OFFSET
1,2
COMMENTS
Numerators in alternating Sum_{n>=0} 19^n/100^(n+1).
Related to decimal expansion of fraction of 1/119 and Pell numbers. [In which way? - Joerg Arndt, May 14 2011]
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..150
Index entries for linear recurrences with constant coefficients, signature (81,1900).
FORMULA
a(n+1) = a(n)*100 +- 19^n with a(0)=0 and a(1)= 1.
From Colin Barker, Oct 02 2015: (Start)
a(n) = 81*a(n-1) + 1900*a(n-2) for n > 2.
G.f.: -x / ((19*x+1)*(100*x-1)).
(End)
MATHEMATICA
LinearRecurrence[{81, 1900}, {1, 81}, 20] (* Harvey P. Dale, Nov 12 2022 *)
PROG
(Magma) [(1/119)*(100^n -(-19)^n): n in [1..20]]; // Vincenzo Librandi, May 17 2011
(PARI) Vec(-x/((19*x+1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Oct 02 2015
CROSSREFS
KEYWORD
nonn,frac,easy
AUTHOR
Mark Dols, May 29 2010
STATUS
approved