OFFSET
1,1
COMMENTS
The last n decimal digits of 2^a(n) form the number 2^n.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-11,5).
FORMULA
a(n) = A005054(n) + n.
From Colin Barker, Jun 26 2019: (Start)
G.f.: x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)).
a(n) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3) for n>3.
(End)
Conjectures confirmed by Robert Israel, Jun 28 2019
EXAMPLE
a(1) = 5, 2^5 = 32, the last digit of 32 is 2, which is 2^1.
a(2) = 22, 2^22 = 4194304, the last 2 digits of 4194304 are 04, which is 2^2.
MAPLE
seq(4*5^(n-1) + n, n=1..30); # Robert Israel, Jun 28 2019
MATHEMATICA
Table[4*5^(n-1)+n, {n, 30}] (* or *) LinearRecurrence[{7, -11, 5}, {5, 22, 103}, 30] (* Harvey P. Dale, Jun 27 2020 *)
PROG
(PARI) Vec(x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)) + O(x^25)) \\ Colin Barker, Jun 29 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clive Tooth, Jun 25 2019
STATUS
approved