A087451 -id:A087451 - cp's OEIS frontend

A015441 Generalized Fibonacci numbers.

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159

Offset: 0

Olivier Gérard

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook), p. 317-318
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
John Elias, Illustration: A Starhex honeycomb variation
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 48.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A000079, A003462, A016153, A109466, A122117. Cf. A001656, A087451. Cf. A000400.

I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)

PARI
```
{a(n) = (3^n - (-2)^n) / 5};
```

[lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From  Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A201455 a(n) = 3a(n-1) + 4a(n-2) for n>1, a(0)=2, a(1)=3.

Original entry on oeis.org

2, 3, 17, 63, 257, 1023, 4097, 16383, 65537, 262143, 1048577, 4194303, 16777217, 67108863, 268435457, 1073741823, 4294967297, 17179869183, 68719476737, 274877906943, 1099511627777, 4398046511103, 17592186044417, 70368744177663, 281474976710657

Offset: 0

Bruno Berselli, Jan 09 2013

This is the Lucas sequence V(3,-4).

Inverse binomial transform of this sequence is A087451.

Bruno Berselli, Table of n, a(n) for n = 0..200
Weerayuth Nilsrakoo and Achariya Nilsrakoo, On One-Parameter Generalization of Jacobsthal Numbers, WSEAS Trans. Math. (2025) Vol. 24, 51-61. See p. 3.
Wikipedia, Lucas sequence: Specific names.
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. for the same recurrence with initial values (i,i+1): A015521 (Lucas sequence U(3,-4); i=0), A122117 (i=1), A189738 (i=3).
Cf. for similar closed form: A014551 (2^n+(-1)^n), A102345 (3^n+(-1)^n), A087404 (5^n+(-1)^n).
Cf. A052539, A024036.

[n le 1 select n+2 else 3*Self(n)+4*Self(n-1): n in [0..25]];

RecurrenceTable[{a[n] == 3 a[n - 1] + 4 a[n - 2], a[0] == 2, a[1] == 3}, a[n], {n, 25}]

a[0]:2$ a[1]:3$ a[n]:=3*a[n-1]+4*a[n-2]$ makelist(a[n], n, 0, 25);

Vec((2-3*x)/((1+x)*(1-4*x)) + O(x^30)) \\ Michel Marcus, Jun 26 2015

G.f.: (2-3*x)/((1+x)*(1-4*x)).
a(n) = 4^n+(-1)^n.
a(n) = A086341(A047524(n)) for n>0, a(0)=2.
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 25*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
a(n) = (2/4^n) * Sum_{k = 0..n} binomial(4*n+1, 4*k). - Peter Bala, Feb 06 2019

A283653 Numbers k such that 3^k + (-2)^k is prime.

Original entry on oeis.org

0, 2, 3, 4, 5, 17, 29, 31, 53, 59, 101, 277, 647, 1061, 2381, 2833, 3613, 3853, 3929, 5297, 7417, 90217, 122219, 173191, 256199, 336353, 485977, 591827, 1059503

Offset: 1

Juri-Stepan Gerasimov, Mar 12 2017

Numbers j such that both 3^j + (-2)^j and 3^j + (-4)^j are primes: 0, 3, 4, 17, 59, ...
See Michael Somos comment in A082101.
Probably this is just A057468 with 0,2,4 added, because we already know that if another even number belong to this sequence it must be greater than log_3(10^16000000) = about 3.3*10^7. This is because 3^n+2^n can be a prime with n>0 only if n is a power of 2. - Giovanni Resta, Mar 12 2017

			4 is in this sequence because 3^4 + (-2)^4 = 97 is prime.

Cf. A174326. Subsequence of A087451. Supersequence of A057468.

Cf. A082101.

[n: n in [0..1000] | IsPrime(3^n+(-2)^n)];

Select[Range[0, 10000], PrimeQ[3^# + (-2)^#] &] (* G. C. Greubel, Jul 29 2018 *)

is(n)=isprime(3^n+(-2)^n) \\ Charles R Greathouse IV, Mar 16 2017

A140796 a(n)=a(n-1)+6a(n-2), n>2.

Original entry on oeis.org

1, 5, 14, 44, 128, 392, 1160, 3512, 10472, 31544, 94376, 283640, 849896, 2551736, 7651112, 22961528, 68868200, 206637368, 619846568, 1859670776, 5578750184, 16736774840, 50209275944, 150629924984, 451885580648, 1355665130552

Offset: 0

Paul Curtz, Jul 15 2008

nonn

The binomial transform is A037481.

The recurrence of the definition is also satisfied by A087451, A102901 and A140725.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1, 6).

Cf. A140725

Join[{1},LinearRecurrence[{1,6},{5,14},30]] (* Harvey P. Dale, Nov 20 2011 *)

a(n+1)-3a(n) = (-1)^n*A000079(n-1), n>0.
d(n+1)-3d(n) = (-1)^(n+1)*A000079(n-1), n>0, where d(n) is the sequence of pair sums d(n)= a(n)+a(n+1)=6, 19, 58, 172,...
O.g.f.: (1+x)(3x+1)/((2x+1)(1-3x)). - R. J. Mathar, Jul 29 2008
a(n) = (-1)^(n+1)*2^n/10+8*3^n/5, n>0. - R. J. Mathar, Jul 29 2008
a(n) = A140725(n)+A140725(n+1). - Philippe Deléham, Nov 17 2013

Edited and extended by R. J. Mathar, Jul 29 2008

A087452 G.f.: (2-x)/((1+3x)(1-4x)); e.g.f.: exp(4x) + exp(-3x); a(n) = 4^n + (-3)^n.

Original entry on oeis.org

2, 1, 25, 37, 337, 781, 4825, 14197, 72097, 242461, 1107625, 4017157, 17308657, 65514541, 273218425, 1059392917, 4338014017, 17050729021, 69106897225, 273715645477, 1102998412177, 4387586157901, 17623567104025, 70274600998837, 281757406247137

Offset: 0

Paul Barry, Sep 06 2003

Generalized Lucas-Jacobsthal numbers.

Index entries for linear recurrences with constant coefficients, signature (1,12).

Cf. A014551, A053404, A087451.

CoefficientList[Series[(2 - z)/((1 + 3 z) (1 - 4 z)), {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)
LinearRecurrence[{1,12},{2,1},30] (* Harvey P. Dale, Nov 06 2022 *)

polsym(x^2-x-12,50) \\ Charles R Greathouse IV, Jun 11 2011

a(n) = (-3)^n+4^n.
a(n) = a(n-1) + 12*a(n-2) for n > 1; a(0)=2, a(1)=1. - Philippe Deléham, Sep 19 2009
a(n) = 2*A053404(n) - A053404(n-1), n > 0. - Ralf Stephan, Jul 21 2013

A317793 a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2.

Original entry on oeis.org

1, 15, 22, 177, 406, 2445, 7162, 36177, 121486, 554325, 2009602, 8656377, 32761366, 136617405, 529712842, 2169039777, 8525430046, 34553579685, 136858084882, 551499730377, 2193794127526, 8811785649165, 35137304693722, 140878711512177, 562526325893806

Offset: 1

Jinyuan Wang, Aug 07 2018

This sequence is an extension of A014551; the sequences A014551(n) = 2^n + (-1)^n, a(n) = 4^n + (-3)^n + 2^n + (-1)^n and b(n) = 6^n + (-5)^n + 4^n + (-3)^n + 2^n + (-1)^n, ... can be considered to be of the same type.

For k>0, a(4k-2)/5, a(2k)/3 and a(2k+1)/2 are integers.

Index entries for linear recurrences with constant coefficients, signature (2,13,-14,-24).

Cf. A014551, A087451.

[(4^n+(-3)^n+2^n+(-1)^n)/2: n in [1..30]]; // Vincenzo Librandi, Aug 08 2018

CoefficientList[ Series[(-48x^3 - 21x^2 + 13x + 1)/(24x^4 + 14x^3 - 13x^2 - 2x + 1), {x, 0, 25}], x]  (* or *)LinearRecurrence[{2, 13, -14, -24}, {1, 15, 22, 177}, 26] (* Robert G. Wilson v, Aug 07 2018 *)

Vec(x*(1 + 13*x - 21*x^2 - 48*x^3) / ((1 + x)*(1 - 2*x)*(1 + 3*x)*(1 - 4*x)) + O(x^40)) \\ Colin Barker, Aug 07 2018

a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2 for n > 0.
From Colin Barker, Aug 07 2018: (Start)
G.f.: x*(1 + 13*x - 21*x^2 - 48*x^3) / ((1 + x)*(1 - 2*x)*(1 + 3*x)*(1 - 4*x)).
a(n) = 2*a(n-1) + 13*a(n-2) - 14*a(n-3) - 24*a(n-4) for n>4.
(End)
E.g.f.: (cosh(3*x/2) + cosh(7*x/2))*(cosh(x/2) + sinh(x/2)) - 2. - Stefano Spezia, Mar 20 2022

More terms from Colin Barker, Aug 07 2018

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A015441 Generalized Fibonacci numbers.

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A201455 a(n) = 3*a(n-1) + 4*a(n-2) for n>1, a(0)=2, a(1)=3.

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A283653 Numbers k such that 3^k + (-2)^k is prime.

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A140796 a(n)=a(n-1)+6a(n-2), n>2.

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A087452 G.f.: (2-x)/((1+3x)(1-4x)); e.g.f.: exp(4x) + exp(-3x); a(n) = 4^n + (-3)^n.

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A317793 a(n) = (4^n + (-3)^n + 2^n + (-1)^n)/2.

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A201455 a(n) = 3a(n-1) + 4a(n-2) for n>1, a(0)=2, a(1)=3.