OFFSET
0,2
COMMENTS
Inverse binomial transform of A090040. [Paul Curtz, Jan 11 2009]
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). [Milan Janjic, Feb 21 2010]
For an integer x, consider the sequence P(x) of polynomials p_{1}, p_{2}, p_{3}, . . . defined by p_{1} = x-1, p_{n+1} = x*p_{1} - 1. P(5) = This sequence. P(1), P(2), P(3), P(4) are A000004, A123412, A007051, A007583 resp. [K.V.Iyer, Jun 22 2010]
It appears that if s(n) is a first order rational sequence of the form s(0)=2, s(n)= (3*s(n-1)+2)/(2*s(n-1)+3), n>0, then s(n)=2*a(n)/(2*a(n)-1), n>0.
An Engel expansion of 5/3 to the base b := 5/4 as defined in A181565, with the associated series expansion 5/3 = b + b^2/4 + b^3/(4*19) + b^4/(4*19*94) + .... Cf. A007051. - Peter Bala, Oct 29 2013
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-5).
FORMULA
a(n) = (3*5^n+1)/4.
G.f.: (1-2*x)/((1-5*x)(1-x)).
E.g.f.: (3*exp(5*x) + exp(x))/4.
a(n) = 5*a(n-1)-1 with n>0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 6*a(n-1)-5*a(n-2). - Vincenzo Librandi, Nov 04 2011
a(n) = 5^n - Sum_{i=0..n-1} 5^i. - Bruno Berselli, Jun 20 2013
MAPLE
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*5-1 od: seq(a[n], n=1..22); # Zerinvary Lajos, Feb 22 2008
MATHEMATICA
CoefficientList[Series[(1-2x)/((1-5x)(1-x)), {x, 0, 30}], x] (* or *) LinearRecurrence[{6, -5}, {1, 4}, 30] (* Harvey P. Dale, Jul 27 2022 *)
PROG
(Magma) [(3*5^n+1)/4: n in [0..30]]; // Vincenzo Librandi, Nov 04 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Apr 21 2003
STATUS
approved