A122189 Heptanacci numbers: each term is the sum of the preceding 7 terms, with a(0),...,a(6) = 0,0,0,0,0,0,1.
0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, 1004, 2000, 3984, 7936, 15808, 31489, 62725, 124946, 248888, 495776, 987568, 1967200, 3918592, 7805695, 15548665, 30972384, 61695880, 122895984, 244804400, 487641600, 971364608, 1934923521
Offset: 0
Links
- Robert Price, Table of n, a(n) for n = 0..1000
- Tomás Aguilar-Fraga, Jennifer Elder, Rebecca E. Garcia, Kimberly P. Hadaway, Pamela E. Harris, Kimberly J. Harry, Imhotep B. Hogan, Jakeyl Johnson, Jan Kretschmann, Kobe Lawson-Chavanu, J. Carlos Martínez Mori, Casandra D. Monroe, Daniel Quiñonez, Dirk Tolson III, and Dwight Anderson Williams II, Interval and L-interval Rational Parking Functions, arXiv:2311.14055 [math.CO], 2023. See p. 14.
- 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.
- Taras Goy and Mark Shattuck, Some Toeplitz-Hessenberg Determinant Identities for the Tetranacci Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.6.8.
- Tian-Xiao He, Impulse Response Sequences and Construction of Number Sequence Identities, J. Int. Seq. 16 (2013) #13.8.2.
- F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, Fibonacci Quart. 49 (2011), no. 3, 231-243.
- Omar Khadir, László Németh, and László Szalay, Tiling of dominoes with ranked colors, Results in Math. (2024) Vol. 79, Art. No. 253. See p. 2.
- Benjamin E. Merkel, Probabilities of Consecutive Events in Coin Flipping, Master's Thesis, Univ. Cincinatti, May 11 2011.
- László Németh and László Szalay, Explicit solution of system of two higher-order recurrences, arXiv:2408.12196 [math.NT], 2024. See p. 10.
- Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1).
Crossrefs
Programs
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Maple
for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i),i=0..floor((n-6)/8))-sum((-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i),i=0..floor((n-7)/8)):od:seq(k(n),n=0..50); a:=taylor((z^6-z^7)/(1-2*z+z^8),z=0,51);for p from 0 to 50 do j(p):=coeff(a,z,p):od :seq(j(p),p=0..50); # Richard Choulet, Feb 22 2010
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Mathematica
LinearRecurrence[{1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *) a={0,0,0,0,0,0,1} For[n=7, n≤100, n++, sum=Plus@@a; Print[sum]; a=RotateLeft[a]; a[[7]]=sum] (* Robert Price, Dec 04 2014 *) -
PARI
a(n)=([0,1,0,0,0,0,0; 0,0,1,0,0,0,0; 0,0,0,1,0,0,0; 0,0,0,0,1,0,0; 0,0,0,0,0,1,0; 0,0,0,0,0,0,1; 1,1,1,1,1,1,1]^n*[0;0;0;0;0;0;1])[1,1] \\ Charles R Greathouse IV, Jun 20 2015
Formula
G.f.: x^6/(1-x-x^2-x^3-x^4-x^5-x^6-x^7). - R. J. Mathar, Feb 13 2009
G.f.: Sum_{n >= 0} x^(n+5) * [ Product_{k = 1..n} (k + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + x^6)/(1 + k*x + k*x^2 + k*x^3 + k*x^4 + k*x^5 + k*x^6) ]. - Peter Bala, Jan 04 2015
Another form of the g.f.: f(z) = (z^6-z^7)/(1-2*z+z^8), then a(n) = Sum_{i=0..floor((n-6)/8)} (-1)^i*binomial(n-6-7*i,i)*2^(n-6-8*i) - Sum_{i=0..floor((n-7)/8)} (-1)^i*binomial(n-7-7*i,i)*2^(n-7-8*i) with Sum_{i=m..n} alpha(i) = 0 for m>n. - Richard Choulet, Feb 22 2010
Sum_{k=0..6*n} a(k+b)*A063265(n,k) = a(7*n+b), b>=0.
a(n) = 2*a(n-1) - a(n-8). - Joerg Arndt, Sep 24 2020
Extensions
Edited by N. J. A. Sloane, Nov 20 2007
Wrong Binet-type formula removed by R. J. Mathar, Feb 13 2009
Comments