OFFSET
0,3
COMMENTS
Binomial transform of expansion of cosh(sqrt(7)x) (A000420 with interpolated zeros: 1, 0, 7, 0, 49, 0, 343, 0, ...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators 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
a(n) is the number of compositions of n when there are 1 type of 1 and 7 types of other natural numbers. - Milan Janjic, Aug 13 2010
REFERENCES
John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000
Project Euler, Problem 752, sequence alpha(n).
Index entries for linear recurrences with constant coefficients, signature (2,6).
FORMULA
G.f.: (1-x)/(1-2*x-6*x^2).
a(n) = (1+sqrt(7))^n/2 + (1-sqrt(7))^n/2.
E.g.f.: exp(x)*cosh(sqrt(7)x).
a(n) = Sum_{k=0..n} A098158(n,k)*7^(n-k). - Philippe Deléham, Dec 26 2007
If p[1]=1, and p[i]=7, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n) = det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, 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
MATHEMATICA
CoefficientList[Series[(1+6x)/(1-2x-6x^2), {x, 0, 25}], x]
LinearRecurrence[{2, 6}, {1, 1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)
a[n_] := Simplify[((1 + Sqrt[7])^n + (1 - Sqrt[7])^n)/2]; Array[a, 25, 0] (* Robert G. Wilson v, Sep 18 2013 *)
PROG
(Sage) [lucas_number2(n, 2, -6)/2 for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009
(PARI) x='x+O('x^30); Vec((1-x)/(1-2*x-6*x^2)) \\ G. C. Greubel, Jan 08 2018
(Magma) I:=[1, 1]; [n le 2 select I[n] else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
STATUS
approved