login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A083099
a(n) = 2*a(n-1) + 6*a(n-2), a(0) = 0, a(1) = 1.
32
0, 1, 2, 10, 32, 124, 440, 1624, 5888, 21520, 78368, 285856, 1041920, 3798976, 13849472, 50492800, 184082432, 671121664, 2446737920, 8920205824, 32520839168, 118562913280, 432250861568, 1575879202816, 5745263575040
OFFSET
0,3
COMMENTS
a(n+1) = a(n) + A083098(n+1). A083098(n+1)/a(n) converges to sqrt(7).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 7 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(7). - Cino Hilliard, Sep 25 2005
Pisano period lengths: 1, 1, 2, 1, 12, 2, 7, 1, 6, 12, 60, 2,168, 7, 12, 1,288, 6, 18, 12, ... - R. J. Mathar, Aug 10 2012
a(n) is divisible by 2^ceiling(n/2), see formula below. - Ralf Stephan, Dec 24 2013
Connect the center of a regular hexagon with side length 1 with its six vertices. a(n) is the number of paths of length n from the center to any of its vertices. Number of paths of length n from the center to itself is 6*a(n-1). - Jianing Song, Apr 20 2019
REFERENCES
John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
LINKS
Project Euler, Problem 752, sequence beta(n).
FORMULA
G.f.: x/(1 - 2*x - 6*x^2).
From Paul Barry, Sep 29 2004: (Start)
E.g.f.: (d/dx)(exp(x)*sinh(sqrt(7)*x)/sqrt(7));
a(n-1) = Sum_{k=0..n} binomial(n, 2k+1)*7^k. (End)
Simplified formula: a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(28). - Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(7*k-1)/(x*(7*k+6) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(2n) = 2^n * A154245(n), a(2n+1) = 2^n * (5*A154245(n) - 9*A154245(n-1)). - Ralf Stephan, Dec 24 2013
a(n) = Sum_{k=1,3,5,...<=n} binomial(n,k)*7^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(n) = i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - G. C. Greubel, Jun 01 2023
MAPLE
A083099 := proc(n)
option remember;
if n <= 1 then
n;
else
2*procname(n-1)+6*procname(n-2) ;
end if;
end proc: # R. J. Mathar, Sep 23 2016
MATHEMATICA
CoefficientList[Series[x/(1-2x-6x^2), {x, 0, 25}], x] (* Adapted for offset 0 by Vincenzo Librandi, Feb 07 2014 *)
Expand[Table[((1 + Sqrt[7])^n - (1 - Sqrt[7])^n)7/(14Sqrt[7]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
LinearRecurrence[{2, 6}, {0, 1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)
PROG
(Sage) [lucas_number1(n, 2, -6) for n in range(0, 25)] # Zerinvary Lajos, Apr 22 2009
(PARI) a(n)=([0, 1; 6, 2]^n*[0; 1])[1, 1] \\ Charles R Greathouse IV, May 10 2016
(PARI) my(x='x+O('x^30)); concat([0], Vec(x/(1-2*x-6*x^2))) \\ G. C. Greubel, Jan 24 2018
(Magma) [n le 2 select n-1 else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
(SageMath)
A083099=BinaryRecurrenceSequence(2, 6, 0, 1)
[A083099(n) for n in range(41)] # G. C. Greubel, Jun 01 2023
CROSSREFS
The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.
Sequence in context: A034555 A084154 A265836 * A032095 A328039 A264960
KEYWORD
nonn,easy
AUTHOR
Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
STATUS
approved