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A015448 a(0) = 1, a(1) = 1, and a(n) = 4*a(n-1) + a(n-2) for n >= 2. 72
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

a(n) = A167808(3*n-1) for n > 0. - Reinhard Zumkeller, Nov 12 2009

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

LINKS

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)).

Edyta Hetmaniok, Bozena Piatek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Math. 15 (2017), 477-485.

I. M. Gessel, Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.

M. 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

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).

FORMULA

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) = 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

MAPLE

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

MATHEMATICA

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 *)

PROG

(Maxima)

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 */

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

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

CROSSREFS

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

Sequence in context: A240461 A273860 A099843 * A273796 A035011 A113987

Adjacent sequences:  A015445 A015446 A015447 * A015449 A015450 A015451

KEYWORD

nonn,easy

AUTHOR

Olivier Gérard

STATUS

approved

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Last modified October 16 08:45 EDT 2019. Contains 328056 sequences. (Running on oeis4.)