A023424 -id:A023424 - cp's OEIS frontend

A127221 a(n) = 2^npentanacci(n) or (2^n)A023424(n-1).

2, 12, 56, 240, 992, 3648, 14464, 57088, 224768, 883712, 3471360, 13651968, 53682176, 211075072, 829915136, 3263102976, 12830244864, 50447253504, 198353354752, 779904614400, 3066503888896, 12057176965120, 47407572189184, 186401664532480, 732912043425792

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,4,8,16,32).

Cf. A087131, A127210, A127211, A127212, A127213, A127214, A127216, A001648, A127220, A127222.

Cf. A023424, A074048.

I:=[2, 12, 56, 240, 992]; [n le 5 select I[n] else 2*Self(n-1) + 4*Self(n-2) + 8*Self(n-3) + 16*Self(n-4) + 32*Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 19 2017

Table[Tr[MatrixPower[2*{{1, 1, 1, 1, 1}, {1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{2, 4, 8, 16, 32}, {2, 12, 56, 240, 992}, 50] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec(-2*x*(1 +4*x +12*x^2 +32*x^3 +80*x^4)/(-1 +2*x +4*x^2 +8*x^3 +16*x^4 +32*x^5)) \\ G. C. Greubel, Dec 19 2017

a(n) = Trace of matrix [({2,2,2,2,2},{2,0,0,0,0},{0,2,0,0,0},{0,0,2,0,0},{0,0,0,2,0})^n].
a(n) = 2^n * Trace of matrix [({1,1,1,1,1},{1,0,0,0,0},{0,1,0,0,0},{0,0,1,0,0},{0,0,0,1,0})^n].
G.f.: -2*x*(1 +4*x +12*x^2 +32*x^3 +80*x^4)/(-1 +2*x +4*x^2 +8*x^3 +16*x^4 +32*x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; corrected by R. J. Mathar, Sep 16 2009
a(n) = 2*a(n-1)+4*a(n-2)+8*a(n-3)+16*a(n-4)+32*a(n-5). - Colin Barker, Sep 02 2013

Definition corrected by R. J. Mathar, Sep 17 2009

More terms from Colin Barker, Sep 02 2013

A127222 a(n) = 3^npentanacci(n) or (3^n)A023424(n-1).

Original entry on oeis.org

3, 27, 189, 1215, 7533, 41553, 247131, 1463103, 8640837, 50959287, 300264165, 1771292853, 10447598619, 61618989627, 363414767589, 2143339285311, 12641143135581, 74555586323649, 439717218548643, 2593383067853775, 15295369041550269, 90209719910309895

Offset: 1

Artur Jasinski, Jan 09 2007

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,9,27,81,243).

Cf. A087131, A127210, A127211, A127212, A127213, A127214, A127216, A001648, A127220, A127221.

Cf. A023424, A074048.

I:=[3, 27, 189, 1215, 7533]; [n le 5 select I[n] else 3*Self(n-1) + 9*Self(n-2) + 27*Self(n-3) + 81*Self(n-4) + 243*Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 19 2017

Table[Tr[MatrixPower[3*{{1, 1, 1, 1, 1}, {1, 0, 0, 0, 0}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}}, x]], {x, 1, 20}]
LinearRecurrence[{3, 9, 27, 81, 243}, {3, 27, 189, 1215, 7533}, 50] (* G. C. Greubel, Dec 19 2017 *)

x='x+O('x^30); Vec(-3*x*(1 +6*x +27*x^2 +108*x^3 +405*x^4)/(-1 +3*x +9*x^2 +27*x^3 +81*x^4 +243*x^5)) \\ G. C. Greubel, Dec 19 2017

a(n) = Trace of matrix [({3,3,3,3,3},{3,0,0,0,0},{0,3,0,0,0},{0,0,3,0,0},{0,0,0,3,0})^n].
a(n) = 3^n * Trace of matrix [({1,1,1,1,1},{1,0,0,0,0},{0,1,0,0,0},{0,0,1,0,0},{0,0,0,1,0})^n].
G.f.: -3*x*(1 +6*x +27*x^2 +108*x^3 +405*x^4)/(-1 +3*x +9*x^2 +27*x^3 +81*x^4 +243*x^5). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 28 2009
a(n) = 3*a(n-1)+9*a(n-2)+27*a(n-3)+81*a(n-4)+243*a(n-5). - Colin Barker, Sep 02 2013

G.f. proposed by Maksym Voznyy checked and corrected by R. J. Mathar, Sep 16 2009
Definition corrected by R. J. Mathar, Sep 17 2009
More terms from Colin Barker, Sep 02 2013

A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15.

Original entry on oeis.org

5, 1, 3, 7, 15, 31, 57, 113, 223, 439, 863, 1695, 3333, 6553, 12883, 25327, 49791, 97887, 192441, 378329, 743775, 1462223, 2874655, 5651423, 11110405, 21842481, 42941187, 84420151, 165965647, 326279871, 641449337, 1261056193, 2479171199, 4873922247

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Aug 14 2002

These pentanacci numbers follow the same pattern as Lucas, generalized tribonacci(A001644) and generalized tetranacci (A073817) numbers: Binet's formula is a(n)=r1^n+r^2^n+r3^n+r4^n+r5^n, with r1, r2, r3, r4, r5 roots of the characteristic polynomial. a(n) is also the trace of A^n, where A is the pentamatrix ((1,1,0,0,0),(1,0,1,0,0),(1,0,0,1,0),(1,0,0,0,1),(1,0,0,0,0)).
For n >= 5, a(n) is the number of cyclic sequences consisting of n zeros and ones that do not contain five consecutive ones provided the positions of the zeros and ones are fixed on a circle. This is proved in Charalambides (1991) and Zhang and Hadjicostas (2015). (For n=1,2,3,4 the statement is still true provided we allow the sequence to wrap around itself on a circle). - Petros Hadjicostas, Dec 18 2016
a(3407) has 1001 decimal digits. - Michael De Vlieger, Dec 28 2016

Michael De Vlieger, Table of n, a(n) for n = 0..3406
Martin Burtscher, Igor Szczyrba, and Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
Spiros D. Dafnis, Andreas N. Philippou, and Ioannis E. Livieris, An Alternating Sum of Fibonacci and Lucas Numbers of Order k, Mathematics (2020) Vol. 9, 1487.
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
S. Saito, T. Tanaka, and N. Wakabayashi, Combinatorial Remarks on the Cyclic Sum Formula for Multiple Zeta Values, Journal of Integer Sequences, 14 (2011), # 11.2.4, Table 3.
Yüksel Soykan, On A Generalized Pentanacci Sequence, Asian Research Journal of Mathematics (2019) Vol. 14, No. 3, 1-9.
Yüksel Soykan, Sum Formulas for Generalized Fifth-Order Linear Recurrence Sequences, Journal of Advances in Mathematics and Computer Science (2019) Vol. 34, No. 5, 1-14.
Eric Weisstein's World of Mathematics, Fibonacci n-Step Number.
L. Zhang and P. Hadjicostas, On sequences of independent Bernoulli trials avoiding the pattern '11..1', Math. Scientist, 40 (2015), 89-96.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Cf. A000078, A001630, A001644, A000032, A073817, A106297 (Pisano Periods).
Essentially the same as A023424.
Cf. A123126, A123127, A074062.
Cf. A106273.

CoefficientList[Series[(5-4*x-3*x^2-2*x^3-x^4)/(1-x-x^2-x^3-x^4-x^5), {x, 0, 30}], x]
LinearRecurrence[{1, 1, 1, 1, 1}, {5, 1, 3, 7, 15}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

polsym(polrecip(1-x-x^2-x^3-x^4-x^5),33) \\ Joerg Arndt, Jan 28 2019

a(n) = a(n-1) +a(n-2) +a(n-3) +a(n-4) +a(n-5).
G.f.: (5-4*x-3*x^2-2*x^3-x^4) / (1-x-x^2-x^3-x^4-x^5).
a(n) = 2*a(n-1) -a(n-6), n>5. [Vincenzo Librandi, Dec 20 2010]
For k>0 and n>=0, a(n+5*k) = a(k)*a(n+4*k) - A123127(k-1)*a(n+3*k) + A123126(k-1)*a(n+2*k) - A074062(k)*a(n+k) + a(n). For example, if k=4, n=3, we have a(n+5*k) = a(23) = 5651423, a(4)*a(19) - A123127(3)*a(15) + A123126(3)*a(1695) - A074062(4)*a(7) + a(3) = (15)*(378329) - (1)*(25327) + (1)*(1695) - (-1)*(113) + (7) = 5651423. - Kai Wang, Sep 13 2020
From Kai Wang, Dec 16 2020: (Start)
For k >= 0,
    | a(k+4) a(k+5) a(k+6) a(k+7) a(k+8) |
    | a(k+3) a(k+4) a(k+5) a(k+6) a(k+7) |
det | a(k+2) a(k+3) a(k+4) a(k+5) a(k+6) | = 9584 = A106273(5).
    | a(k+1) a(k+2) a(k+3) a(k+4) a(k+5) |
    | a(k)   a(k+1) a(k+2) a(k+3) a(k+4) |
(End)

A251653 5-step Fibonacci sequence starting with 0,0,1,0,0.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 2, 4, 7, 14, 28, 55, 108, 212, 417, 820, 1612, 3169, 6230, 12248, 24079, 47338, 93064, 182959, 359688, 707128, 1390177, 2733016, 5372968, 10562977, 20766266, 40825404, 80260631, 157788246, 310203524, 609844071, 1198921876, 2357018348, 4633776065, 9109763884, 17909324244, 35208804417

Offset: 0

Arie Bos, Dec 06 2014

Doubling the entries > 1 as 1, 2, 2, 4, 4, 7, 7, 14, 14, 28, 28, 55, 55,... (offset 0) gives Nyblom's palindromic binary strings having no 5-runs of 1's. - R. J. Mathar, Mar 28 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
M. A. Nyblom, Counting Palindromic Binary Strings Without r-Runs of Ones, J. Int. Seq. 16 (2013) #13.8.7, P_5(n)
H. Prodinger, Counting Palindromes According to r-Runs of Ones Using Generating Functions, J. Int. Seq. 17 (2014) # 14.6.2, even length, r=4.
Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Other 5-step sequences are A000322, A001591, A023424, A074048, A124312, A124313, A122997, A135055, A135056, A145029

(see www.jsoftware.com) First construct the generating matrix
1  1  1  1  1
1  2  2  2  2
2  3  4  4  4
4  6  7  8  8
8 12 14 15 16
Given that matrix one can produce the first 5*200 numbers by
, M(+/ . *)^:(i.250) 0 0 1 0 0x

LinearRecurrence[{1, 1, 1, 1, 1}, {0, 0, 1, 0, 0}, 100] (* G. C. Greubel, May 27 2016 *)

a(n+5) = a(n) + a(n+1) + a(n+2) + a(n+3) + a(n+4).

G.f.: x^2*(x^2 + x - 1)/(x^5 + x^4 + x^3 + x^2 + x - 1). - Chai Wah Wu, May 27 2016

cp's OEIS Frontend

A127221 a(n) = 2^n*pentanacci(n) or (2^n)*A023424(n-1).

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A127222 a(n) = 3^n*pentanacci(n) or (3^n)*A023424(n-1).

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A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15.

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A251653 5-step Fibonacci sequence starting with 0,0,1,0,0.

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A127221 a(n) = 2^npentanacci(n) or (2^n)A023424(n-1).

A127222 a(n) = 3^npentanacci(n) or (3^n)A023424(n-1).