A308807 a(n) = 4*5^(n-1) + n.

5, 22, 103, 504, 2505, 12506, 62507, 312508, 1562509, 7812510, 39062511, 195312512, 976562513, 4882812514, 24414062515, 122070312516, 610351562517, 3051757812518, 15258789062519, 76293945312520, 381469726562521, 1907348632812522, 9536743164062523

Offset: 1

Author

Clive Tooth, Jun 25 2019

Keywords

• nonn
• easy

Comments

The last n decimal digits of 2^a(n) form the number 2^n.

Examples

			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.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (7,-11,5).

Programs

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 *)

PARI

Vec(x*(5 - 13*x + 4*x^2) / ((1 - x)^2*(1 - 5*x)) + O(x^25)) \\ Colin Barker, Jun 29 2019

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