A084134 -id:A084134 - cp's OEIS frontend

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

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

A191348 Array read by antidiagonals: ((ceiling(sqrt(n)) + sqrt(n))^k + (ceiling(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 4, 6, 2, 1, 0, 8, 20, 7, 2, 1, 0, 16, 68, 26, 8, 3, 1, 0, 32, 232, 97, 32, 14, 3, 1, 0, 64, 792, 362, 128, 72, 15, 3, 1, 0, 128, 2704, 1351, 512, 376, 81, 16, 3, 1, 0

Offset: 0

Charles L. Hohn, May 31 2011

			1, 0,  0,   0,    0,     0,      0,      0,       0,        0,         0, ...
1, 1,  2,   4,    8,    16,     32,     64,     128,      256,       512, ...
1, 2,  6,  20,   68,   232,    792,   2704,    9232,    31520,    107616, ...
1, 2,  7,  26,   97,   362,   1351,   5042,   18817,    70226,    262087, ...
1, 2,  8,  32,  128,   512,   2048,   8192,   32768,   131072,    524288, ...
1, 3, 14,  72,  376,  1968,  10304,  53952,  282496,  1479168,   7745024, ...
1, 3, 15,  81,  441,  2403,  13095,  71361,  388881,  2119203,  11548575, ...
1, 3, 16,  90,  508,  2868,  16192,  91416,  516112,  2913840,  16450816, ...
1, 3, 17,  99,  577,  3363,  19601, 114243,  665857,  3880899,  22619537, ...
1, 3, 18, 108,  648,  3888,  23328, 139968,  839808,  5038848,  30233088, ...
1, 4, 26, 184, 1316,  9424,  67496, 483424, 3462416, 24798784, 177615776, ...
1, 4, 27, 196, 1433, 10484,  76707, 561236, 4106353, 30044644, 219825387, ...
1, 4, 28, 208, 1552, 11584,  86464, 645376, 4817152, 35955712, 268377088, ...
1, 4, 29, 220, 1673, 12724,  96773, 736012, 5597777, 42574180, 323800109, ...
1, 4, 30, 232, 1796, 13904, 107640, 833312, 6451216, 49943104, 386642400, ...
...

Row 1 is A000007, row 2 is A011782, row 3 is A006012, row 4 is A001075, row 5 is A081294, row 6 is A098648, row 7 is A084120, row 8 is A146963, row 9 is A001541, row 10 is A081341, row 11 is A084134, row 13 is A090965.
Row 3*2 is A056236, row 4*2 is A003500, row 5*2 is A155543, row 9*2 is A003499.
Cf. A191347 which uses floor() in place of ceiling().

T(n, k) = if (k==0, 1, if (k==1, ceil(sqrt(n)), T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2));
matrix(9, 9, n, k, T(n-1, k-1)) \\ Charles L. Hohn, Aug 23 2019

For each row n >= 0 let T(n,0)=1 and T(n,1) = ceiling(sqrt(n)), then for each column k >= 2: T(n,k) = T(n,k-2)*(n-T(n,1)^2) + T(n,k-1)*T(n,1)*2. - Charles L. Hohn, Aug 23 2019

A084135 a(n) = 10a(n-1) - 15a(n-2), a(0)=1, a(1)=5.

Original entry on oeis.org

1, 5, 35, 275, 2225, 18125, 147875, 1206875, 9850625, 80403125, 656271875, 5356671875, 43722640625, 356876328125, 2912923671875, 23776091796875, 194067062890625, 1584029251953125, 12929286576171875, 105532426982421875

Offset: 0

Paul Barry, May 16 2003

Binomial transform of A084134.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-15).

Cf. A084134.

[n le 2 select 5^(n-1) else 10*Self(n-1) -15*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 13 2022

LinearRecurrence[{10,-15},{1,5},30] (* Harvey P. Dale, Oct 10 2012 *)

a(n)=if(n<0,0,polsym(x^2-10*x+15,n)[1+n]/2)

A084135=BinaryRecurrenceSequence(10,-15,1,5)
[A084135(n) for n in range(41)] # G. C. Greubel, Oct 13 2022

a(n) = (5+sqrt(10))^n/2 + (5-sqrt(10))^n/2.
G.f.: (1-5*x)/(1 - 10*x + 15*x^2).
E.g.f.: exp(5*x)*cosh(sqrt(10)*x).

cp's OEIS Frontend

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

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A191348 Array read by antidiagonals: ((ceiling(sqrt(n)) + sqrt(n))^k + (ceiling(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

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A084135 a(n) = 10a(n-1) - 15a(n-2), a(0)=1, a(1)=5.

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cp's OEIS Frontend

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

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Magma

Maple

Mathematica

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A191348 Array read by antidiagonals: ((ceiling(sqrt(n)) + sqrt(n))^k + (ceiling(sqrt(n)) - sqrt(n))^k)/2 for columns k >= 0 and rows n >= 0.

Views

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Examples

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PARI

Formula

A084135 a(n) = 10*a(n-1) - 15*a(n-2), a(0)=1, a(1)=5.

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Magma

Mathematica

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Formula

A084135 a(n) = 10a(n-1) - 15a(n-2), a(0)=1, a(1)=5.