A188646 -id:A188646 - cp's OEIS frontend

A001570 Numbers k such that k^2 is centered hexagonal.

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621

Offset: 1

N. J. A. Sloane

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
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).

G. C. Greubel, Table of n, a(n) for n = 1..870 (terms 1..101 from T. D. Noe)
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.
G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes, part C (in French).
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
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.
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics, Hex Number
Wikipedia, Beal's conjecture
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
Cf. A076139, A076140, A102871.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

[((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // G. C. Greubel, Nov 04 2017

A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
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 A041017 *)
a[12, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)

{a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/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, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A188647 Array read by antidiagonals of a(n) = a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2+1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 5, 1, 29, 17, 1, 169, 305, 37, 1, 985, 5473, 1405, 65, 1, 5741, 98209, 53353, 4289, 101, 1, 33461, 1762289, 2026009, 283009, 10301, 145, 1, 195025, 31622993, 76934989, 18674305, 1050601, 21169, 197, 1, 1136689, 567451585, 2921503573, 1232221121, 107151001, 3090529, 39005, 257, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=1 results shown here; y=-1 results are A188646.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/sqrt(n^2+1)) * T_{2*k+1}(sqrt(n^2+1)), with T the Chebyshev polynomial of the first kind. - Seiichi Manyama, Jan 02 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   5,     29,       169,          985, ...
   2 | 1,  17,    305,      5473,        98209, ...
   3 | 1,  37,   1405,     53353,      2026009, ...
   4 | 1,  65,   4289,    283009,     18674305, ...
   5 | 1, 101,  10301,   1050601,    107151001, ...
   6 | 1, 145,  21169,   3090529,    451196065, ...
   7 | 1, 197,  39005,   7722793,   1529074009, ...
   8 | 1, 257,  66305,  17106433,   4413393409, ...
   9 | 1, 325, 105949,  34539049,  11259624025, ...
  10 | 1, 401, 161201,  64802401,  26050404001, ...
  11 | 1, 485, 235709, 114554089,  55673051545, ...
  12 | 1, 577, 333505, 192765313, 111418017409, ...
  13 | 1, 677, 459005, 311204713, 210996336409, ...
  14 | 1, 785, 617009, 484968289, 381184458145, ...
  15 | 1, 901, 812701, 733055401, 661215159001, ...
  ...

Row 1 is A001653, row 2 is A007805, row 3 is A097315, row 4 is A078988, row 5 is A097727, row 6 is A097730, row 7 is A097733, row 8 is A097736, row 9 is A097739, row 10 is A097742, row 11 is A097767, row 12 is A097770, row 13 is A097773.
Column 1 is A053755.
A(n,n) gives A323012.
Cf. A188645, A188646 (f(x, y) as above with y=-1).

A(n,k) = 2 * A188645(n,k) - A(n,k-1).

A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j)*(n^2+1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 02 2019

Edited and extended by Seiichi Manyama, Jan 02 2019

A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

Original entry on oeis.org

1, 33, 1121, 38081, 1293633, 43945441, 1492851361, 50713000833, 1722749176961, 58522759015841, 1988051057361633, 67535213191279681, 2294209197446147521, 77935577499977736033, 2647515425801796877601

Offset: 0

Wolfdieter Lang, Nov 29 2002

(3*a(n))^2 - 2*(2*b(n))^2 = 1 with companion sequence b(n)= A046176(n+1), n>=0 (special solutions of Pell equation).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Z. Cerin and G. M. Gianella, On sums of squares of Pell-Lucas Numbers, INTEGERS 6 (2006) #A15
Tanya Khovanova, Recursive Sequences
S. Vidhyalakshmi, V. Krithika, and K. Agalya, On The Negative Pell Equation y^2 = 72x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER) Volume 4, Issue 2, February (2016).
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A056771 (even part).
Row 34 of array A094954.
Row 3 of array A188646.
Cf. similar sequences listed in A238379.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329. This is the case k=3.

I:=[1,33]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

LinearRecurrence[{34,-1},{1,33},20] (* Vincenzo Librandi, Nov 22 2011 *)
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 A041027 *)
a[18, 20] (* Gerry Martens, Jun 07 2015 *)

makelist(expand(((1+sqrt(2))^(4*n+2)+(1-sqrt(2))^(4*n+2))/6),n,0,14);  /* _Bruno Berselli, Nov 22 2011 */

Vec((1-x)/(1-34*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 22 2011

a(n) = 34*a(n-1) - a(n-2), a(-1)=1, a(0)=1.
a(n) = T(2*n+1, 3)/3 = S(n, 34) - S(n-1, 34), 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, 34)= A029547(n), T(n, 3)=A001541(n).
G.f.: (1-x)/(1-34*x+x^2).
a(n) = sqrt(8*A046176(n+1)^2 + 1)/3.
a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(2*n)-a(n-1), where k = (sqrt(2)+1)^4 = 17+12*sqrt(2) and a(0)=1. - Charles L. Hohn, Apr 05 2011
a(n) = a(-n-1) = A029547(n)-A029547(n-1) = ((1+sqrt(2))^(4n+2)+(1-sqrt(2))^(4n+2))/6. - Bruno Berselli, Nov 22 2011

A082109 Third row of number array A082105.

Original entry on oeis.org

1, 13, 33, 61, 97, 141, 193, 253, 321, 397, 481, 573, 673, 781, 897, 1021, 1153, 1293, 1441, 1597, 1761, 1933, 2113, 2301, 2497, 2701, 2913, 3133, 3361, 3597, 3841, 4093, 4353, 4621, 4897, 5181, 5473, 5773, 6081, 6397, 6721, 7053, 7393, 7741, 8097, 8461

Offset: 0

Paul Barry, Apr 03 2003

G. C. Greubel, Table of n, a(n) for n = 0..1000
Russ Cox et al., incorrect A082109 comment about A000217, SeqFan, 2024.
Takao Komatsu, Ritika Goel, and Neha Gupta, The Frobenius number for the triple of the 2-step star numbers, arXiv:2409.14788 [math.CO], 2024. See p. 2.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A082105, A082108, A000217.

Column 2 of array A188646.

[4*n^2+8*n+1: n in [0..60]]; // G. C. Greubel, Dec 22 2022

LinearRecurrence[{3,-3,1}, {1,13,33}, 51] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)

a(n)=4*n^2+8*n+1 \\ Charles R Greathouse IV, Jun 17 2017

[4*n^2+8*n+1 for n in range(61)] # G. C. Greubel, Dec 22 2022

a(n) = 4*n^2 + 8*n + 1.
a(n) = a(n-1) + 8*n + 4, with a(0)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: (1 + 10*x - 3*x^2)/(1-x)^3. - Bruno Berselli, Apr 18 2011
E.g.f.: (1 + 12*x + 4*x^2)*exp(x). - G. C. Greubel, Dec 22 2022
From Amiram Eldar, Jan 18 2023: (Start)
Sum_{n>=0} 1/a(n) = 1/6 - cot(sqrt(3)*Pi/2)*sqrt(3)*Pi/12.
Sum_{n>=0} (-1)^n/a(n) = cosec(sqrt(3)*Pi/2)*sqrt(3)*Pi/12 - 1/6. (End)

A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 97, 17, 1, 1, 1351, 577, 31, 1, 1, 18817, 19601, 1921, 49, 1, 1, 262087, 665857, 119071, 4801, 71, 1, 1, 3650401, 22619537, 7380481, 470449, 10081, 97, 1, 1, 50843527, 768398401, 457470751, 46099201, 1431431, 18817, 127, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given the function f(x,y) = (sqrt(x^2+y) + x)^2 and constant k=f(x,y), then for all integers x >= 1 and y=[+-]1, k may be irrational, but (k^n + k^(-n))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{2*k}(x), evaluated at x=n. - Seiichi Manyama, Dec 30 2018

			Row 2 gives {( (2+sqrt(3))^(2*n) + (2-sqrt(3))^(2*n) )/2}.
Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,   7,     97,      1351,        18817, ...
   3 | 1,  17,    577,     19601,       665857, ...
   4 | 1,  31,   1921,    119071,      7380481, ...
   5 | 1,  49,   4801,    470449,     46099201, ...
   6 | 1,  71,  10081,   1431431,    203253121, ...
   7 | 1,  97,  18817,   3650401,    708158977, ...
   8 | 1, 127,  32257,   8193151,   2081028097, ...
   9 | 1, 161,  51841,  16692641,   5374978561, ...
  10 | 1, 199,  79201,  31521799,  12545596801, ...
  11 | 1, 241, 116161,  55989361,  26986755841, ...
  12 | 1, 287, 164737,  94558751,  54276558337, ...
  13 | 1, 337, 227137, 153090001, 103182433537, ...
  14 | 1, 391, 305761, 239104711, 186979578241, ...
  15 | 1, 449, 403201, 362074049, 325142092801, ...
  ...

Row 2 is A011943, row 3 is A056771, row 8 is A175633, (row 2)*2 is A067902, (row 9)*2 is A089775.
Column 0-5 give A000012, A056220, A144130, A243132, A243134, A243136.
(column 1)*2 is A060626.
Cf. A188645 (f(x, y) as above with y=1).
Diagonals give A173129, A322899.
Cf. A188646, A322836.

max = 9; y = -1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188646(n,k-1) + A188646(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited by Seiichi Manyama, Dec 30 2018

More terms from Seiichi Manyama, Jan 01 2019

A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

Original entry on oeis.org

1, 61, 3781, 234361, 14526601, 900414901, 55811197261, 3459393815281, 214426605350161, 13290990137894701, 823826961944121301, 51063980650397625961, 3165142973362708688281, 196187800367837541047461, 12160478479832564836254301, 753753477949251182306719201

Offset: 0

Bruno Berselli, Apr 05 2018

Centered hexagonal numbers (A003215) with index in A145607. Example: 35 is a member of A145607, therefore A003215(35) = 3781 is a term of this sequence.
Also, centered 10-gonal numbers (A062786) with index in A182432. Example: 28 is a member of A182432 and A062786(28) = 3781.
a(n) is a solution to the Pell equation (4*a(n))^2 - 15*b(n)^2 = 1. The corresponding b(n) are A258684(n). - Klaus Purath, Jul 19 2025

Colin Barker, Table of n, a(n) for n = 0..500
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (62,-1).

Cf. A003215, A145607; A062786, A182432.
Fourth row of the array A188646.
First bisection of A041449, A042859.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k: A000012 (k=1), A001570 (k=2), A077420 (k=3), this sequence (k=4), A302330 (k=5), A302331 (k=6), A302332 (k=7), A253880 (k=8).

Mathematica
```
LinearRecurrence[{62, -1}, {1, 61}, 20]
```

x='x+O('x^99); Vec((1-x)/(1-62*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 62*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(4))/4.
a(n) = ((4 + sqrt(15))^(2*n + 1) + 1/(4 + sqrt(15))^(2*n + 1))/8.
a(n) = (1/4)*T(2*n+1, 4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022
E.g.f.: exp(31*x)*(4*cosh(8*sqrt(15)*x) + sqrt(15)*sinh(8*sqrt(15)*x))/4. - Stefano Spezia, Jul 25 2025

A302332 a(0)=1, a(1)=193; for n>1, a(n) = 194*a(n-1) - a(n-2).

Original entry on oeis.org

1, 193, 37441, 7263361, 1409054593, 273349327681, 53028360515521, 10287228590683393, 1995669318232062721, 387149560508429484481, 75105019069317087926593, 14569986549887006628274561, 2826502285659009968797338241, 548326873431298046940055344193, 106372586943386162096401939435201

Offset: 0

Bruno Berselli, Apr 05 2018

Let G and H be sequences of the form G(i) = 4*G(i-1) - G(i-2) and H(j) = 14*H(j-1) - H(j-2) for arbitrary integers i, j and without regard to initial values of G or H, then a(n) = (G(i) + G(i+8*n+4))/(14*G(i+4*n+2)) = (H(j) + H(j+4*n+2))/(14*H(j+2*n+1)) with the exception of G(i+4*n+2) or H(j+2*n+1) != 0. - Klaus Purath, Aug 31 2020
From Klaus Purath, Aug 20 2025: (Start)
Solutions to the Pell equation (7*a(n))^2 - 3*(4*b(n))^2 = 1. The corresponding b(n) are given by A084232.
For any two consecutive terms (x,y), x^2 - 194*x*y + y^2 + 192 = 0. By analogy to this, for three consecutive terms (x, y, z), y^2 - x*z + 192 = 0. (End)

Colin Barker, Table of n, a(n) for n = 0..400
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (194,-1).

Seventh row of the array A188646.
First bisection of A041269, A042127.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329.
Cf. A084232.

LinearRecurrence[{194, -1}, {1, 193}, 20]

x='x+O('x^99); Vec((1-x)/(1-194*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 194*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(7))/7.
a(n) = ((7 + 4*sqrt(3))^(2*n + 1) + 1/(7 + 4*sqrt(3))^(2*n + 1))/14.
a(n) = (a(n-1)^2 + 192)/a(n-2). - Klaus Purath, Aug 31 2020
a(n) = (1/7)*T(2*n+1, 7), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022

A302330 a(0)=1, a(1)=97; for n>1, a(n) = 98*a(n-1) - a(n-2).

Original entry on oeis.org

1, 97, 9505, 931393, 91267009, 8943235489, 876345810913, 85872946233985, 8414672385119617, 824552020795488481, 80797683365572751521, 7917348417805334160577, 775819347261557174985025, 76022378683214797814371873, 7449417291607788628633458529

Offset: 0

Bruno Berselli, Apr 05 2018

Colin Barker, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (98,-1).

Fifth row of the array A188646.
First bisection of A041275, A042151.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329.

Mathematica
```
LinearRecurrence[{98, -1}, {1, 97}, 20]
```

x='x+O('x^99); Vec((1-x)/(1-98*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 98*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(5))/5.
a(n) = ((5 + 2*sqrt(6))^(2*n + 1) + 1/(5 + 2*sqrt(6))^(2*n + 1))/10.
a(n) = (1/5)*T(2*n+1, 5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022

A302331 a(0)=1, a(1)=141; for n>1, a(n) = 142*a(n-1) - a(n-2).

Original entry on oeis.org

1, 141, 20021, 2842841, 403663401, 57317360101, 8138661470941, 1155632611513521, 164091692173449041, 23299864656018250301, 3308416689462418093701, 469771870039007351055241, 66704297128849581431750521, 9471540420426601555957518741, 1344892035403448571364535910701

Offset: 0

Bruno Berselli, Apr 05 2018

Colin Barker, Table of n, a(n) for n = 0..450
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (142,-1).

Sixth row of the array A188646.

Similar sequences of the type cosh((2*n+1)*arccosh(k))/k are listed in A302329.

LinearRecurrence[{142, -1}, {1, 141}, 20]
CoefficientList[Series[(1-x)/(1-142x+x^2),{x,0,20}],x] (* Harvey P. Dale, Jun 21 2021 *)

x='x+O('x^99); Vec((1-x)/(1-142*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 142*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(6))/6.
a(n) = ((6 + sqrt(35))^(2*n + 1) + 1/(6 + sqrt(35))^(2*n + 1))/12.
a(n) = (1/6)*T(2*n+1, 6), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022

A322904 a(n) = Sum_{k=0..n} binomial(2n+1,2k+1)(n^2-1)^(n-k)n^(2*k).

Original entry on oeis.org

1, 1, 181, 38081, 14526601, 8943235489, 8138661470941, 10287228590683393, 17254778510170993681, 37095265466946847758401, 99474891266913130060486021, 325534304813775692747248543681, 1276941308627620432293188401109401, 5914558735952850788377566338591400673

Offset: 0

Seiichi Manyama, Dec 30 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..193
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Diagonal of A188646.

Cf. A253880, A302329, A302330, A302331, A302332.

[&+[Binomial(2*n+1,2*k+1)*(n^2-1)^(n-k)*n^(2*k): k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Jan 03 2019

a[0] = 1; a[n_] := 1/n ChebyshevT[2n+1, n];
Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jan 02 2019 *)

{a(n) = sum(k=0, n, binomial(2*n+1, 2*k+1)*(n^2-1)^(n-k)*n^(2*k))}

a(n) = if (n==0, 1, polchebyshev(2*n+1, 1, n)/n); \\ Michel Marcus, Jan 02 2019

For n > 0, a(n) = (1/n) * T_{2*n+1}(n) where T_{n}(x) is a Chebyshev polynomial of the first kind.
For n > 0, a(n) = (1/n) * cosh((2*n+1)*arccosh(n)).
a(n) ~ 4^n * n^(2*n). - Vaclav Kotesovec, Jan 03 2019

cp's OEIS Frontend

A001570 Numbers k such that k^2 is centered hexagonal.

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A188647 Array read by antidiagonals of a(n) = a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2+1)+x)^2 for integers x>=1.

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A077420 Bisection of Chebyshev sequence T(n,3) (odd part) with Diophantine property.

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Maxima

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A082109 Third row of number array A082105.

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A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

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A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

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A302332 a(0)=1, a(1)=193; for n>1, a(n) = 194*a(n-1) - a(n-2).

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A322904 a(n) = Sum_{k=0..n} binomial(2n+1,2k+1)(n^2-1)^(n-k)n^(2*k).