A069835 -id:A069835 - cp's OEIS frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). - Gary W. Adamson, Jul 25 2016
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
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.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Mathematics Stack Exchange.
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
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A081577 Pascal-(1,2,1) array read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 7, 7, 1, 1, 10, 22, 10, 1, 1, 13, 46, 46, 13, 1, 1, 16, 79, 136, 79, 16, 1, 1, 19, 121, 307, 307, 121, 19, 1, 1, 22, 172, 586, 886, 586, 172, 22, 1, 1, 25, 232, 1000, 2086, 2086, 1000, 232, 25, 1, 1, 28, 301, 1576, 4258, 5944, 4258, 1576, 301, 28, 1

Offset: 0

Paul Barry, Mar 23 2003

One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016777, A038764, A081583, A081584. Coefficients of the row polynomials in the Newton basis are given by A013610.
As a number triangle, this is the Riordan array (1/(1-x), x(1+2x)/(1-x)). It has row sums A002605 and diagonal sums A077947. - Paul Barry, Jan 24 2005
All entries are == 1 mod 3. - Roger L. Bagula, Oct 04 2008
Row sums are A002605. - Roger L. Bagula, Dec 09 2008
As a number triangle T, T(2n,n)=A069835(n). - Philippe Deléham, Jan 10 2014

			Square array begins as:
  1,  1,  1,   1,   1, ... A000012;
  1,  4,  7,  10,  13, ... A016777;
  1,  7, 22,  46,  79, ... A038764;
  1, 10, 46, 136, 307, ... A081583;
  1, 13, 79, 307, 886, ... A081584;
From _Roger L. Bagula_, Dec 09 2008: (Start)
As a triangle this begins:
  1;
  1,  1;
  1,  4,   1;
  1,  7,   7,    1;
  1, 10,  22,   10,    1;
  1, 13,  46,   46,   13,    1;
  1, 16,  79,  136,   79,   16,    1;
  1, 19, 121,  307,  307,  121,   19,    1;
  1, 22, 172,  586,  886,  586,  172,   22,   1;
  1, 25, 232, 1000, 2086, 2086, 1000,  232,  25,  1;
  1, 28, 301, 1576, 4258, 5944, 4258, 1576, 301, 28, 1; (End)

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, J. Integer Sequ., Vol. 9 (2006), Article 06.2.4.
Paul Barry, The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
Shanghua Zheng, Li Guo, and Huizhen Qiu, Extended Rota-Baxter algebras, diagonally colored Delannoy paths and Hopf algebras, arXiv:2401.11363 [math.RA], 2024. See pp. 44-45.

Cf. A002605, A081578, A081579, A081580.

Cf. Pascal-(1,a,1) array: A123562 (a=-3), A098593 (=-2), A000012 (a=-1), A007318 (a=0), A008288 (a=1), A081577(a=2), A081578 (a=3), A081579 (a=4), A081580 (a=5), A081581 (a=6), A081582 (a=7), A143683(a=8). [From Roger L. Bagula, Dec 09 2008], Philippe Deléham, Jan 10 2014, Mar 16 2014.

a081577 n k = a081577_tabl !! n !! k
a081577_row n = a081577_tabl !! n
a081577_tabl = map fst $ iterate
    (\(us, vs) -> (vs, zipWith (+) (map (* 2) ([0] ++ us ++ [0])) $
                       zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
-- Reinhard Zumkeller, Mar 16 2014

A081577:= func< n,k | (&+[Binomial(k,j)*Binomial(n-j,k)*2^j: j in [0..n-k]]) >;
[A081577(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 25 2021

a[0]={1}; a[1]={1, 1}; a[n_]:= a[n]= 2*Join[{0}, a[n-2], {0}] + Join[{0}, a[n-1]] + Join[a[n-1], {0}]; Table[a[n], {n,0,10}]//Flatten (* Roger L. Bagula, Dec 09 2008 *)
Table[Hypergeometric2F1[-k, k-n, 1, 3], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, May 24 2013 *)

flatten([[hypergeometric([-k, k-n], [1], 3).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 25 2021

Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 2*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+2*x)^k/(1-x)^(k+1).
G.f.: 1/(1-x-y-2*x*y). - Ralf Stephan, Apr 28 2004
T(n,k) = Sum_{j=0..n} binomial(k,j-k)*binomial(n+k-j,k)*2^(j-k). - Paul Barry, Oct 23 2006
a(n) = 2*{0, a(n-2), 0} + {0, a(n-1)} + {a(n-1), 0}. - Roger L. Bagula, Dec 09 2008
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 3). - Jean-François Alcover, May 24 2013
The e.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(3*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 6*x + 9*x^2/2) = 1 + 7*x + 22*x^2/2! + 46*x^3/3! + 79*x^4/4! + 121*x^5/5! + .... - Peter Bala, Mar 05 2017
Sum_{k=0..n} T(n,k) = A002605(n). - G. C. Greubel, May 25 2021

A307883 Square array read by descending antidiagonals: T(n, k) where column k is the expansion of 1/sqrt(1 - 2(k+1)x + ((k-1)*x)^2).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 13, 20, 1, 1, 5, 22, 63, 70, 1, 1, 6, 33, 136, 321, 252, 1, 1, 7, 46, 245, 886, 1683, 924, 1, 1, 8, 61, 396, 1921, 5944, 8989, 3432, 1, 1, 9, 78, 595, 3606, 15525, 40636, 48639, 12870, 1, 1, 10, 97, 848, 6145, 33876, 127905, 281488, 265729, 48620, 1

Offset: 0

Seiichi Manyama, May 02 2019

Column k is the diagonal of the rational function 1 / ((1-x)*(1-y) - k*x*y). - Seiichi Manyama, Jul 11 2020

More generally, column k is the diagonal of the rational function r / ((1-r*x)*(1-r*y) + r-1 - (k+r-1)*r*x*y) for any nonzero real number r. - Seiichi Manyama, Jul 22 2020

			Square array begins:
  1,   1,    1,     1,      1,      1,      1, ...
  1,   2,    3,     4,      5,      6,      7, ...
  1,   6,   13,    22,     33,     46,     61, ...
  1,  20,   63,   136,    245,    396,    595, ...
  1,  70,  321,   886,   1921,   3606,   6145, ...
  1, 252, 1683,  5944,  15525,  33876,  65527, ...
  1, 924, 8989, 40636, 127905, 324556, 712909, ...
Seen as a triangle T(n, k):
  [0] 1;
  [1] 1, 1;
  [2] 1, 2,  1;
  [3] 1, 3,  6,   1;
  [4] 1, 4, 13,  20,    1;
  [5] 1, 5, 22,  63,   70,     1;
  [6] 1, 6, 33, 136,  321,   252,     1;
  [7] 1, 7, 46, 245,  886,  1683,   924,     1;
  [8] 1, 8, 61, 396, 1921,  5944,  8989,  3432,     1;
  [9] 1, 9, 78, 595, 3606, 15525, 40636, 48639, 12870, 1;

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..6 give A000012, A000984, A001850, A069835, A084771, A084772, A098659.
Main diagonal gives A187021.
T(n,n+1) gives A335309.
Cf. A307855, A307884, A307910.

# Seen as a triangle read by rows:
T := (n, k) -> simplify(hypergeom([-k, -k], [1], n - k)):
seq(lprint(seq(T(n, k), k = 0..n)), n = 0..9);  # Peter Luschny, May 13 2024

T[n_, k_] := Sum[If[k == j == 0, 1, k^j] * Binomial[n, j]^2, {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 13 2021 *)
(* Seen as a triangle read by rows: *)
T[n_, k_] := HypergeometricPFQ[{-k, -k}, {1}, n - k];
Flatten[Table[T[n, k], {n, 0, 10}, {k, 0, n}]] (* Detlef Meya, May 13 2024 *)

T(n,k) is the coefficient of x^n in the expansion of (1 + (k+1)*x + k*x^2)^n.
T(n,k) = Sum_{j=0..n} k^j * binomial(n,j)^2.
T(n,k) = Sum_{j=0..n} (k-1)^(n-j) * binomial(n,j) * binomial(n+j,j).
n * T(n,k) = (k+1) * (2*n-1) * T(n-1,k) - (k-1)^2 * (n-1) * T(n-2,k).
T(n,k) = hypergeom([-k, -k], [1], n - k), (triangular form). - Detlef Meya, May 13 2024

A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 208, 1552, 11584, 86464, 645376, 4817152, 35955712, 268377088, 2003193856, 14952042496, 111603564544, 833020346368, 6217748512768, 46409906716672, 346408259682304, 2585626450591744, 19299378566004736

Offset: 0

Philippe Deléham, Feb 29 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-4).

Cf. A001075.

Sum_{k>=0} A086645(n,k)*m^k for m = 0, 1, 2, 4 gives A000007, A081294, A001541, A083884.

a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-4*a[n-2]; od; a; # G. C. Greubel, Feb 03 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-4*x)/(1-8*x+4*x^2) )); // G. C. Greubel, Feb 03 2019

LinearRecurrence[{8,-4}, {1,4}, 20] (* G. C. Greubel, Feb 03 2019 *)

my(x='x+O('x^20)); Vec((1-4*x)/(1-8*x+4*x^2)) \\ G. C. Greubel, Feb 03 2019

[lucas_number2(n,8,4)/2 for n in range(0,21)] # Zerinvary Lajos, Jul 08 2008

a(n) = Sum_{k>=0} binomial(2*n, 2*k)*3^k = Sum_{k>=0} A086645(n, k)*3^k.
a(n) = 2^n*A001075(n).
G.f.: (1-4*x)/(1-8*x+4*x^2). - Philippe Deléham, Sep 07 2009
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k-4)/(x*(3*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*12^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 12*x^2))^n.
G.f.: A(x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

Corrected by T. D. Noe, Nov 07 2006

A084603 Coefficients of 1/sqrt(1 - 2x - 11x^2); also, a(n) is the central coefficient of (1 + x + 3*x^2)^n.

Original entry on oeis.org

1, 1, 7, 19, 91, 331, 1441, 5797, 24739, 103411, 441397, 1876777, 8047909, 34533253, 148803487, 642228139, 2778852979, 12043194163, 52286516821, 227323871929, 989675651041, 4313712072241, 18822940658947, 82215245701519

Offset: 0

Paul D. Hanna, Jun 01 2003

5th binomial transform of 2^n*LegendreP(n,-2) (signed version of A069835). - Paul Barry, Sep 03 2004
Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the U (or D) steps come in three colors. - N-E. Fahssi, Feb 05 2008
Diagonal of the rational function 1 / (1 - 3*x^2 - y^2 - x*y). - Ilya Gutkovskiy, Apr 22 2025

Seiichi Manyama, Table of n, a(n) for n = 0..1542 (terms 0..200 from Vincenzo Librandi)
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A002426, A084600-A084602, A084604-A084615, A025237.

Table[Sum[Binomial[n-k,k]*Binomial[n,k]*3^k,{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

for(n=0,30,t=polcoeff((1+x+3*x^2)^n,n,x); print1(t","))

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*binomial(n, k)3^k. - Paul Barry, Aug 26 2004
Binomial transform is A084609. Hankel transform is 6^n*3^C(n,2). - Paul Barry, Sep 16 2006
a(n) = (1/Pi)*Integral_{x=1-2*sqrt(3)..1+2*sqrt(3)} x^n/sqrt(-x^2 + 2*x + 11). - Paul Barry, Sep 16 2006
From Paul Barry, Sep 16 2006: (Start)
a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k,k)*3^k;
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*3^k. (End)
From N-E. Fahssi, Feb 05 2008: (Start)
a(n) is also the central coefficient of (3+x+x^2)^n;
a(n) = Sum_{k=0..n} 2^(n-k)*C(n,k)*T(k,n), where T(k,n) is the triangle of trinomial coefficients = coefficient of x^n of (1+x+x^2)^k: A027907. (End)
D-finite with recurrence: a(n+2) = ( (2*n+3)*a(n+1) + 11*(n+1)*a(n) )/(n+2); a(0)=a(1)=1. - Sergei N. Gladkovskii, Aug 01 2012
a(n) ~ sqrt(18+3*sqrt(3))*(1+2*sqrt(3))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
E.g.f.: exp(x)*BesselI(0, 2*sqrt(3)*x). - Paul D. Hanna, Nov 09 2014, after Vladeta Jovovic in A084601
From Peter Bala, Jan 07 2022: (Start)
O.g.f. A(x) = 1 + x*d/dx(log(B(x))), where B(x) = (1 - x - sqrt(1 - 2*x - 11*x^2))/(6*x^2) is the o.g.f. of A025237.
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. (End)

A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2k+1) = A152842(2n,2(n-k)) and T(n,2k) = A152842(2n,2(n-k)+1) for n >= k > 0.

Original entry on oeis.org

0, 1, 0, 3, 4, 1, 0, 9, 24, 22, 8, 1, 0, 27, 108, 171, 136, 57, 12, 1, 0, 81, 432, 972, 1200, 886, 400, 108, 16, 1, 0, 243, 1620, 4725, 7920, 8430, 5944, 2810, 880, 175, 20, 1, 0, 729, 5832, 20898, 44280, 61695, 59472, 40636, 19824, 6855, 1640, 258, 24, 1

Offset: 0

Franck Maminirina Ramaharo, Feb 26 2018

T(n,k) is the number of state diagrams having k components of n connected summed trefoil knots.

Row sums gives A001018.

			The triangle T(n, k) begins:
n\k 0     1      2      3       4       5       6      7        8       9
0:  0     1
1:  0     3      4      1
2:  0     9     24     22       8       1
3:  0    27    108    171     136      57      12       1
4:  0    81    432    972    1200     886     400     108      16       1

V. I. Arnold, Topological Invariants of Plane Curves and Caustics, American Math. Soc., 1994.

Michael De Vlieger, Table of n, a(n) for n = 0..10301 (rows 0 <= n <= 100, flattened.)
Ryo Hanaki, Pseudo diagrams of knots, links and spatial graphs, Osaka Journal of Mathematics, Vol. 47 (2010), 863-883.
Louis H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
Carolina Medina, Jorge Ramírez-Alfonsín and Gelasio Salazar, On the number of unknot diagrams, arXiv:1710.06470 [math.CO], 2017.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Row 2: row 5 of A158454.
Row 3: row 2 of A220665.
Row 4: row 5 of A219234.

row[n_] := CoefficientList[x*(x^2 + 4*x + 3)^n, x]; Array[row, 7, 0] // Flatten (* Jean-François Alcover, Mar 16 2018 *)

g(x, y) := taylor(x/(1 - y*(x^2 + 4*x + 3)), y, 0, 10)$
a : makelist(ratcoef(g(x, y), y, n), n, 0, 10)$
T : []$
for i:1 thru 11 do
  T : append(T, makelist(ratcoef(a[i], x, n), n, 0, 2*i - 1))$
T;

T(n, k) = polcoeff(x*(x^2 + 4*x + 3)^n, k);
tabf(nn) = for (n=0, nn, for (k=0, 2*n+1, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 03 2018

T(n,k) = coefficients of x*(x^2 + 4*x + 3)^n.
T(n,k) = T(n-1,k-2) + 4*T(n-1,k-1) + 3*T(n-1,k), with T(n,0) = 0, T(n,1) = 3^n and  T(n,2) =  4*n*3^(n-1).
T(n,n+k+1) = A152842(2*n,n+k) and T(n,n-k) = A152842(2*n,n+k+1), for n >= k >= 0.
T(n,1) =  A000244(n).
T(n,2) =  A120908(n).
T(n,n+1) = A069835(n).
T(n,2*n-1) = A139272(n).
T(n,2*n) = A008586(n).
T(n,2*n-2) = A140138(4*n) = A185872(2n,2) for n >= 1.
G.f.: x/(1 - y*(x^2 + 4*x + 3)).

Typo in row 6 corrected by Jean-François Alcover, Mar 16 2018

cp's OEIS Frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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A081577 Pascal-(1,2,1) array read by antidiagonals.

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A307883 Square array read by descending antidiagonals: T(n, k) where column k is the expansion of 1/sqrt(1 - 2*(k+1)*x + ((k-1)*x)^2).

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A090965 a(n) = 8*a(n-1) - 4*a(n-2), where a(0) = 1, a(1) = 4.

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A084603 Coefficients of 1/sqrt(1 - 2*x - 11*x^2); also, a(n) is the central coefficient of (1 + x + 3*x^2)^n.

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A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2*k+1) = A152842(2*n,2*(n-k)) and T(n,2*k) = A152842(2*n,2*(n-k)+1) for n >= k > 0.

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A307883 Square array read by descending antidiagonals: T(n, k) where column k is the expansion of 1/sqrt(1 - 2(k+1)x + ((k-1)*x)^2).

A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

A084603 Coefficients of 1/sqrt(1 - 2x - 11x^2); also, a(n) is the central coefficient of (1 + x + 3*x^2)^n.

A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2k+1) = A152842(2n,2(n-k)) and T(n,2k) = A152842(2n,2(n-k)+1) for n >= k > 0.

A331515 Expansion of 1/(1 - 8x + 4x^2)^(3/2).

A331792 Expansion of ((1 - 4x)/sqrt(1 - 8x + 4x^2) - 1)/(6x^2).