A096951 -id:A096951 - cp's OEIS frontend

A007689 a(n) = 2^n + 3^n.

2, 5, 13, 35, 97, 275, 793, 2315, 6817, 20195, 60073, 179195, 535537, 1602515, 4799353, 14381675, 43112257, 129271235, 387682633, 1162785755, 3487832977, 10462450355, 31385253913, 94151567435, 282446313697, 847322163875

Offset: 0

N. J. A. Sloane, Robert G. Wilson v

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 14.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 92.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
I. Amburg, K. Dasaratha, L. Flapan, T. Garrity, C. Lee, C. Mihailak, N. Neumann-Chun, S. Peluse, and M. Stoffregen, Stern Sequences for a Family of Multidimensional Continued Fractions: TRIP-Stern Sequences, arXiv:1509.05239 [math.CO], 2015-2017.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 169
Index entries for linear recurrences with constant coefficients, signature (5,-6).

For odd-indexed members divided by 5 see A096951.
Binomial transform of A000051.
Cf. A000079, A000244, A001550, A063376, A063481.
Cf. A074600 - A074624, A082101 (primes).

a007689 n = a000079 n + a000244 n  -- Reinhard Zumkeller, Apr 28 2013

[2^n+3^n: n in [0..30]]; // G. C. Greubel, Mar 11 2023

A007689:=n->2^n + 3^n: seq(A007689(n), n=0..50); # Wesley Ivan Hurt, Jan 24 2017

Table[2^n + 3^n, {n, 0, 25}]
a=2;Numerator[Table[a=2*a-((a+1)/2),{n,0,7!}]] (*10 times (or more) faster for large numbers.*) (* Vladimir Joseph Stephan Orlovsky, Apr 19 2010 *)
LinearRecurrence[{5,-6},{2,5},30] (* nearly 20 times faster than the above program for large numbers. *) (* Harvey P. Dale, Oct 20 2013 *)

a(n)=2^n+3^n \\ Charles R Greathouse IV, Jun 15 2011

[lucas_number2(n,5,6)for n in range(0,27)] # Zerinvary Lajos, Jul 08 2008

E.g.f.: exp(2*x)*(1+exp(x)).
G.f.: (2-5*x)/((1-2*x)*(1-3*x)).
a(n) = 5*a(n-1) - 6*a(n-2).
Sum_{j=0..n-1} a(j) = (1/2)*(3^n - 1) + (2^n - 1). [Jolley] - Gary W. Adamson, Dec 20 2006
Equals double binomial transform of [2, 1, 1, 1, ...]. - Gary W. Adamson, Apr 23 2008
If p[i] = Fibonacci(2i-5) and if A is the Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)= det A. - Milan Janjic, May 08 2010
a(n) = 2*a(n-1) + 3^(n-1), with a(0)=2. - Vincenzo Librandi, Nov 18 2010
a(n) = A001550(n) - 1 = A000079(n) + A000244(n). - Reinhard Zumkeller, Mar 01 2012

Additional comments from Michael Somos, Jun 10 2000

A015441 Generalized Fibonacci numbers.

Original entry on oeis.org

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)

A096952 Numerators of upper bounds for Lagrange remainder in Taylor's expansion of log((1+x)/(1-x)) for x=1/3, multiplied by 6/5.

Original entry on oeis.org

1, 7, 11, 463, 4039, 35839, 320503, 575267, 25854247, 232557151, 298927153, 18830313487, 6778577311, 1525146340543, 13726182847159, 123535108753519, 1111813831298023, 2001263178349523, 90056808665990167, 810511140554958031

Offset: 0

Wolfdieter Lang, Jul 16 2004

An upper bound for the Lagrange remainder in the expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2), is R(2*n) = (1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1), n >= 0.
The denominators are found in A096953.
log(2) = 2*Sum_{k>=1} ((1/3)^(2*k-1))/(2*k-1), from the Taylor series of log((1+x)/(1-x)) for x=1/3.

			n=3: R(2*3)=(5/6)* a(3)/A096953(3) = (5/6)*463/326592 = 2315/1959552 = 0.001181..., therefore log(2) - 2*Sum_{k=1..3} ((1/3)^(2*k-1))/(2*k-1) < 0.001181... . In fact, the partial sum is 0.0001430654... .

M. Barner and F. Flohr, Analysis I, de Gruyter, 5. Auflage, 2000; p. 293.

W. Lang, More comments.

a(n) = numerator(A(n)), where A(n) = (6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1) = A096951(n)/((2*n+1)*6^(2*n)).

A138233 a(n) = 2^(2n+1) + 3^(2n+1).

Original entry on oeis.org

5, 35, 275, 2315, 20195, 179195, 1602515, 14381675, 129271235, 1162785755, 10462450355, 94151567435, 847322163875, 7625731702715, 68630914235795, 617675543767595, 5559069156490115, 50031579458738075, 450284043329950835

Offset: 0

Reinhard Zumkeller, Mar 07 2008

nonn

Subsequence of A008587.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (13,-36).

Cf. A008587, A096951, A138199.

[2^(2*n+1) + 3^(2*n+1): n in [0..30]]; // G. C. Greubel, Mar 11 2023

LinearRecurrence[{13, -36},{5, 35},19] (* Ray Chandler, Jul 14 2017 *)
2^#+3^#&/@(2*Range[0,20]+1) (* Harvey P. Dale, Sep 25 2019 *)

[2^(2*n+1) + 3^(2*n+1) for n in range(31)] # G. C. Greubel, Mar 11 2023

a(n) = 5*A096951(n).
a(n+1) = 4*a(n) + 5*3^(2*n+1), a(0) = 5.
O.g.f.: 5*x*(7-36*x)/((1-4*x)*(1-9*x)). - R. J. Mathar, Apr 24 2008
E.g.f.: 2*exp(4*x) + 3*exp(9*x). - G. C. Greubel, Mar 11 2023

A165259 Sum of odd powers of 4 and 9 divided by 13.

Original entry on oeis.org

1, 61, 4621, 369181, 29821741, 2414250301, 195533302861, 15837861987421, 1282861452271981, 103911691734684541, 8416845656119913101, 681764476155480405661, 55222922216750191970221

Offset: 0

Jaume Oliver Lafont, Sep 11 2009

nonn

G. C. Greubel, Table of n, a(n) for n = 0..520
Index entries for linear recurrences with constant coefficients, signature (97,-1296).

Cf. A007689, A082101, A096951.

[(4^(2*n+1)+9^(2*n+1))/13: n in [0..30]]; // G. C. Greubel, Mar 11 2023

Table[(4^n+9^n)/13,{n,1,31,2}] (* or *) LinearRecurrence[{97,-1296},{1,61},20] (* Harvey P. Dale, Jun 23 2013 *)

PARI
```
a(n)=(4^(2*n+1)+9^(2*n+1))/13
```

[(4^(2*n+1)+9^(2*n+1))/13 for n in range(31)] # G. C. Greubel, Mar 11 2023

G.f.: (1-36*x)/((1-16*x)*(1-81*x))
a(n) = 97*a(n-1) - 1296*a(n-2). - Harvey P. Dale, Jun 23 2013
E.g.f.: (1/13)*(4*exp(16*x) + 9*exp(81*x)). - G. C. Greubel, Mar 11 2023

A096953 Denominators of upper bounds for Lagrange-remainder in Taylor's expansion of log((1+x)/(1-x)) multiplied by 6/5.

Original entry on oeis.org

1, 108, 1296, 326592, 15116544, 665127936, 28298170368, 235092492288, 47958868426752, 1929639176699904, 10968475320188928, 3027299188372144128, 4738381338321616896, 4605706660848611622912, 178087324219479649419264, 6853291511342734094893056

Offset: 0

Wolfdieter Lang, Jul 16 2004

An upper bound for the Lagrange-remainder in the expansion of log((1+x)/(1-x)) for x=1/3, i.e., for log(2), is R(2*n):=(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1).

			n=4: R(2*4)=(5/6)* A096952(4)/a(4) = (5/6)*4039/15116544 = 20195/90699264 = 0.0002226589..., therefore log(2)-2*sum(((1/3)^(2*k-1))/(2*k-1),k=1..4) < 0.0002226589... In fact, the partial sum is 0.0000124233...

M. Barner and F. Flohr, Analysis I, de Gruyter, 5te Auflage, 2000; p. 293.

W. Lang, More comments.

Numerators are given in A096952.

[Denominator((6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1)): n in [0..20]]; // Vincenzo Librandi, Jul 06 2015

vector(30, n, n--; denominator((6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1))) \\ Michel Marcus, Jul 06 2015

a(n)=denominator(A(n)), where A(n):=(6/5)*(1/2^(2*n+1) + 1/3^(2*n+1))/(2*n+1) = A096951(n)/((2*n+1)*6^(2*n)).

A165443 a(n) = ( 16^(2n+1) + 81^(2n+1) )/97.

Original entry on oeis.org

1, 5521, 35957041, 235845988561, 1547368082644081, 10152277523461827601, 66609091687940958003121, 437022250271846649679394641, 2867302983958645970747063186161, 18812374877733491600234823630721681

Offset: 0

Jaume Oliver Lafont, Sep 19 2009

The general form of the g.f. for (A^(2*n+1)+B^(2*n+1))/(A+B) is (1-A*B*x)/((1-A^2*x)(1-B^2*x)).

			a(0) = (16^1 + 81^1)/97 = 97/97 = 1.
a(1) = (16^3 + 81^3)/97 = 535537/97 = 5521.

G. C. Greubel, Table of n, a(n) for n = 0..260
Index entries for linear recurrences with constant coefficients, signature (6817,-1679616).

Cf. A007689, A082101, A096951, A165259.

List([0..10],n->(16^(2*n+1)+81^(2*n+1))/97); # Muniru A Asiru, Oct 21 2018

[(2^(8*n+4) + 3^(8*n+4))/97: n in [0..20]]; // G. C. Greubel, Oct 20 2018

seq(coeff(series((1-16*81*x)/((1-16^2*x)*(1-81^2*x)),x,n+1), x, n), n = 0 .. 10); # Muniru A Asiru, Oct 21 2018

f[n_]:=Module[{c=2n+1},(16^c+81^c)/97]; Array[f,20,0] (* Harvey P. Dale, Oct 02 2012 *)

PARI
```
a(n)=(16^(2*n+1)+81^(2*n+1))/97
```

for n in range(0, 10): print(int((16**(2*n+1)+81**(2*n+1))/97), end=', ') # Stefano Spezia, Oct 21 2018

G.f.: (1 - 16*81*x)/((1 - 16^2*x)*(1 - 81^2*x)).

a(n) = (16^2+81^2)*a(n-1) - 16^2*81^2*a(n-2).

Definition replaced with formula by R. J. Mathar, Sep 21 2009

cp's OEIS Frontend

A007689 a(n) = 2^n + 3^n.

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A015441 Generalized Fibonacci numbers.

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A096952 Numerators of upper bounds for Lagrange remainder in Taylor's expansion of log((1+x)/(1-x)) for x=1/3, multiplied by 6/5.

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A138233 a(n) = 2^(2*n+1) + 3^(2*n+1).

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A165259 Sum of odd powers of 4 and 9 divided by 13.

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Magma

Mathematica

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A096953 Denominators of upper bounds for Lagrange-remainder in Taylor's expansion of log((1+x)/(1-x)) multiplied by 6/5.

Views

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Keywords

Comments

Examples

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Links

Crossrefs

Programs

Magma

PARI

Formula

A165443 a(n) = ( 16^(2*n+1) + 81^(2*n+1) )/97.

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A138233 a(n) = 2^(2n+1) + 3^(2n+1).

A165443 a(n) = ( 16^(2n+1) + 81^(2n+1) )/97.