
  1,  -1, -10,   9,  36, -28, -56,  35,  35, -15,  -6,   1;
  ...

Friedrich L. Bauer, 'De Moivre und Lagrange: Cosinus eines rationalen Vielfachen von Pi', Informatik Spektrum 28 (Springer, 2005).
Jay Kappraff, S. Jablan, G. Adamson, & R. Sazdonovich: "Golden Fields, Generalized Fibonacci Sequences, & Chaotic Matrices"; FORMA, Vol 19, No 4, (2005).

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Henry W. Gould, A Variant of Pascal's Triangle, Corrections, The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257-271.
L. Edson Jeffery, Unit-primitive matrices.
Ju, Hyeong-Kwan On the sequence generated by a certain type of matrices. Honam Math. J. 39, No. 4, 665-675 (2017).
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
Frank Ruskey and Carla Savage, Gray codes for set partitions and restricted growth tails, Australasian Journal of Combinatorics, Volume 10(1994), pp. 85-96. See Table 1 p. 95.

Cf. A049310, A039961, A124645 (matrix inverse).
Triangle sums (see the comments): A193884 (Kn11), A154955 (Kn21), A087960 (Kn22), A000007 (Kn3), A010892 (Fi1), A134668 (Fi2), A078031 (Ca2), A193669 (Gi1), A001519 (Gi3), A193885 (Ze1), A050935 (Ze3). - Johannes W. Meijer, Aug 08 2011
Cf. A003558.
Cf. A033999, A059841.

a108299 n k = a108299_tabl !! n !! k
a108299_row n = a108299_tabl !! n
a108299_tabl = [1] : iterate (\row ->
   zipWith (+) (zipWith (*) ([0] ++ row) a033999_list)
               (zipWith (*) (row ++ [0]) a059841_list)) [1,-1]
-- Reinhard Zumkeller, May 06 2012

A108299 := proc(n,k): binomial(n-floor((k+1)/2), floor(k/2))*(-1)^floor((k+1)/2) end: seq(seq(A108299 (n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

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]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 16 2013 *)

{T(n,k)=polcoeff(polcoeff((1-x*y)/(1-x+x^2*y^2+x^2*O(x^n)),n,x)+y*O(y^k),k,y)} (Hanna)

T(n,k) = binomial(n-floor((k+1)/2),floor(k/2))*(-1)^floor((k+1)/2).
T(n+1, k) = if sign(T(n, k-1))=sign(T(n, k)) then T(n, k-1)+T(n, k) else -T(n, k-1) for 0 < k < n, T(n, 0) = 1, T(n, n) = (-1)^floor((n+1)/2).
G.f.: A(x, y) = (1 - x*y)/(1 - x + x^2*y^2). - Paul D. Hanna, Jun 12 2005
The generating polynomial (in z) of row n >= 0 is (u^(2*n+1) + v^(2*n+1))/(u + v), where u and v are defined by u^2 + v^2 = 1 and u*v = z. - Emeric Deutsch, Jun 16 2011
From Johannes W. Meijer, Aug 08 2011: (Start)
abs(T(n,k)) = A065941(n,k) = abs(A187660(n,n-k));
T(n,n-k) = A130777(n,k); abs(T(n,n-k)) = A046854(n,k) = abs(A066170(n,k)). (End)

Corrected and edited by Philippe Deléham, Oct 20 2008

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

A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 11, 13, 1, 1, 5, 19, 41, 34, 1, 1, 6, 29, 91, 153, 89, 1, 1, 7, 41, 169, 436, 571, 233, 1, 1, 8, 55, 281, 985, 2089, 2131, 610, 1, 1, 9, 71, 433, 1926, 5741, 10009, 7953, 1597, 1, 1, 10, 89, 631, 3409, 13201, 33461, 47956, 29681

Offset: 1

Ralf Stephan, May 31 2004

Also, values of polynomials with coefficients in A098493 (see Fink et al.). See A098495 for negative k.

Number of dimer tilings of the graph S_{k-1} X P_{2n-2}.

			1,1,1,1,1,1,1,1,1,1,1,1,1,1, ...
1,2,5,13,34,89,233,610,1597, ...
1,3,11,41,153,571,2131,7953, ...
1,4,19,91,436,2089,10009,47956, ...
1,5,29,169,985,5741,33461,195025, ...
1,6,41,281,1926,13201,90481,620166, ...

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.
Elizabeth Wilmer, A note on Stephan's conjecture 87
Elizabeth Wilmer, A note on Stephan's conjecture 87 [cached copy]

Rows are first differences of rows in array A073134.
Rows 2-14 are A000012, A001519, A079935/A001835, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570. Other rows: A007805 (k=18), A075839 (k=20), A077420 (k=34), A078988 (k=66).
Columns include A028387. Diagonals include A094955, A094956. Antidiagonal sums are A094957.

max = 14; row[k_] := Rest[ CoefficientList[ Series[ x*(1-x)/(1-k*x+x^2), {x, 0, max}], x]]; t = Table[ row[k], {k, 2, max+1}]; Flatten[ Table[ t[[k-n+1, n]], {k, 1, max}, {n, 1, k}]] (* Jean-François Alcover, Dec 27 2011 *)

PARI
```
T(k,n)=polcoeff(x*(1-x)/(1-k*x+x*x),n)
```

Recurrence: T(k, 1) = 1, T(k, 2) = k-1, T(k, n) = kT(k, n-1) - T(k, n-2).
For n>3, T(k, n) = [k(k-2) + T(k, n-1)T(k, n-2)] / T(k, n-3).
T(k, n+1) = S(n, k) - S(n-1, k) = U(n, k/2) - U(n-1, k/2), with S, U = Chebyshev polynomials of second kind.
T(k+2, n+1) = Sum[i=0..n, k^(n-i) * C(2n-i, i)] (from comments by Benoit Cloitre).

A077417 Chebyshev T-sequence with Diophantine property.

Original entry on oeis.org

1, 11, 131, 1561, 18601, 221651, 2641211, 31472881, 375033361, 4468927451, 53252096051, 634556225161, 7561422605881, 90102515045411, 1073668757939051, 12793922580223201, 152453402204739361, 1816646903876649131, 21647309444315050211

Offset: 0

Wolfdieter Lang, Nov 29 2002

7*a(n)^2 - 5*b(n)^2 = 2 with companion sequence b(n) = A077416(n), n>=0.
a(n) = L(n,12), where L is defined as in A108299; see also A077416 for L(n,-12). - Reinhard Zumkeller, Jun 01 2005
[a(n), A004191(n)] = the 2 X 2 matrix [1,10; 1,11]^(n+1) * [1,0]. - Gary W. Adamson, Mar 19 2008
Hankel transform of A174227. - Paul Barry, Mar 12 2010
Alternate denominators of the continued fraction convergents to sqrt(35), see A041059. - James R. Buddenhagen, May 20 2010
For positive n, a(n) equals the permanent of the (2n)X(2n) tridiagonal matrix with sqrt(10)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Positive values of x (or y) satisfying x^2 - 12xy + y^2 + 10 = 0. - Colin Barker, Feb 09 2014
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jun 29 2019

			G.f. = 1 + 11*x + 131*x^2 + 1561*x^3 + 18601*x^4 221651*x^5 + 2641211*x^6 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
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, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
Index entries for linear recurrences with constant coefficients, signature (12,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A072256(n) with companion A054320(n-1), n>=1.
Row 12 of array A094954.
Cf. A004191.
Cf. A041059. [James R. Buddenhagen, May 20 2010]
Cf. similar sequences listed in A238379.

I:=[1,11]; [n le 2 select I[n] else 12*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014

CoefficientList[Series[(1 - x)/(1 - 12 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
LinearRecurrence[{12,-1},{1,11},30] (* Harvey P. Dale, Apr 09 2015 *)
a[ n_] := With[{x = Sqrt[7/2]}, ChebyshevT[2 n + 1, x]/x] // Expand; (* Michael Somos, Jun 29 2019 *)

my(x='x+O('x^30)); Vec((1-x)/(1-12*x+x^2)) \\ G. C. Greubel, Jan 18 2018

{a(n) = my(x = quadgen(56)/2); simplify(polchebyshev(2*n + 1, 1, x)/x)}; /* Michael Somos, Jun 29 2019 */

a(n) = 12*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = S(n, 12) - S(n-1, 12) = T(2*n+1, sqrt(14)/2)/(sqrt(14)/2) with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 12)=A004191(n).
G.f.: (1-x)/(1-12*x+x^2).
a(n) = (ap^(2*n+1) + am^(2*n+1))/sqrt(14) with ap := (sqrt(7)+sqrt(5))/sqrt(2) and am := (sqrt(7)-sqrt(5))/sqrt(2).
a(n) = sqrt((5*A077416(n)^2 + 2)/7).
a(n)*a(n+3) = 120 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
E.g.f.: exp(6*x)*(7*cosh(sqrt(35)*x) + sqrt(35)*sinh(sqrt(35)*x))/7. - Stefano Spezia, Aug 29 2025

More terms from Vincenzo Librandi, Feb 10 2014

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

Original entry on oeis.org

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 equations, three sequences a(n), b(n), c(n) and a constant A:
    A    * c(n)+1 = a(n)^2,
   (A+1) * c(n)+1 = b(n)^2, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
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 A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A077288 First member of the Diophantine pair (m,k) that satisfies 6(m^2 + m) = k^2 + k: a(n) = m.

Original entry on oeis.org

0, 1, 3, 14, 34, 143, 341, 1420, 3380, 14061, 33463, 139194, 331254, 1377883, 3279081, 13639640, 32459560, 135018521, 321316523, 1336545574, 3180705674, 13230437223, 31485740221, 130967826660, 311676696540, 1296447829381, 3085281225183, 12833510467154

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

Also nonnegative m such that 24*m^2 + 24*m + 1 is a square. - Gerald McGarvey, Apr 02 2005

			a(3) = 2*3 - 1 + 9 = 14, a(4) = 2*14 - 3 + 9 = 34, etc.
G.f. = x + 3*x^2 + 14*x^3 + 34*x^4 + 143*x^5 + 341*x^6 + 1420*x^7 + 3380*x^8 + ... - _Michael Somos_, Jul 15 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 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 linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

The k values are in A077291
Cf. A077289, A077290, A077291.
Cf. A053141.

m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))); // G. C. Greubel, Jul 15 2018

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 10*a(n - 2) - a(n - 4) + 4}, a(n), remember); map(f, [$ (0 .. 100)]); - Vladimir Pletser, Jul 24 2020

CoefficientList[Series[x*(1 + x)^2/((1 - x)*(1 - 10 x^2 + x^4)), {x, 0, 40}],x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 3, 14, 34}, 50] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ x (1 + x)^2 / ((1 - x) (1 - 10 x^2 + x^4)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

my(x='x+O('x^30)); concat([0], Vec(x*(1+x)^2/((1-x)*(1-10*x^2+x^4)))) \\ G. C. Greubel, Jul 15 2018

Let b(n) be A072256. Then a(2*n+2) = 2*a(2*n+1) - a(2*n) + b(n+1), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + b(n+2), with a(0)=0, a(1)=1.
G.f.: x*(1+x)^2/((1-x)*(1-10*x^2+x^4)).
a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018
a(n) = 10*a(n-2) - a(n-4) + 4, n > 4. - Vladimir Pletser, Feb 29 2020
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
2*a(n) + 1 = A080806(n+1). - R. J. Mathar, Oct 01 2021

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, -1, -1, 1, -2, 1, 1, 1, -3, 5, -1, 0, 1, -4, 11, -13, 1, -1, 1, -5, 19, -41, 34, -1, 1, 1, -6, 29, -91, 153, -89, 1, 0, 1, -7, 41, -169, 436, -571, 233, -1, -1, 1, -8, 55, -281, 985, -2089, 2131, -610, 1, 1, 1, -9, 71, -433, 1926, -5741, 10009, -7953, 1597, -1, 0

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Feb 01 2018

This array is used to compute A269252: A269252(n) = least k such that |A(n,k)| is a prime, or -1 if no such k exists.
For detailed theory, see [Hone].
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
1   0  -1     1     0      -1       1         0        -1           1
1  -1   1    -1     1      -1       1        -1         1          -1
1  -2   5   -13    34     -89     233      -610      1597       -4181
1  -3  11   -41   153    -571    2131     -7953     29681     -110771
1  -4  19   -91   436   -2089   10009    -47956    229771    -1100899
1  -5  29  -169   985   -5741   33461   -195025   1136689    -6625109
1  -6  41  -281  1926  -13201   90481   -620166   4250681   -29134601
1  -7  55  -433  3409  -26839  211303  -1663585  13097377  -103115431
1  -8  71  -631  5608  -49841  442961  -3936808  34988311  -310957991
1  -9  89  -881  8721  -86329  854569  -8459361  83739041  -828931049

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269251, A269252, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Cf. A094954 (unsigned version of this array, but missing the first row).
Cf. Rows: A057078, A033999, A099496, A079935 (or A001835), A004253, A001653, A049685, A070997, A070998, A138288 (or A072256), ...
Cf. Columns: A000012, A001477 (A000027), A110331 (A165900), A123972, A192398, ...

(* Array: *)
Grid[Table[LinearRecurrence[{-n, -1}, {1, 1 - n}, 10], {n, 10}]]
(*Array antidiagonals flattened (gives this sequence):*)
A299045[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] (-n)^(k - j), {j, 0, k}]; Flatten[Table[A299045[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*(-n)^(k-j))}; /* Michael Somos, Jun 19 2023 */

G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.

A(n, k) = B(-n, k) where B = A294099. - Michael Somos, Jun 19 2023

A165253 Triangle T(n,k), read by rows given by [1,0,1,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 10, 15, 7, 1, 0, 1, 15, 35, 28, 9, 1, 0, 1, 21, 70, 84, 45, 11, 1, 0, 1, 28, 126, 210, 165, 66, 13, 1, 0, 1, 36, 210, 462, 495, 286, 91, 15, 1, 0, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 0, 1, 55, 495, 1716, 3003, 3003, 1820, 680

Offset: 0

Philippe Deléham, Sep 10 2009

Mirror image of triangle in A121314.

			Triangle begins:
  1;
  1,    0;
  1,    1,    0;
  1,    3,    1,    0;
  1,    6,    5,    1,    0;
  1,   10,   15,    7,    1,    0;
  1,   15,   35,   28,    9,    1,    0;
  1,   21,   70,   84,   45,   11,    1,    0;
  1,   28,  126,  210,  165,   66,   13,    1,    0;
  1,   36,  210,  462,  495,  286,   91,   15,    1,    0,
  1,   45,  330,  924, 1287, 1001,  455,  120,   17,    1,    0;

Indranil Ghosh, Rows 0..125 of triangle, flattened

Cf. A054142, A085478, A121314.

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{1, 0, 1}, Table[0, {m}]], Join[{0, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

T(0,0)=1, T(n,k) = binomial(n-1+k,2k) for n >= 1.
Sum {k=0..n} T(n,k)*x^k = A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A001519(n), A047849(n), A165310(n), A165311(n), A165312(n), A165314(n), A165322(n), A165323(n), A165324(n) for x= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 26 2009
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=T(1,0)=1, T(1,1)=0. - Philippe Deléham, Feb 18 2012
G.f.: (1-x-y*x)/((1-x)^2-y*x). - Philippe Deléham, Feb 19 2012

A077289 Triangular numbers that are 1/6 of another triangular number.

Original entry on oeis.org

0, 1, 6, 105, 595, 10296, 58311, 1008910, 5713890, 98862891, 559902916, 9687554415, 54864771885, 949281469786, 5376187741821, 93019896484620, 526811533926580, 9115000574022981, 51622154137063026, 893177036357767525, 5058444293898249975, 87522234562487194476

Offset: 0

Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002

The triangular numbers multiplied by 6 are in A077290.

			b(3)=14 so a(3) = 14*15/2 = 105, etc.

Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Index entries for linear recurrences with constant coefficients, signature (1,98,-98,-1,1).

Cf. A077288, A077290, A077291.

f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 98*a(n-2)-a(n-4)+7}, a(n), remember); map(f, [`$`(0 .. 1000)])[]; # Vladimir Pletser, Feb 19 2021

tr6Q[n_]:= IntegerQ[1/2 (Sqrt[1+48n]-1)]; Select[Accumulate[ Range[0,1380000]],tr6Q]  (* Harvey P. Dale, Apr 21 2011 *)

T(n)=n*(n+1)\2;
istriang(n)=issquare(8*n+1);
for(n=0, 10^10, t=T(n); if ( t%6==0 && istriang(t\6), print1(t\6, ", ") ) );
\\ Joerg Arndt, Jul 03 2013

concat(0, Vec(-x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015

Let b(n) be A077288. Then a(n)=b(n)*(b(n)+1)/2.
G.f.: -x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)). - Colin Barker, Jul 02 2013
a(n) = 98*a(n-2) - a(n-4) + 7. - Vladimir Pletser, Feb 19 2021
96*a(n) = 9*A072256(n+1) -2*(-1)^n*A054320(n) -7. - R. J. Mathar, Oct 01 2021

cp's OEIS Frontend

A004189 a(n) = 10*a(n-1) - a(n-2); a(0) = 0, a(1) = 1.

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A108299 Triangle read by rows, 0 <= k <= n: T(n,k) = binomial(n-[(k+1)/2],[k/2])*(-1)^[(k+1)/2].

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

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A094954 Array T(k,n) read by antidiagonals. G.f.: x(1-x)/(1-kx+x^2), k>1.

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A077417 Chebyshev T-sequence with Diophantine property.

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A157014 Expansion of x*(1-x)/(1 - 22*x + x^2).

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.