A057076 -id:A057076 - cp's OEIS frontend

A059100 a(n) = n^2 + 2.

2, 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403, 2502, 2603

Offset: 0

Henry Bottomley, Feb 13 2001

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002
Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008
1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013
a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013
This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014
Also the number of maximum matchings in the n-gear graph. - Eric W. Weisstein, Dec 31 2017
Also the Wiener index of the n-dipyramidal graph. - Eric W. Weisstein, Jun 14 2018
Numbers of the form n^2+2 have no factors that are congruent to 7 (mod 8). - Gordon E. Michaels, Sep 12 2019
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {n, 2n}]. - Magus K. Chu, Sep 10 2022

			For n = 2, a(2) = 6 since there are 6 election results in a 4-candidate, 2-voter election that have candidates c1 and c2 tied. Letting <i,j> denote voter 1 voting for candidate i and voter 2 voting for candidate j, the six election results are <1,2>, <2,1>, <3,3>, <3,4>, <4,3>, and <4,4>. - _Dennis P. Walsh_, May 08 2013

Harry J. Smith, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 52, 56.
Hesam Dashti, A New Upper Bound on the Length of Shortest Permutation Strings; An Algorithm for Generating Permutation Strings, arXiv:1009.5053 [math.CO], 2010. - _Jonathan Vos Post_, Sep 28 2010
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Alexander Soifer, Coffee Hour and the Conway-Soifer Cover-Up, In: How Does One Cut a Triangle? (2009), pp. 147-156. See also here
Dennis P. Walsh, Notes on a tied election.
Eric Weisstein's World of Mathematics, Dipyramidal Graph.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Matching.
Eric Weisstein's World of Mathematics, Maximum Independent Edge Set.
Eric Weisstein's World of Mathematics, Near-Square Prime.
Eric Weisstein's World of Mathematics, Pappus Chain.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000032, A000196, A000290, A001541, A002522, A005248, A008865, A056105, A056109, A056899, A057076, A069987, A081908, A114964, A156798, A166464.
Apart from initial terms, same as A010000.
2nd row/column of A295707.

a059100 = (+ 2) . (^ 2)
a059100_list = scanl (+) (2) [1, 3 ..]
-- Reinhard Zumkeller, Feb 09 2015

with(combinat, fibonacci):seq(fibonacci(3, i)+1, i=0..49); # Zerinvary Lajos, Mar 20 2008

Table[n^2 + 2, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)
LinearRecurrence[{3, -3, 1}, {2, 3, 6}, 50] (* Vincenzo Librandi, Feb 15 2012 *)
Range[0, 20]^2 + 2 (* Eric W. Weisstein, Dec 31 2017 *)
CoefficientList[Series[(-2 + 3 x - 3 x^2)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Dec 31 2017 *)

a(n) = { n^2+2 } \\ Harry J. Smith, Jun 24 2009

[lucas_number1(3,n,-2) for n in range(0, 50)] # Zerinvary Lajos, May 16 2009

G.f.: (2 - 3*x + 3*x^2)/(1 - x)^3. - R. J. Mathar, Oct 05 2008
a(n) = ((n - 2)^2 + 2*(n + 1)^2)/3. - Reinhard Zumkeller, Feb 13 2009
a(n) = A000196(A156798(n) - A000290(n)). - Reinhard Zumkeller, Feb 16 2009
a(n) = 2*n + a(n-1) - 1 with a(0) = 2. - Vincenzo Librandi, Aug 07 2010
a(n+3) = (A166464(n+5) - A166464(n))/20. - Paul Curtz, Nov 07 2012
From Paul Curtz, Nov 07 2012: (Start)
a(3*n) mod 9 = 2.
a(3*n+1) = 3*A056109(n).
a(3*n+2) = 3*A056105(n+1). (End)
Sum_{n >= 1} 1/a(n) = Pi * coth(sqrt(2)*Pi) / 2^(3/2) - 1/4. - Vaclav Kotesovec, May 01 2018
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*(csch(sqrt(2)*Pi)))/4.
Product_{n>=0} (1 + 1/a(n)) = sqrt(3/2)*csch(sqrt(2)*Pi)*sinh(sqrt(3)*Pi).
Product_{n>=0} (1 - 1/a(n)) = csch(sqrt(2)*Pi)*sinh(Pi)/sqrt(2). (End)
E.g.f.: exp(x)*(2 + x + x^2). - Stefano Spezia, Aug 07 2024

A006497 a(n) = 3*a(n-1) + a(n-2) with a(0) = 2, a(1) = 3.

Original entry on oeis.org

2, 3, 11, 36, 119, 393, 1298, 4287, 14159, 46764, 154451, 510117, 1684802, 5564523, 18378371, 60699636, 200477279, 662131473, 2186871698, 7222746567, 23855111399, 78788080764, 260219353691, 859446141837, 2838557779202

Offset: 0

N. J. A. Sloane

For more information about this type of recurrence follow the Khovanova link and see A086902 and A054413. - Johannes W. Meijer, Jun 12 2010

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its [sic] Properties, Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92, Sequence L_{3,n}.
A. F. Horadam, Generating identities for generalized Fibonacci and Lucas triples, Fib. Quart., 15 (1977), 289-292.
Haruo Hosoya, What Can Mathematical Chemistry Contribute to the Development of Mathematics?, HYLE--International Journal for Philosophy of Chemistry, Vol. 19, No.1 (2013), pp. 87-105.
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.
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
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (3,1).

Cf. A006190, A100230, A001622, A014176, A080039, A098316.

a006497 n = a006497_list !! n
a006497_list = 2 : 3 : zipWith (+) (map (* 3) $ tail a006497_list) a006497_list
-- Reinhard Zumkeller, Feb 19 2011

[ n eq 1 select 2 else n eq 2 select 3 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 20 2011

a:= n-> (<<0|1>, <1|3>>^n. <<2, 3>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Jan 26 2018

Table[LucasL[n, 3], {n, 0, 30}] (* Zerinvary Lajos, Jul 09 2009 *)
LucasL[Range[0, 30], 3] (* Eric W. Weisstein, Apr 17 2018 *)
LinearRecurrence[{3,1},{2,3},30] (* Harvey P. Dale, Feb 17 2020 *)

my(x='x+O('x^30)); Vec((2-3*x)/(1-3*x-x^2)) \\ G. C. Greubel, Jul 05 2017

apply( {A006497(n)=[2,3]*([0,1;1,3]^n)[,1]}, [0..30]) \\ M. F. Hasler, Mar 06 2020

[lucas_number2(n,3,-1) for n in range(0, 30)] # Zerinvary Lajos, Apr 30 2009

G.f.: (2-3*x)/(1-3*x-x^2). - Simon Plouffe in his 1992 dissertation
From Gary W. Adamson, Jun 15 2003: (Start)
a(n) = ((3 + sqrt(13))/2)^n + ((3 - sqrt(13))/2)^n. See bronze mean (A098316).
A006190(n-2) + A006190(n) = a(n-1).
a(n)^2 - 13*A006190(n)^2 = 4(-1)^n. (End)
From Paul Barry, Nov 15 2003: (Start)
E.g.f.: 2*exp(3*x/2)*cosh(sqrt(13)*x/2).
a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2*k)* (13)^k * 3^(n-2*k).
a(n) = 2*T(n, 3i/2)*(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. (End)
From Hieronymus Fischer, Jan 02 2009: (Start)
fract(((3+sqrt(13))/2)^n) = (1/2)*(1+(-1)^n) - (-1)^n*((3+sqrt(13))/2)^(-n) = (1/2)*(1+(-1)^n) - ((3-sqrt(13))/2)^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
a(n) = round(((3+sqrt(13))/2)^n) for n > 0. (End)
From Johannes W. Meijer, Jun 12 2010: (Start)
a(2n+1) = 3*A097783(n), a(2n) = A057076(n).
a(3n+1) = A041018(5n), a(3n+2) = A041018(5n+3) and a(3n+3) = 2*A041018(5n+4).
Limit_{k -> infinity} a(n+k)/a(k) = (a(n) + A006190(n)*sqrt(13))/2.
Limit_{n -> infinity} a(n)/A006190(n) = sqrt(13).
(End)
a(n) = sqrt(13*(A006190(n))^2 + 4*(-1)^n). - Vladimir Shevelev, Mar 13 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(13*k-9))/((x*(13*k+4)) - 6/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 13*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = Lucas(n,3), Lucas polynomials, L(n,x), evaluated at x=3. - G. C. Greubel, Jun 06 2019
a(n) = 2 * Sum_{k=0..n-2} A168561(n-2,k)*3^k + 3 * Sum_{k=0..n-1} A168561(n-1,k)*3^k, n>0. - R. J. Mathar, Feb 14 2024
a(n) = 2*A006190(n+1) - 3*A006190(n). - R. J. Mathar, Feb 14 2024
a(2*n+1) = 3 + 3*Sum_{k=1..n} a(2*k). - Greg Dresden and Canran Wang, Jul 11 2024
From Peter Bala, Jul 14 2025: (Start)
The following series telescope (Cf. A000032):
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

Definition completed by M. F. Hasler, Mar 06 2020

A004190 Expansion of 1/(1 - 11*x + x^2).

Original entry on oeis.org

1, 11, 120, 1309, 14279, 155760, 1699081, 18534131, 202176360, 2205405829, 24057287759, 262424759520, 2862615066961, 31226340977051, 340627135680600, 3715672151509549, 40531766530924439, 442133759688659280, 4822939590044327641, 52610201730798944771, 573889279448744064840

Offset: 0

N. J. A. Sloane

Chebyshev or generalized Fibonacci sequence.
This is the m=13 member of the m-family of sequences S(n,m-2) = S(2*n+1,sqrt(m))/sqrt(m). The m=4..12 (nonnegative) sequences are A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913 and A004189. The m=1..3 (signed) sequences are A049347, A056594, A010892.
All positive integer solutions of Pell equation b(n)^2 - 117*a(n)^2 = +4 together with b(n+1)=A057076(n+1), n >= 0. - Wolfdieter Lang, Aug 31 2004
For positive n, a(n) equals the permanent of the tridiagonal matrix of order n with 11'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) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,10}. - Milan Janjic, Jan 25 2015

			G.f. = 1 + 11*x + 120*x^2 + 1309*x^3 + 14279*x^4 + 155760*x^5 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..900
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.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
D. Birmajer, J. B. Gil, and M. D. Weiner, On the Enumeration of Restricted Words over a Finite Alphabet, J. Int. Seq. 19 (2016) # 16.1.3, example 12
S. Falcon, Generalized Fibonacci Sequences Generated from a k-Fibonacci Sequence, Journal of Mathematics Research Vol. 4, No. 2; April 2012. - From _N. J. A. Sloane_, Sep 22 2012
R. Flórez, R. A. Higuita, and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
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=11, 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
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eq.(44), lhs, m=13.
Index entries for sequences related to Chebyshev polynomials..
Index entries for linear recurrences with constant coefficients, signature (11,-1).

Cf. A000027, A001090, A001109, A001353, A001906, A004187, A004189, A004254, A006190, A010892, A018913, A049310, A049347, A056594, A057076, A075835, A101950.

with(combinat):seq(fibonacci(2*n+2, 3)/3, n=0..20); # Zerinvary Lajos, Apr 20 2008

Join[{a=1,b=11},Table[c=11*b-a; a=b; b=c, {n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 20 2011 *)
CoefficientList[Series[1/(1-11*x+x^2),{x,0,30}],x] (* Vincenzo Librandi, Jun 13 2012 *)
Table[Fibonacci[2n + 2, 3]/3, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
a[ n_] := ChebyshevU[n, 11/2]; (* Michael Somos, Jul 14 2018 *)

Vec(1/(1-11*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

{a(n) = polchebyshev(n, 2, 11/2)}; /* Michael Somos, Jul 14 2018 */

[lucas_number1(n,11,1) for n in range(1, 20)] # Zerinvary Lajos, Jun 25 2008

Recursion: a(n) = 11*a(n-1) - a(n-2), n >= 1; a(-1)=0, a(0)=1.
a(n) = S(2*n+1, sqrt(13))/sqrt(13) = S(n, 11); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
G.f.: 1/(1 - 11*x + x^2).
a(n) = ((11+3*sqrt(13))^(n+1) - (11-3*sqrt(13))^(n+1))/(2^(n+1)*3*sqrt(13)). - Rolf Pleisch, May 22 2011
a(n) = Sum_{k=0..n} A101950(n,k)*10^k. - Philippe Deléham, Feb 10 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n>=0} (1 + 1/a(n)) = 1/3*(3 + sqrt(13)).
Product_{n>=1} (1 - 1/a(n)) = 3/22*(3 + sqrt(13)). (End)
a(n) = sqrt((A057076(n+1)^2 - 4)/117).
a(n) = A075835(n+1)/3 = A006190(2*n+2)/3. - Vladimir Reshetnikov, Sep 16 2016
a(n) = -a(-2-n) for all n in Z. - Michael Somos, Jul 14 2018
E.g.f.: exp(11*x/2)*(39*cosh(3*sqrt(13)*x/2) + 11*sqrt(13)*sinh(3*sqrt(13)*x/2))/39. - Stefano Spezia, Aug 07 2024

Wolfdieter Lang, Oct 31 2002

A092936 Area of n-th triple of hexagons around a triangle.

Original entry on oeis.org

1, 9, 100, 1089, 11881, 129600, 1413721, 15421329, 168220900, 1835008569, 20016873361, 218350598400, 2381839709041, 25981886201049, 283418908502500, 3091626107326449, 33724468272088441, 367877524885646400

Offset: 1

Peter J. C. Moses, Apr 18 2004

This is the unsigned member r=-9 of the family of Chebyshev sequences S_r(n) defined in A092184: ((-1)^(n+1))*a(n) = S_{-9}(n), n>=0.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using (1/2,1/2)-fences, red half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal), green half-squares, and blue half-squares. A (w,g)-fence is a tile composed of two w X 1 pieces separated by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,3/4)-fences, red (1/4,1/4)-fences, green (1/4,1/4)-fences, and blue (1/4,1/4)-fences. - Michael A. Allen, Dec 30 2022

			a(5) = 10*(1089+100)-9 = 11881. From A006190, a(5) = (3*33+10)^2 = 11881.

Muniru A Asiru, Table of n, a(n) for n = 1..200
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Sergio Falcon, Some series of reciprocal k-Fibonacci numbers, Asian Journal of Mathematics and Computer Research, Vol. 11, No. 3 (2016), pp. 184-191; ResearchGate link.
Index entries for linear recurrences with constant coefficients, signature (10,10,-1).

Cf. A005386, A006190, A057076, A092184, A176019.

a:=[1,9,100];; for n in [4..18] do a[n]:=10*(a[n-1]+a[n-2])-a[n-3]; od; a; # Muniru A Asiru, Feb 20 2018

seq(fibonacci(n,3)^2,n=1..18); # Zerinvary Lajos, Apr 05 2008

CoefficientList[Series[(1-x)*x/(1-10*x-10*x^2+x^3), {x, 0, 20}], x]
(CoefficientList[Series[x/(1-3*x-x^2), {x, 0, 20}], x])^2
Table[Round[((3+Sqrt[13])^n)^2/(13*4^n)], {n, 0, 20}]
LinearRecurrence[{10, 10, -1}, {1, 9, 100}, 18] (* Georg Fischer, Feb 22 2019 *)

a(n) = A006190(n)^2.
a(n) = 10*(a(n-1)+a(n-2)) - a(n-3).
G.f.: (1-x)*x/(1-10*x-10*x^2+x^3).
a(n) = ((3-sqrt(13))^n-(3+sqrt(13))^n)^2/(13*4^n).
a(n) = 2*(T(n, 11/2)-(-1)^n)/13 with twice the Chebyshev polynomials of the first kind evaluated at x = 11/2: 2*T(n, 11/2) = A057076(n) = ((11+3*sqrt(13))^n + (11-3*sqrt(13))^n)/2^n. - Wolfdieter Lang, Oct 18 2004
From Michael A. Allen, Dec 30 2022: (Start)
a(n+1) = 11*a(n) - a(n-1) + 2*(-1)^n.
a(n+1) = (1 + (-1)^n)/2 + 9*Sum_{k=1..n} ( k*a(n+1-k) ). (End)
Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(13))/6 (A176019) (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

A098298 Member r=13 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 13, 144, 1573, 17161, 187200, 2042041, 22275253, 242985744, 2650567933, 28913261521, 315395308800, 3440435135281, 37529391179293, 409382867836944, 4465682155027093, 48713120837461081, 531378647057044800, 5796451996790031721, 63229593317633304133

Offset: 0

Wolfdieter Lang, Oct 18 2004

Colin Barker, Table of n, a(n) for n = 0..950
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (12,-12,1).

Cf. A098296, A098297.

a:=[0,1,13];; for n in [4..30] do a[n]:=12*a[n-1]-12*a[n-2]+ a[n-3]; od; a; # G. C. Greubel, May 24 2019

[n le 2 select n-1 else 11*Self(n-1)- Self(n-2) + 2: n in [1..30]]; // Vincenzo Librandi, Mar 06 2016

LinearRecurrence[{12,-12,1},{0,1,13},30] (* Harvey P. Dale, May 11 2012 *)
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == 11 a[n-1] - a[n-2] + 2}, a, {n, 30}] (* Vincenzo Librandi, Mar 06 2016 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-11*x+x^2)) + O(x^25))) \\ Colin Barker, Mar 06 2016

(x*(1+x)/((1-x)*(1-11*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019

a(n) = 2*(T(n, 11/2) - 1)/9 with twice Chebyshev's polynomials of the first kind evaluated at x=11/2: 2*T(n, 11/2) = A057076(n) = ((11 + sqrt(117))^n + (11 - sqrt(117))^n)/2^n.
a(n) = 11*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 12*a(n-1) - 12*a(n-2) + a(n-3), n >= 3, a(0)=0, a(1)=1, a(2)=13.
G.f.: x*(1+x)/((1-x)*(1-11*x+x^2)) = x*(1+x)/(1-12*x+12*x^2-x^3) (from the Stephan link, see A092184).

A319749 a(n) is the numerator of the Heron sequence with h(0)=3.

Original entry on oeis.org

3, 11, 119, 14159, 200477279, 40191139395243839, 1615327685887921300502934267457919, 2609283532796026943395592527806764363779539144932833602430435810559

Offset: 0

Paul Weisenhorn, Sep 27 2018

The denominator of the Heron sequence is in A319750.
The following relationship holds between the numerator of the Heron sequence and the numerator of the continued fraction A041018(n)/A041019(n) convergent to sqrt(13).
n even: a(n)=A041018((5*2^n-5)/3).
n  odd: a(n)=A041018((5*2^n-1)/3).
More generally, all numbers c(n)=A078370(n)=(2n+1)^2+4 have the same relationship between the numerator of the Heron sequence and the numerator of the continued fraction convergent to 2n+1.
sqrt(c(n)) has the continued fraction 2n+1; n,1,1,n,4n+2.
hn(n)^2-c(n)*hd(n)^2=4 for n>1.
From Peter Bala, Mar 29 2022: (Start)
Applying Heron's method (sometimes called the Babylonian method) to approximate the square root of the function x^2 + 4, starting with a guess equal to x, produces the sequence of rational functions [x, 2*T(1,(x^2+2)/2)/x, 2*T(2,(x^2+2)/2)/( 2*x*T(1,(x^2+2)/2) ), 2*T(4,(x^2+2)/2)/( 4*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2) ), 2*T(8,(x^2+2)/2)/( 8*x*T(1,(x^2+2)/2)*T(2,(x^2+2)/2)*T(4,(x^2+2)/2) ), ...], where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. The present sequence is the case x = 3. Cf. A001566 and A058635 (case x = 1), A081459 and A081460 (essentially the case x = 4). (End)

			A078370(2)=29.
hn(0)=A041046(0)=5; hn(1)=A041046(3)=27; hn(2)=A041046(5)=727;
hn(3)=A041046(13)=528527.

P. Liardet and P. Stambul, Séries d'Engel et fractions continuées, Journal de Théorie des Nombres de Bordeaux 12 (2000), 37-68.
Wikipedia, Engel Expansion
Wikipedia, Methods of computing square roots
Index entries for sequences of form a(n+1)=a(n)^2 + ...

2*T(2^n,x/2) modulo differences of offset: A001566 (x = 3 and x = 7), A003010 (x = 4), A003487 (x = 5), A003423 (x = 6), A346625 (x = 8), A135927 (x = 10), A228933 (x = 18).

Cf. A041018, A041019, A057076, A078370, A081459, A010122, A319750, A041046.

hn[0]:=3:  hd[0]:=1:
for n from 1 to 6 do
hn[n]:=(hn[n-1]^2+13*hd[n-1]^2)/2:
hd[n]:=hn[n-1]*hd[n-1]:
   printf("%5d%40d%40d\n", n, hn[n], hd[n]):
end do:
#alternative program
a := n -> if n = 0 then 3 else simplify( 2*ChebyshevT(2^(n-1), 11/2) ) end if:
seq(a(n), n = 0..7); # Peter Bala, Mar 16 2022

def aupton(nn):
    hn, hd, alst = 3, 1, [3]
    for n in range(nn):
        hn, hd = (hn**2 + 13*hd**2)//2, hn*hd
        alst.append(hn)
    return alst
print(aupton(7)) # Michael S. Branicky, Mar 16 2022

h(n) = hn(n)/hd(n); hn(0)=3; hd(0)=1.
hn(n+1) = (hn(n)^2+13*hd(n)^2)/2.
hd(n+1) = hn(n)*hd(n).
A041018(n) = A010122(n)*A041018(n-1) + A041018(n-2).
A041019(n) = A010122(n)*A041019(n-1) + A041019(n-2).
From Peter Bala, Mar 16 2022: (Start)
a(n) = 2*T(2^(n-1),11/2) for n >= 1, where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = 2*T(2^n, 3*sqrt(-1)/2) for n >= 2.
a(n) = ((11 + 3*sqrt(13))/2)^(2^(n-1)) + ((11 - 3*sqrt(13))/2)^(2^(n-1)) for n >= 1.
a(n+1) = a(n)^2 - 2 for n >= 1.
a(n) = A057076(2^(n-1)) for n >= 1.
Engel expansion of (1/6)*(13 - 3*sqrt(13)); that is, (1/6)*(13 - 3*sqrt(13)) = 1/3 + 1/(3*11) + 1/(3*11*119) + .... (Define L(n) = (1/2)*(n - sqrt(n^2 - 4)) for n >= 2 and show L(n) = 1/n + L(n^2-2)/n. Iterate this relation with n = 11. See also Liardet and Stambul, Section 4.)
sqrt(13) = 6*Product_{n >= 0} (1 - 1/a(n)).
sqrt(13) = (9/5)*Product_{n >= 0} (1 + 2/a(n)). See A001566. (End)

a(6) and a(7) added by Peter Bala, Mar 16 2022

cp's OEIS Frontend

A059100 a(n) = n^2 + 2.

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A006497 a(n) = 3*a(n-1) + a(n-2) with a(0) = 2, a(1) = 3.

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A004190 Expansion of 1/(1 - 11*x + x^2).

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A092936 Area of n-th triple of hexagons around a triangle.

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A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

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A098298 Member r=13 of the family of Chebyshev sequences S_r(n) defined in A092184.

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