A078986 -id:A078986 - cp's OEIS frontend

A078987 Chebyshev U(n,x) polynomial evaluated at x=19.

1, 38, 1443, 54796, 2080805, 79015794, 3000519367, 113940720152, 4326746846409, 164302439443390, 6239165952002411, 236924003736648228, 8996872976040630253, 341644249085807301386, 12973484592284636822415, 492650770257730391950384, 18707755785201470257292177

Offset: 0

Wolfdieter Lang, Jan 10 2003

A078986(n+1)^2 - 10*(6*a(n))^2 = +1, n>=0 (Pell equation +1, see A033313 and A033317).

a(n) equals the number of 01-avoiding words of length n on alphabet {0,1,...,37}. - Milan Janjic, Jan 26 2015

Colin Barker, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
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).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A097314, A097315.

Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (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), this sequence (m=19), A097316 (m=33).

m:=19;; a:=[1,2*m];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 22 2019

m:=19; I:=[1, 2*m]; [n le 2 select I[n] else 2*m*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 22 2019

seq( simplify(ChebyshevU(n, 19)), n=0..20); # G. C. Greubel, Dec 22 2019

lst={};Do[AppendTo[lst, GegenbauerC[n, 1, 19]], {n, 0, 8^2}];lst (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
ChebyshevU[Range[21] -1, 19] (* G. C. Greubel, Dec 22 2019 *)

a(n)=subst(polchebyshev(n,2),x,19) \\ Charles R Greathouse IV, Feb 10 2012

Vec(1/(1-38*x+x^2) + O(x^50)) \\ Colin Barker, Jun 15 2015

[lucas_number1(n,38,1) for n in range(1, 16)] # Zerinvary Lajos, Nov 07 2009

[chebyshev_U(n,19) for n in (0..20)] # G. C. Greubel, Dec 22 2019

a(n) = 38*a(n-1) - a(n-2), n>=1, a(-1)=0, a(0)=1.
a(n) = S(n, 38) with S(n, x) = U(n, x/2), Chebyshev's polynomials of the second kind. See A049310.
G.f.: 1/(1-38*x+x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k, k)*38^(n-2*k).
a(n) = ((19+6*sqrt(10))^(n+1) - (19-6*sqrt(10))^(n+1))/(12*sqrt(10)).
a(n) = Sum_{k=0..n} A101950(n,k)*37^k. - Philippe Deléham, Feb 10 2012
Product_{n>=0} (1 + 1/a(n)) = 1/3*(3 + sqrt(10)). - Peter Bala, Dec 23 2012
Product_{n>=1} (1 - 1/a(n)) = 3/19*(3 + sqrt(10)). - Peter Bala, Dec 23 2012
From Andrea Pinos, Jan 02 2023: (Start)
a(n) = (A097314(n+1) - A097315(n+1))/2.
a(n) = (A097314(n) + A097315(n))/2. (End)

A005667 Numerators of continued fraction convergents to sqrt(10).

Original entry on oeis.org

1, 3, 19, 117, 721, 4443, 27379, 168717, 1039681, 6406803, 39480499, 243289797, 1499219281, 9238605483, 56930852179, 350823718557, 2161873163521, 13322062699683, 82094249361619, 505887558869397, 3117419602578001, 19210405174337403, 118379850648602419

Offset: 0

N. J. A. Sloane, R. K. Guy

a(2*n+1) with b(2*n+1) := A005668(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation a^2 - 10*b^2 = -1, a(2*n) with b(2*n) := A005668(2*n), n >= 1, give all (positive integer) solutions to Pell equation a^2 - 10*b^2 = +1 (cf. Emerson reference).
Bisection: a(2*n) = T(n,19) = A078986(n), n >= 0 and a(2*n+1) = 3*S(2*n, 2*sqrt(10)), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. See A053120, resp. A049310.
The initial 1 corresponds to a denominator 0 in A005668. But according to standard conventions, a continued fraction starts with b(0) = integer part of the number, and the sequence of convergents p(n)/q(n) start with (p(0),q(0)) = (b(0),1). A fraction 1/0 has no mathematical meaning, the only justification is that initial terms p(-1) = 1, q(-1) = 0 are consistent with the recurrent relations p(n) = b(n)*p(n-1) + b(n-2) and the same for q(n). - M. F. Hasler, Nov 02 2019

			G.f. = 1 + 3*x + 19*x^2 + 117*x^3 + 721*x^4 + 4443*x^5 + 27379*x^6 + ... - _Michael Somos_, Jul 14 2018

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
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242, Thm. 1, p. 233.
R. K. Guy, Letter to N. J. A. Sloane, 1987
Tian-Xiao He and Peter J.-S. Shiue, Identities for linear recursive sequences of order 2, Elect. Res. Archive (2021) Vol. 29, No. 5, 3489-3507.
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 sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,1).

Cf. A010467, A040006, A084134, A005668 (denominators).

I:=[1, 3]; [n le 2 select I[n] else 6*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013

A005667:=(-1+3*z)/(-1+6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

Join[{1},Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[10],n]]],{n,1,30}]] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
CoefficientList[Series[(1-3x)/(1-6x-x^2), {x,0,30}], x] (* Vincenzo Librandi, Jun 09 2013 *)
Join[{1},Numerator[Convergents[Sqrt[10],30]]] (* or *) LinearRecurrence[ {6,1},{1,3},30] (* Harvey P. Dale, Aug 22 2016 *)
a[ n_] := (-I)^n ChebyshevT[ n, 3 I]; (* Michael Somos, Jul 14 2018 *)
LucasL[Range[0,30], 6]/2 (* G. C. Greubel, Jun 06 2019 *)

a(n)=([0,1;1,6]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Jun 11 2015

((1-3*x)/(1-6*x-x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jun 06 2019

a(n) = 6*a(n-1) + a(n-2).
G.f.: (1-3*x)/(1-6*x-x^2).
a(n) = ((-i)^n)*T(n, 3*i) with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1.
From Paul Barry, Nov 15 2003: (Start)
Binomial transform of A084132.
E.g.f.: exp(3*x)*cosh(sqrt(10)*x).
a(n) = ((3+sqrt(10))^n + (3-sqrt(10))^n)/2.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k) * 10^k * 3^(n-2*k). (End)
a(n) = (-1)^n * a(-n) for all n in Z. - Michael Somos, Jul 14 2018 [This refers to the sequence extended to negative indices according to the recurrence relation, but not to the sequence as it is currently defined. - M. F. Hasler, Nov 02 2019]
a(n) = Lucas(n,6)/2, Lucas polynomial, L(n,x), evaluated at x = 6. - G. C. Greubel, Jun 06 2019

Chebyshev comments from Wolfdieter Lang, Jan 10 2003

A188645 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, 3, 1, 17, 9, 1, 99, 161, 19, 1, 577, 2889, 721, 33, 1, 3363, 51841, 27379, 2177, 51, 1, 19601, 930249, 1039681, 143649, 5201, 73, 1, 114243, 16692641, 39480499, 9478657, 530451, 10657, 99, 1, 665857, 299537289, 1499219281, 625447713, 54100801, 1555849, 19601, 129, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given 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 A188644.

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_{k}(x), evaluated at x=2*n^2+1. - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   3,     17,        99,          577, ...
   2 | 1,   9,    161,      2889,        51841, ...
   3 | 1,  19,    721,     27379,      1039681, ...
   4 | 1,  33,   2177,    143649,      9478657, ...
   5 | 1,  51,   5201,    530451,     54100801, ...
   6 | 1,  73,  10657,   1555849,    227143297, ...
   7 | 1,  99,  19601,   3880899,    768398401, ...
   8 | 1, 129,  33281,   8586369,   2215249921, ...
   9 | 1, 163,  53137,  17322499,   5647081537, ...
  10 | 1, 201,  80801,  32481801,  13057603201, ...
  11 | 1, 243, 118097,  57394899,  27893802817, ...
  12 | 1, 289, 167041,  96549409,  55805391361, ...
  13 | 1, 339, 229841, 155831859, 105653770561, ...
  14 | 1, 393, 308897, 242792649, 190834713217, ...
  15 | 1, 451, 406801, 366934051, 330974107201, ...
  ...

Row 1 is A001541, row 2 is A023039, row 3 is A078986, row 4 is A099370, row 5 is A099397, row 6 is A174747, row 8 is A176368, (row 1)*2 is A003499, (row 2)*2 is A087215.
Column 1 is A058331, (column 1)*2 is A005899.
A188644 (f(x, y) as above with y=-1).
Diagonal gives A173128.
Cf. A188647.

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) = (A188647(n,k-1) + A188647(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 and extended by Seiichi Manyama, Jan 01 2019

A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1

Offset: 0

Seiichi Manyama, Dec 28 2018

			Square array begins:
   1, 1,    1,     1,      1,      1,       1, ...
   0, 1,    2,     3,      4,      5,       6, ...
  -1, 1,    7,    17,     31,     49,      71, ...
   0, 1,   26,    99,    244,    485,     846, ...
   1, 1,   97,   577,   1921,   4801,   10081, ...
   0, 1,  362,  3363,  15124,  47525,  120126, ...
  -1, 1, 1351, 19601, 119071, 470449, 1431431, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Mirror of A101124.
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890.
Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136.
Main diagonal gives A115066.
Cf. A323182 (Chebyshev polynomial of the second kind).

Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)

T(n,k) = polchebyshev(n,1,k);
matrix(7, 7, n, k, T(n-1,k-1)) \\ Michel Marcus, Dec 28 2018

T(n, k) = round(cos(n*acos(k)));\\ Seiichi Manyama, Mar 05 2021

T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021

A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = cos(n*arccos(k)). - Seiichi Manyama, Mar 05 2021
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021

A212331 a(n) = 5n(n+5)/2.

Original entry on oeis.org

0, 15, 35, 60, 90, 125, 165, 210, 260, 315, 375, 440, 510, 585, 665, 750, 840, 935, 1035, 1140, 1250, 1365, 1485, 1610, 1740, 1875, 2015, 2160, 2310, 2465, 2625, 2790, 2960, 3135, 3315, 3500, 3690, 3885, 4085, 4290, 4500, 4715, 4935, 5160, 5390, 5625, 5865

Offset: 0

Bruno Berselli, May 30 2012

Numbers of the form n*t(n+5,h)-(n+5)*t(n,h), where t(k,h) = k*(k+2*h+1)/2 for any h. Likewise:
A000217(n) = n*t(n+1,h)-(n+1)*t(n,h),
A005563(n) = n*t(n+2,h)-(n+2)*t(n,h),
A140091(n) = n*t(n+3,h)-(n+3)*t(n,h),
A067728(n) = n*t(n+4,h)-(n+4)*t(n,h) (n>0),
A140681(n) = n*t(n+6,h)-(n+6)*t(n,h).
This is the case r=7 in the formula:
u(r,n) = (P(r, P(n+r, r+6)) - P(n+r, P(r, r+6))) / ((r+5)*(r+6)/2)^2, where P(s, m) is the m-th s-gonal number.
Also, a(k) is a square for k = (5/2)*(A078986(n)-1).
Sum of reciprocals of a(n), for n>0: 137/750.
Also, numbers h such that 8*h/5+25 is a square.
The table given below as example gives the dimensions D(h, n) of the irreducible SU(3) multiplets (h,n). See the triangle A098737 with offset 0, and the comments there, also with a link and the Coleman reference. - Wolfdieter Lang, Dec 18 2020

			From the first and second comment derives the following table:
----------------------------------------------------------------
h \ n | 0   1    2    3    4    5    6    7    8    9    10
------|---------------------------------------------------------
0     | 0,  1,   3,   6,  10,  15,  21,  28,  36,  45,   55, ...  (A000217)
1     | 0,  3,   8,  15,  24,  35,  48,  63,  80,  99,  120, ...  (A005563)
2     | 0,  6,  15,  27,  42,  60,  81, 105, 132, 162,  195, ...  (A140091)
3     | 0, 10,  24,  42,  64,  90, 120, 154, 192, 234,  280, ...  (A067728)
4     | 0, 15,  35,  60,  90, 125, 165, 210, 260, 315,  375, ...  (A212331)
5     | 0, 21,  48,  81, 120, 165, 216, 273, 336, 405,  480, ...  (A140681)
6     | 0, 28,  63, 105, 154, 210, 273, 343, 420, 504,  595, ...
7     | 0, 36,  80, 132, 192, 260, 336, 420, 512, 612,  720, ...
8     | 0, 45,  99, 162, 234, 315, 405, 504, 612, 729,  855, ...
9     | 0, 55, 120, 195, 280, 375, 480, 595, 720, 855, 1000, ...
with the formula n*(h+1)*(h+n+1)/2. See also A098737.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000537, A005563, A055998, A067728, A098737, A140091, A140681.

Magma
```
[5*n*(n+5)/2: n in [0..46]];
```
Mathematica
```
Table[(5/2) n (n + 5), {n, 0, 46}]
```

a(n)=5*n*(n+5)/2 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 5*x*(3-2*x)/(1-x)^3.
a(n) = a(-n-5) = 5*A055998(n).
E.g.f.: (5/2)*x*(x + 6)*exp(x). - G. C. Greubel, Jul 21 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/25 - 47/750. - Amiram Eldar, Feb 26 2022

Extended by Bruno Berselli, Aug 05 2015

A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

Original entry on oeis.org

1, 10, 361, 13690, 519841, 19740250, 749609641, 28465426090, 1080936581761, 41047124680810, 1558709801289001, 59189925324301210, 2247658452522156961, 85351831270517663290, 3241121929827149048041, 123077281502161146162250

Offset: 0

Kival Ngaokrajang, Sep 18 2014

Refer to comment of A240926. Consider a circle C of radius 10/9 (in some length units) with a chord of length 4/3. This has been chosen such that the larger sagitta has length 2. The smaller sagitta has length 2/9. The input, besides the circle C is the circle C_0 with radius R_0 = 1, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). If one considers the curvature of touching circles inscribed in the smaller segment, the sequence would be A247512. See an illustration given in the link.
a(n) also seems to be A078986(i)^2 and 10*A097315(j)^2 interleaved; where i = n/2 for n even, j = n/2 - 1/2 for n odd; as following:
   1          = 1^2
   10         = 10*1^2
   361        = 19^2
   13690      = 10*37^2
   519841     = 721^2
   19740250   = 10*1405^2
   749609641  = 27379^2
   ...
A078986; Chebyshev... polynomial: 1, 19, 721, 27379, ...
A097315; Pell equation...       : 1, 37, 1405, 53353, ...

Colin Barker, Table of n, a(n) for n = 0..600
Kival Ngaokrajang, Illustration of initial terms
Wolfdieter Lang, Curvature computation for A247335 and A247512.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Cf. A240926, A078986, A097315, A247512.

I:=[39,-39,1]; [n le 3 select I[n] else Self(n-1) - 10*Self(n-2) + 361*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{39,-39,1}, {1, 10, 361}, 50] (* or *) Table[Round[((19 + 6*Sqrt[10])^(-n)*(1 + (19 + 6*Sqrt[10])^n)^2)]/4, {n, 0, 30}] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.9;print1(1,", ");r1=r;
for (n=1,50,
if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
ac=sqrt(ab^2-r^2);
if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
b=acos(r/ab)-z;
r=r*(1-cos(b))/(1+cos(b));
an=floor(9/(10*r));
print1(if(an>9,an,10),", ")
)
}

Vec(-(10*x^2-29*x+1)/((x-1)*(x^2-38*x+1)) + O(x^20)) \\ Colin Barker, Mar 03 2016

Conjectures from Colin Barker, Sep 18 2014: (Start)
a(n) = 39*a(n-1) - 39*a(n-2) + a(n-3).
G.f.: -(10*x^2-29*x+1) / ((x-1)*(x^2-38*x+1)). (End)
From Wolfdieter Lang, Sep 30 2014 (Start)
See the W. Lang link for proofs of the following statements.
One step nonlinear recurrence: a(n) = -9 + 19*a(n-1) + 60*sqrt(a(n-1)*(a(n-1) - 1)/10), n>=1, with a(0) = 1.
a(n) =  (1 + A078986(n))/2  = (2 + S(n, 38) - S(n-2, 38))/4 =
  (1 + S(n, 38) -19*S(n-1, 38))/2 for n>=0, with Chebyshev's S-polynomials (see A049310).  S(n, 38) = A078987(n).
The G.f. conjectured by Colin Barker above follows from the one for Chebyshev's T(n, 19) = A078986(n):  (1/(1-x) + (1-19*x)/(1-38*x+x^2))/2 = (1-29*x+10*x^2)/((1-x)* (1-38*x+x^2)).
The four term recurrence conjectured by Colin Barker above follows from the expanded g.f. denominator: (1-x)* (1-38*x+x^2) = 1- 39*x + 39*x^2 - x^3.
(End)
a(n) = ((19+6*sqrt(10))^(-n)*(1+(19+6*sqrt(10))^n)^2)/4. - Colin Barker, Mar 03 2016

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A084070 a(n) = 38*a(n-1) - a(n-2), with a(0)=0, a(1)=6.

Original entry on oeis.org

0, 6, 228, 8658, 328776, 12484830, 474094764, 18003116202, 683644320912, 25960481078454, 985814636660340, 37434995712014466, 1421544022419889368, 53981237856243781518, 2049865494514843808316, 77840907553707820934490, 2955904621546382351702304

Offset: 0

Benoit Cloitre, May 10 2003

nonn

This sequence gives the values of y in solutions of the Diophantine equation x^2 - 10*y^2 = 1. The corresponding x values are in A078986. - Vincenzo Librandi, Aug 08 2010 [edited by Jon E. Schoenfield, May 04 2014]

			G.f. = 6*x + 228*x^2 + 8658*x^3 + 328776*x^4 + ... - _Michael Somos_, Feb 24 2023

Indranil Ghosh, Table of n, a(n) for n = 0..632
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (38,-1).

Cf. A001078, A001109, A001353, A001653, A005668, A060645, A084068, A084069, A221874.

Cf. A078986. - Vincenzo Librandi, Apr 14 2010

a:=[0,6];; for n in [3..20] do a[n]:=38*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 12 2020

I:=[0,6]; [n le 2 select I[n] else 38*Self(n-1) - Self(n-2): n in [1..20]]; // G. C. Greubel, Jan 12 2020

seq( simplify(6*ChebyshevU(n-1, 19)), n=0..20); # G. C. Greubel, Jan 12 2020

LinearRecurrence[{38,-1},{0,6},30] (* Harvey P. Dale, Nov 01 2011 *)
6*ChebyshevU[Range[20]-2, 19] (* G. C. Greubel, Jan 12 2020 *)

u=0; v=6; for(n=2,20, w=38*v-u; u=v; v=w; print1(w,","))

vector(21, n, 6*polchebyshev(n-2, 2, 19) ) \\ G. C. Greubel, Jan 12 2020

[6*chebyshev_U(n-1, 19) for n in (0..20)] # G. C. Greubel, Jan 12 2020

Numbers k such that 10*k^2 = floor(k*sqrt(10)*ceiling(k*sqrt(10))).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 37*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 39*(a(n-1) - a(n-2)) + a(n-3). (End)
From R. J. Mathar, Feb 19 2008: (Start)
O.g.f.: 6*x/(1 - 38*x + x^2).
a(n) = 6*A078987(n-1). (End)
a(n) = 6*ChebyshevU(n-1, 19). - G. C. Greubel, Jan 12 2020
a(n) = A005668(2*n). - Michael Somos, Feb 24 2023

A099366 Squares of A005668.

Original entry on oeis.org

0, 1, 36, 1369, 51984, 1974025, 74960964, 2846542609, 108093658176, 4104712468081, 155870980128900, 5918992532430121, 224765845252215696, 8535183127051766329, 324112192982714904804, 12307728150216114616225

Offset: 0

Wolfdieter Lang, Oct 18 2004

See the comment in A099279. This is example a=6.

a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 6 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 6 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Apr 21 2023

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 (37,37,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. other squares of k-metallonacci numbers (for k=1 to 10): A007598, A079291, A092936, A099279, A099365, this sequence, A099367, A099369, A099372, A099374.

with (combinat):seq(fibonacci(n,6)^2,n=0..15); # Zerinvary Lajos, Apr 09 2008

LinearRecurrence[{37,37,-1},{0,1,36},20] (* Harvey P. Dale, Sep 23 2018 *)

a(n) = A005668(n)^2.
a(n) = 37*a(n-1) + 37*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=36.
a(n) = 38*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = (T(n, 19) - (-1)^n)/20 with the Chebyshev polynomials of the first kind: T(n, 19) = A078986(n).
G.f.: x*(1-x)/((1 - 38*x + x^2)*(1+x)) = x*(1-x)/(1 - 37*x - 37*x^2 + x^3).
a(n) = (1 - (-1)^n)/2 + 36*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Apr 21 2023
Product_{n>=2} (1 + (-1)^n/a(n)) = (3 + sqrt(10))/6 (Falcon, 2016, p. 189, eq. (3.1)). - Amiram Eldar, Dec 03 2024

A249457 The numerator of curvatures of touching circles inscribed in a special way in the larger segment of a unit circle divided by a chord of length sqrt(84)/5.

Original entry on oeis.org

10, 100, 2890, 96100, 3237610, 109202500, 3683712490, 124263300100, 4191798484810, 141402777864100, 4769968258260490, 160906295771812900, 5427884341892493610, 183099910962324064900, 6176546013641762558890, 208354665265158340802500, 7028469704892605715408010

Offset: 0

Kival Ngaokrajang, Oct 29 2014

The denominators are conjectured to be A005032.
Refer to comments and links of A240926. Consider a unit circle with a chord of length sqrt(84)/5. This has been chosen such that the larger sagitta has length 7/5. The input, besides the unit circle C, is the circle C_0 with radius R_0 = 7/10, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle is C_n = 1/R_n, n >= 0, and its numerator is conjectured to be a(n).
If one considers the curvature of touching circles inscribed in the smaller segment (sagitta length 3/5), the rational sequence would be A249458/A169634. See an illustration given in the link.
For the proof and the formula for the rational curvatures of the circles in the larger segment see a comment under A249862. C_n = (5/7)*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). - Wolfdieter Lang, Nov 07 2014

Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Sagitta.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (37,-111,27).

Cf. A005032, A049310, A078986, A097315, A169364, A240926, A247335, A247512, A248834, A249458, A249862.

I:=[10,100,2890]; [n le 3 select I[n] else 37*Self(n-1) - 111*Self(n-2) + 27*Self(n-3): n in [1..30]]; // G. C. Greubel, Dec 20 2017

LinearRecurrence[{37, -111, 27},{10, 100, 2890},16] (* Ray Chandler, Aug 11 2015 *)
CoefficientList[Series[10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)

{
r=0.7;dn=7;print1(round(dn/r),", ");r1=r;
for (n=1,40,
     if (n<=1,ab=2-r,ab=sqrt(ac^2+r^2));
     ac=sqrt(ab^2-r^2);
     if (n<=1,z=0,z=(Pi/2)-atan(ac/r)+asin((r1-r)/(r1+r));r1=r);
     b=acos(r/ab)-z;
     r=r*(1-cos(b))/(1+cos(b)); dn=dn*3;
     print1(round(dn/r),", ");
)
}

x='x+O('x^30); Vec(10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x))) \\ G. C. Greubel, Dec 20 2017

Empirical g.f.: -10*(30*x^2-27*x+1) /((3*x - 1)*(9*x^2-34*x+1)). - Colin Barker, Oct 29 2014
From Wolfdieter Lang, Nov 07 2014: (Start)
a(n) = 5*(A249862(n) + 3^n) = 5*3^n*(S(n, 34/3) - (17/3)*S(n-1, 34/3) + 1), n >= 0, with Chebyshev's S polynomials (A049310). See the comments on A249862 for the proof.
O.g.f.: 5*((1 - 17*x)/(1 - 34*x + 9*x^2) + 1/(1-3*x)) = 10*(1 - 27*x + 30*x^2)/((1 - 34*x + 9*x^2)*(1 - 3*x)) proving the conjecture of Colin Barker above. (End)
E.g.f.: 5*exp(3*x)*(1 + exp(14*x)*cosh(2*sqrt(70)*x)). - Stefano Spezia, Mar 24 2023

Edited. Name and comment small changes, keyword easy added. - Wolfdieter Lang, Nov 07 2014

a(16) from Stefano Spezia, Mar 24 2023

cp's OEIS Frontend

A078987 Chebyshev U(n,x) polynomial evaluated at x=19.

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A005667 Numerators of continued fraction convergents to sqrt(10).

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A188645 Array of ((k^n)+(k^(-n)))/2 where k=(sqrt(x^2+1)+x)^2 for integers x>=1.

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A322836 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.

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A212331 a(n) = 5*n*(n+5)/2.

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A247335 The curvature of touching circles inscribed in a special way in the larger segment of circle of radius 10/9 divided by a chord of length 4/3.

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A212331 a(n) = 5n(n+5)/2.

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.