A086344 -id:A086344 - cp's OEIS frontend

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

1, 1, 5, 21, 89, 377, 1597, 6765, 28657, 121393, 514229, 2178309, 9227465, 39088169, 165580141, 701408733, 2971215073, 12586269025, 53316291173, 225851433717, 956722026041, 4052739537881, 17167680177565, 72723460248141, 308061521170129, 1304969544928657, 5527939700884757

Offset: 0

Olivier Gérard

If one deletes the leading 0 in A084326, takes the inverse binomial transform, and adds a(0)=1 in front, one obtains this sequence here. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
For n >= 1, row sums of triangle
  m |k=0     1     2     3     4     5     6     7
====+=============================================
  0 |  1
  1 |  1     4
  2 |  1     4    16
  3 |  1     8    16    64
  4 |  1     8    48    64   256
  5 |  1    12    48   256   256  1024
  6 |  1    12    96   256  1280  1024  4096
  7 |  1    16    96   640  1280  6144  4096 16384
which is triangle for numbers 4^k*C(m,k) with duplicated diagonals. - Vladimir Shevelev, Apr 12 2012
a(n) = a(n;-2) = 3^n*Sum_{k=0..n} binomial(n,k)*F(k+1)*(-2/3)^k, where a(n;d), n=0,1,...,d, denotes the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). We note that (see A033887) F(3n+1) = 3^n*a(n,2/3) = Sum_{k=0..n} binomial(n,k)*F(k-1)*(-2/3)^k, which implies F(3n+1) + 3^(-n)*a(n) = Sum_{k=0..n} binomial(n,k)*L(k)*(-2/3)^k, where L(k) denotes the k-th Lucas number. - Roman Witula, Jul 12 2012
a(n+1) is (for n >= 0) the number of length-n strings of 5 letters {0,1,2,3,4} with no two adjacent nonzero letters identical. The general case (strings of L letters) is the sequence with g.f. (1+x)/(1-(L-1)*x-x^2). - Joerg Arndt, Oct 11 2012
Starting with offset 1 the sequence is the INVERT transform of (1, 4, 4*3, 4*3^2, 4*3^3, ...); i.e., of A003946: (1, 4, 12, 36, 108, ...). - Gary W. Adamson, Aug 06 2016
a(n+1) equals the number of quinary sequences of length n such that no two consecutive terms differ by 3. - David Nacin, May 31 2017

T. D. Noe, Table of n, a(n) for n = 0..200
Joerg Arndt, Matters Computational (The Fxtbook), pp. 313-315.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (see Corollary 1 (vi)).
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
Amelia Gilson, Hadley Killen, Steven J. Miller, Nadia Razek, Joshua M. Siktar, and Liza Sulkin, Zeckendorf's Theorem Using Indices in an Arithmetic Progression, arXiv:2005.10396 [math.NT], 2020.
Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Bahar Kuloğlu, Engin Özkan, and Marin Marin, Fibonacci and Lucas Polynomials in n-gon, An. Şt. Univ. Ovidius Constanţa (Romania 2023) Vol. 31, No 2, 127-140.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Math. 46 (2013), 15-27.
R. Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A001076, A147722 (INVERT transform), A109499 (INVERTi transform), A154626 (Binomial transform), A086344 (inverse binomial transform), A003946, A049310.

Cf. A000032, A000045, A014445, A014448, A033887, A046854, A055830, A055870, A084326, A167808.

[Fibonacci(3*n-1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

with(combinat): a:=n->fibonacci(n,4)-3*fibonacci(n-1,4): seq(a(n), n=1..23); # Zerinvary Lajos, Apr 04 2008

Fibonacci/@(3*Range[0,30]-1) (* Vladimir Joseph Stephan Orlovsky, Mar 01 2010 *)
LinearRecurrence[{4,1},{1,1},30] (* Harvey P. Dale, May 15 2019 *)

a[0]:1$
a[1]:1$
a[n]:=4*a[n-1]+a[n-2]$
A015448(n):=a[n]$
makelist(A015448(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

a(n) = fibonacci(3*n-1); \\ Altug Alkan, Dec 10 2015

a(n) = Fibonacci(3n-1) = ( (1+sqrt(5))*(2-sqrt(5))^n - (1-sqrt(5))*(2+sqrt(5))^n )/ (2*sqrt(5)).
O.g.f.: (1-3*x)/(1-4*x-x^2). - Len Smiley, Dec 09 2001
a(n) = Sum_{k=0..n} 3^k*A055830(n,k). - Philippe Deléham, Oct 18 2006
a(n) = upper left term in the 2 X 2 matrix [1,2; 2,3]^n. - Gary W. Adamson, Mar 02 2008
[a(n), A001076(n)] = [1,4; 1,3]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009
a(n) = Fibonacci(3n+1) mod Fibonacci(3n), n > 0.
a(n) = (A000032(3*n)-Fibonacci(3*n))/2 = (A014448(n)-A014445(n))/2.
For n >= 2, a(n) = F_n(4) + F_(n+1)(4), where F_n(x) is a Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n-1)/2)} binomial(n-i-1,i)*x^(n-2*i-1). - Vladimir Shevelev, Apr 13 2012
a(n) = A001076(n+1) - 3*A001076(n). - R. J. Mathar, Jul 12 2012
From Gary Detlefs and Wolfdieter Lang, Aug 20 2012: (Start)
a(n) = (5*F(n)^3 + 5*F(n-1)^3 + 3*(-1)^n*F(n-2))/2,
a(n) = (F(n+1)^3 + 2*F(n)^3 - F(n-2)^3)/2, n >= 0, with F(-1) = 1 and F(-2) = -1. Second line from first one with 3*(-1)^n* F(n-2) = F(n-1)^3 - 4*F(n-2)^3 - F(n-3)^3 (in Koshy's book, p. 89, 32. (with a - sign) and 33. For the Koshy reference see A000045) and the F^3 recurrence (see row n=4 of A055870, or Koshy p. 87, 1.). First line from the preceding R. J. Mathar formula with F(3*n) = 5*F(n)^3 + 3*(-1)^n*F(n) (Koshy p. 89, 46.) and the above mentioned formula, Koshy's 32. and 33., with n -> n+2 in order to eliminate F(n+1)^3. (End)
For n > 0, a(n) = L(n-1)*L(n)*F(n) + F(n+1)*(-1)^n with L(n)=A000032(n). - J. M. Bergot, Dec 10 2015
For n > 1, a(n)^2 is the denominator of continued fraction [4,4,...,4, 6, 4,4,...4], which has n-1 4's before, and n-1 4's after, the middle 6. - Greg Dresden, Sep 18 2019
From Gary Detlefs and Wolfdieter Lang, Mar 06 2023: (Start)
a(n) = A001076(n) + A001076(n-1), with A001076(-1) = 1. See the R. J. Mathar formula above.
a(n+1) = i^n*(S(n-1,-4*i) - i*S(n-2,-4*i)), for n >= 0, with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified Fibonacci trisection formula for {F(3*n+2)}_{n>=0}. (End)
a(n) = Sum_{k=0..n} A046854(n-1,k)*4^k. - R. J. Mathar, Feb 10 2024
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) - sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, Jun 03 2024

A103435 a(n) = 2^n * Fibonacci(n).

Original entry on oeis.org

0, 2, 4, 16, 48, 160, 512, 1664, 5376, 17408, 56320, 182272, 589824, 1908736, 6176768, 19988480, 64684032, 209321984, 677380096, 2192048128, 7093616640, 22955425792, 74285318144, 240392339456, 777925951488, 2517421260800

Offset: 0

Ralf Stephan, Feb 08 2005

Cardinality of set of bracelets of size at most n that are tiled with two types of colored squares and four types of colored dominoes.
a(n) is also the diagonal element of the matrix A(i,j) whose first row (i=1) and first column (j=1) are the Fibonacci numbers: A(1,k)=A(k,1)=fib(k) and whose generic element is the sum of element in adjacent (preceding) row and column minus the absolute value of their difference. So a(n) = A(n,n) = A(i-1,j)+A(i,j-1)-abs(A(i-1,j)-A(i,j-1)). - Carmine Suriano, May 13 2010
a(n) is the coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1.  The constant terms under this reduction are the absolute values of terms of A086344.  See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011
The exponential convolution of A000032 and A000045. - Vladimir Reshetnikov, Oct 06 2016

			a(5)=160=A(5,5)=A(4,5)+A(5,4)-abs[A(4,5)+A(5,4)]=80+80-0. - _Carmine Suriano_, May 13 2010
G.f. = 2*x + 4*x^2 + 16*x^3 + 48*x^4 + 160*x^5 + 512*x^6 + 1664*x^7 + ...

Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 236, p. 131.

Tom Edgar, Extending Some Fibonacci-Lucas Relations, The Fibonacci Quarterly, Vol. 54, No. 1 (2016), p. 79.
Harris Kwong, An Alternate Proof of Sury's Fibonacci-Lucas Relation, The American Mathematical Monthly, Vol. 121, No. 6 (2014), p. 514.
Diego Marques, A new Fibonacci-Lucas relation, Amer. Math. Monthly, Vol. 122, No. 7 (2015), p. 683.
Ivica Martinjak and Helmut Prodinger, Complementary Families of the Fibonacci-Lucas Relations, arXiv:1508.04949 [math.CO], 2015-2016.
B. Sury, A polynomial parent to a Fibonacci-Lucas relations, Amer. Math. Monthly, Vol. 121, No. 3 (2014), p. 236.
Charles R. Wall, Problem B-607, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 25, No. 4 (1987), p. 370; Product of Exponential Generating Functions, Solution to Problem B-607 by Bob Prielipp, ibid., Vol. 26, No. 4 (1988), pp. 374-375.
Index entries for linear recurrences with constant coefficients, signature (2,4).

First differences of A014334.
Partial sums of A087131.
Cf. A000032, A000045, A006483, A063727, A085449, A269991.

[2^n *Fibonacci(n): n in [0..50]]; // Vincenzo Librandi, Apr 04 2011

Expand[Table[((1 + Sqrt[5])^n - (1 - Sqrt[5])^n)5/(5 Sqrt[5]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
Table[2^n Fibonacci[n],{n,0,40}] (* or *) LinearRecurrence[{2,4},{0,2},40] (* Harvey P. Dale, Oct 14 2020 *)

a(n)=fibonacci(n)<Charles R Greathouse IV, Feb 03 2014

concat(0, Vec(2*x/(1-2*x-4*x^2) + O(x^99))) \\ Altug Alkan, May 11 2016

a(n) = A006483(n) + 1 = 2*A085449(n) = 2*A063727(n-1), n>0.
G.f.: 2*x / (1 - 2*x - 4*x^2).
a(n) = Sum_{i=0..n-1}( 2^i * Lucas(i) ).
a(n) = 2*a(n-1) + 4*a(n-2). - Carmine Suriano, May 13 2010
a(n) = a(-n) * -(-4)^n for all n in Z. - Michael Somos, Sep 20 2014
E.g.f.: 2*sinh(sqrt(5)*x)*exp(x)/sqrt(5). - Ilya Gutkovskiy, May 10 2016
Sum_{n>=1} 1/a(n) = (1/2) * A269991. - Amiram Eldar, Nov 17 2020
a(n) == 2*n (mod 10). - Amiram Eldar, Jan 15 2022
a(n) = Sum_{k=0..n} binomial(n,k) * Fibonacci(k) * Lucas(n-k) (Wall, 1987). - Amiram Eldar, Jan 27 2022

A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.

Original entry on oeis.org

0, 4, 8, 32, 96, 320, 1024, 3328, 10752, 34816, 112640, 364544, 1179648, 3817472, 12353536, 39976960, 129368064, 418643968, 1354760192, 4384096256, 14187233280, 45910851584, 148570636288, 480784678912, 1555851902976, 5034842521600, 16293092655104

Offset: 0

Seiichi Kirikami, Mar 06 2012

a(n)/A063727(n) are convergents for A134972.

Abs(Sum_{i=0..n} C(n,n-i)*a(i)-(sqrt(5)-1)* A033887(n))->0. - Seiichi Kirikami, Jan 20 2016

E. W. Cheney, Introduction to Approximation Theory, McGraw-Hill, Inc., 1966.

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A063727, A033887, A085449, A103435, A134972, A269991.

Cf. A086344 (this sequence with signs).

I:=[0,4]; [n le 2 select I[n] else 2*Self(n-1)+4*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 16 2016

RecurrenceTable[{a[n]==2*a[n-1]+4*a[n-2], a[0]==0, a[1]==4}, a, {n, 30}]
LinearRecurrence[{2, 4}, {0, 4}, 40] (* Vincenzo Librandi, Jan 16 2016 *)

concat(0, Vec(4*x/(1-2*x-4*x^2) + O(x^40))) \\ Michel Marcus, Jan 16 2016

a(n) = (2/sqrt(5))*((1+sqrt(5))^n-(1-sqrt(5))^n).
G.f.: 4*x/(1-2*x-4*x^2). - Bruno Berselli, Mar 08 2012
a(n) = 4*A085449(n) = 2*A103435(n). - Bruno Berselli, Mar 08 2012
Sum_{n>=1} 1/a(n) = (1/4) * A269991. - Amiram Eldar, Feb 01 2021

A208459 Triangle T_x = T(n,k) given by (0, 1/x, 1-1/x, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (x, 1/x-1, -1/x, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938, for x = 0.

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 0, 1, 0, 2, 0, -3, 0, 1, 0, 3, -1, 0, 5, 0, 1, 0, 4, -2, 3, 2, -8, 0, 1, 0, 5, -3, 7, -2, -5, 13, 0, 1, 0, 6, -4, 12, -8, 2, 12, -21, 0, 1, 0, 7, -5, 18, -16, 15, 3, -25, 34

Offset: 0

Philippe Deléham, Feb 27 2012

Triangle T_x : T_1 = A103631, T_2 = A208343, T_3 = A208345.

			Triangle begins :
1
0, 0
0, 1, 1
0, 1, 0, -1
0, 1, 0, 1, 2
0, 1, 0, 2, 0, -3
0, 1, 0, 3, -1, 0, 5
0, 1, 0, 4, -2, 3, 2, -8
0, 1, 0, 5, -3, 7, -2, -5, 13
0, 1, 0, 6, -4, 12, -8, 2, 12, -21
0, 1, 0, 7, -5, 18, -16, 15, 3, -25, 34

Cf. A103631, A208343, A208345, A000045 (Fibonacci)

T(n,k) = T(n-1,k) - T(n-1,k-1) + T(n-2,k-1) + T(n-2,k-2) with T(0,0) = 1 T(1,0) = 0, T(1,1) = 0, T(n,k) = 0 if k<0 or if k>n.
G.f.: (1-x+y*x)/(1-x+y*x- y^2*x^2-y*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = 12*A015548(n-1), 6*A085939(n-1), A106434(n), A000007(n), A000007(n), A077957(n), (-1)^n*A102901(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively.
Sm_{k, 0<=k<=n} T(n,k)*x^(n-k) = A000007(n), A034834(n-1), A077957(n), A052533(n), (-1)^n*A086344(n) for x = -1, 0, 1, 2, 3 respectively.

cp's OEIS Frontend

A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2.

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A103435 a(n) = 2^n * Fibonacci(n).

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A209084 a(n) = 2*a(n-1) + 4*a(n-2) with n>1, a(0)=0, a(1)=4.

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A208459 Triangle T_x = T(n,k) given by (0, 1/x, 1-1/x, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (x, 1/x-1, -1/x, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938, for x = 0.

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A209084 a(n) = 2a(n-1) + 4a(n-2) with n>1, a(0)=0, a(1)=4.