A090301 -id:A090301 - cp's OEIS frontend

A090316 a(n) = 24*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 24.

2, 24, 578, 13896, 334082, 8031864, 193098818, 4642403496, 111610782722, 2683301188824, 64510839314498, 1550943444736776, 37287153512997122, 896442627756667704, 21551910219673022018, 518142287899909196136, 12456966819817493729282, 299485345963519758698904

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 25 2004

Lim_{n->infinity} a(n)/a(n+1) = 0.0415945... = 1/(12+sqrt(145)) = sqrt(145) - 12.

Lim_{n->infinity} a(n+1)/a(n) = 24.0415945... = 12+sqrt(145) = 1/(sqrt(145)-12).

			a(4) =334082 = 24a(3) + a(2) = 24*13896+ 578 = (12+sqrt(145))^4 + (12-sqrt(145))^4 = 334081.99999700672 + 0.00000299327 = 334082.

G. C. Greubel, Table of n, a(n) for n = 0..500
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (24,1).

Cf. A041264, A058168, A056949.

Lucas polynomials Lucas(n,m): A000032 (m=1), A002203 (m=2), A006497 (m=3), A014448 (m=4), A087130 (m=5), A085447 (m=6), A086902 (m=7), A086594 (m=8), A087798 (m=9), A086927 (m=10), A001946 (m=11), A086928 (m=12), A088316 (m=13), A090300 (m=14), A090301 (m=15), A090305 (m=16), A090306 (m=17), A090307 (m=18), A090308 (m=19), A090309 (m=20), A090310 (m=21), A090313 (m=22), A090314 (m=23), this sequence (m=24), A330767 (m=25).

a:=[2,24];; for n in [3..20] do a[n]:=24*a[n-1]+a[n-2]; od; a; # G. C. Greubel, Dec 29 2019

I:=[2,24]; [n le 2 select I[n] else 24*Self(n-1) +Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 29 2019

seq(simplify(2*(-I)^n*ChebyshevT(n, 12*I)), n = 0..20); # G. C. Greubel, Dec 29 2019

LinearRecurrence[{24,1},{2,24},20] (* Harvey P. Dale, Aug 30 2015 *)
LucasL[Range[20]-1,24] (* G. C. Greubel, Dec 29 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 12*I) ) \\ G. C. Greubel, Dec 29 2019

[2*(-I)^n*chebyshev_T(n, 12*I) for n in (0..20)] # G. C. Greubel, Dec 29 2019

a(n) = 24*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 24.
a(n) = (12+sqrt(145))^n + (12-sqrt(145))^n.
(a(n))^2 = a(2n) - 2 if n=1,3,5,..., (a(n))^2 = a(2n)+2 if n=2,4,6,....
G.f.: 2*(1-12*x)/(1-24*x-x^2). - Philippe Deléham, Nov 02 2008
a(n) = 2*(-i)^n * ChebyshevT(n, 12*i) = Lucas(n, 24). - G. C. Greubel, Dec 29 2019
a(n) = 2 * A041264(n-1) for n>0. - Alois P. Heinz, Dec 29 2019

More terms from Ray Chandler, Feb 14 2004

Corrected by T. D. Noe, Nov 07 2006

A154597 a(n) = 15*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 15, 226, 3405, 51301, 772920, 11645101, 175449435, 2643386626, 39826248825, 600037119001, 9040383033840, 136205782626601, 2052127122432855, 30918112619119426, 465823816409224245, 7018275358757483101, 105739954197771470760, 1593117588325329544501

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009

Limit_{n -> infinity} a(n)/a(n-1) = (15 + sqrt(229))/2. - Klaus Brockhaus, Oct 07 2009
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. - Johannes W. Meijer, Jun 12 2010
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 15's along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
a(n) equals the number of words of length n - 1 on alphabet {0,1,...,15} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015
From Michael A. Allen, Apr 30 2023: (Start)
Also called the 15-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 15 kinds of squares available. (End)

G. C. Greubel, Table of n, a(n) for n = 0..840 (a(0) = 0 added by Jianing Song)
Michael A. Allen and Kenneth Edwards, Fence tiling derived identities involving the metallonacci numbers squared or cubed, Fib. Q. 60:5 (2022) 5-17.
Tanya Khovanova, Recursive sequences. [From _Johannes W. Meijer_, Jun 12 2010]
Index entries for linear recurrences with constant coefficients, signature (15,1).

Row n=15 of A073133, A172236 and A352361 and column k=15 of A157103.
First bisection is A098247.
Cf. A166125 (decimal expansion of sqrt(229)), A166126 (decimal expansion of (15 + sqrt(229))/2).
Cf. also A041427, A090301, A098245.
Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), this sequence (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-229); S:=[ ((15+r)^n-(15-r)^n)/(2^n*r): n in [1..17] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Jan 12 2009

[n le 2 select n-1 else 15*Self(n-1) +Self(n-2): n in [1..30]]; // G. C. Greubel, Sep 20 2024

LinearRecurrence[{15,1}, {0,1}, 31] (* Vladimir Joseph Stephan Orlovsky, Oct 27 2009 *)
CoefficientList[Series[x/(1-15*x-x^2), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)

my(x='x+O('x^50)); concat([0], Vec(x/(1-15*x-x^2))) \\ G. C. Greubel, Apr 16 2017

def A154597(n): return (-i)^(n-1)*chebyshev_U(n-1, 15*i/2)
[A154597(n) for n in range(31)] # G. C. Greubel, Sep 20 2024

G.f.: x/(1 - 15*x - x^2). - Klaus Brockhaus, Jan 12 2009, corrected Oct 07 2009
a(n) = ((15 + sqrt(229))^n - (15 - sqrt(229))^n)/(2^n*sqrt(229)).
From Johannes W. Meijer, Jun 12 2010: (Start)
Limit_{k -> infinity} a(n+k)/a(k) = (A090301(n) + a(n)*sqrt(229))/2.
Limit_{n -> infinity} A090301(n)/a(n) = sqrt(229).
a(2n+1) = 15*A098245(n-1).
a(3n+1) = A041427(5n), a(3n+2) = A041427(5n+3), a(3n+3) = 2*A041427(5n+4). (End)
E.g.f.: (2/sqrt(229))*exp(15*x/2)*sinh(sqrt(229)*x/2). - G. C. Greubel, Sep 20 2024

Extended beyond a(7) by Klaus Brockhaus and Philippe Deléham, Jan 12 2009
Name from Philippe Deléham, Jan 12 2009
Edited by Klaus Brockhaus, Oct 07 2009
Missing a(0) added by Jianing Song, Jan 29 2019

A330767 a(n) = 25*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 25.

Original entry on oeis.org

2, 25, 627, 15700, 393127, 9843875, 246490002, 6172093925, 154548838127, 3869893047100, 96901875015627, 2426416768437775, 60757321085960002, 1521359443917437825, 38094743419021905627, 953889944919465078500, 23885343366405648868127, 598087474105060686781675, 14976072195992922818410002

Offset: 0

G. C. Greubel, Dec 29 2019

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (25,1).

a:=[2,25];; for n in [3..25] do a[n]:=25*a[n-1]+a[n-2]; od; a;

I:=[2,25]; [n le 2 select I[n] else 25*Self(n-1) +Self(n-2): n in [1..25]];

seq(simplify(2*(-I)^n*ChebyshevT(n, 25*I/2)), n = 0..25);

Mathematica
```
LucasL[Range[25] -1, 25]
```

vector(26, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 25*I/2) )

[2*(-I)^n*chebyshev_T(n, 25*I/2) for n in (0..25)]

a(n) = ( (25 + sqrt(629))^n + (25 - sqrt(629))^n )/2^n.
G.f.: (2 - 25*x)/(1-25*x-x^2).
a(n) = Lucas(n, 25) = 2*(-i)^n * ChebyshevT(n, 25*i/2).

A089772 a(n) = Lucas(11*n).

Original entry on oeis.org

2, 199, 39603, 7881196, 1568397607, 312119004989, 62113250390418, 12360848946698171, 2459871053643326447, 489526700523968661124, 97418273275323406890123, 19386725908489881939795601, 3858055874062761829426214722

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

Lim_{n-> infinity} a(n+1)/a(n) = 199.00502499874... = (199 + sqrt(39605))/2.

Lim_{n->infinity} a(n)/a(n+1) = 0.00502499874... = 2/(199 + sqrt(39605)) = (sqrt(39605) - 199)/2.

			a(4) = 1568397607 = 199*a(3) + a(2) = 199*7881196 + 39603 = ((199 + sqrt(39605) )/2)^4 + ((199 - sqrt(39605))/2)^4 = 1568397606.9999999993624065... + 0.0000000006375934...

Nathaniel Johnston, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (199,1).

List([0..20], n-> Lucas(1,-1,11*n)[2] ); # G. C. Greubel, Dec 30 2019

[Lucas(11*n): n in [0..20]]; // Vincenzo Librandi, Apr 15 2011

seq(simplify(2*(-I)^n*ChebyshevT(n, 199*I/2)), n = 0..20); # G. C. Greubel, Dec 31 2019

LucasL[11*Range[0,20]] (* or *) LinearRecurrence[{199,1},{2,199},20] (* Harvey P. Dale, Dec 23 2015 *)
LucasL[Range[20]-1,199] (* G. C. Greubel, Dec 31 2019 *)

vector(21, n, 2*(-I)^(n-1)*polchebyshev(n-1, 1, 199*I/2) ) \\ G. C. Greubel, Dec 31 2019

[lucas_number2(11*n,1,-1) for n in (0..20)] # G. C. Greubel, Dec 30 2019

a(n) = 199*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 199.
a(n) = ((199 + sqrt(39605))/2)^n + ((199 - sqrt(39605))/2)^n.
a(n)^2 = a(2n) - 2 if n = 1, 3, 5, ...;
a(n)^2 = a(2n) + 2 if n = 2, 4, 6, ....
G.f.: (2 - 199*x)/(1 - 199*x - x^2). - Philippe Deléham, Nov 02 2008
a(n) = Lucas(n, 199) = 2*(-i)^n * ChebyshevT(n, 199*i/2). - G. C. Greubel, Dec 31 2019
E.g.f.: 2*exp(199*x/2)*cosh(sqrt(39605)*x/2). - Stefano Spezia, Jan 01 2020

A041426 Numerators of continued fraction convergents to sqrt(229).

Original entry on oeis.org

15, 106, 121, 227, 1710, 51527, 362399, 413926, 776325, 5848201, 176222355, 1239404686, 1415627041, 2655031727, 20000849130, 602680505627, 4238764388519, 4841444894146, 9080209282665, 68402909872801, 2061167505466695, 14496575448139666, 16557742953606361

Offset: 0

N. J. A. Sloane

From Johannes W. Meijer, Jun 12 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A090301.
For the terms of the periodical sequence of the continued fraction for sqrt(229) see A040213. We observe that its period is five. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 3420, 0, 0, 0, 0, 1).

Cf. A041427, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Numerator[Convergents[Sqrt[229], 30]] (* Vincenzo Librandi, Nov 01 2013 *)
LinearRecurrence[{0,0,0,0,3420,0,0,0,0,1},{15,106,121,227,1710,51527,362399,413926,776325,5848201},30] (* Harvey P. Dale, Dec 19 2016 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n) = A090301(3n+1), a(5n+1) = (A090301(3n+2) - A090301(3n+1))/2, a(5n+2) = (A090301(3n+2) + A090301(3n+1))/2, a(5n+3) = A090301(3n+2) and a(5n+4) = A090301(3n+3)/2. (End)
G.f.: -(x^9-15*x^8+106*x^7-121*x^6+227*x^5+1710*x^4+227*x^3+121*x^2+106*x+15) / (x^10+3420*x^5-1). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

cp's OEIS Frontend

A090316 a(n) = 24*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 24.

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A154597 a(n) = 15*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.

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A330767 a(n) = 25*a(n-1) + a(n-2), starting with a(0) = 2 and a(1) = 25.

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A089772 a(n) = Lucas(11*n).

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A041426 Numerators of continued fraction convergents to sqrt(229).

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