OFFSET
1,1
COMMENTS
Lim_{n->infinity} a(n)/10^(2n-1) = 0.25, thus the first digits converge toward 24999999999999999999999...
In other words, Sum_{i>=1} 2^n/10^n = Sum_{i>=1} 5^(-n) = 5/(1-5) = 5/4 = 1.25. Excluding the 1 at the beginning of the number gives 0.25. Dividing each term by 2 gives the previous term with 1s attached on each side.
For example, 24998736842 / 2 = 12499368421.
In the set of {a(n)}, the final digits of a(n) eventually tend to be the repeating portion of 1/19 as n approaches infinity: ... 052631578947368421 05263157894736842.
If 8421... is analytically continued, 052631578947436... is obtained because Sum_{i>=1} 1/(2^n*10^n) is 1/19.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (121, -2120, 2000).
FORMULA
a(n) = 2^1*10^0 + 2^2*10^1 + ... + 2^(n-1)*10^(n-2) + 2^n*10^(n-1) + 2^(n-1)*10^n + 2^(n-2)*10^(n+1) + ... + 2^2*10^(2n-3) + 2^1*10^(2n-2).
Conjectures from Colin Barker, Jul 16 2019: (Start)
G.f.: 2*x*(1 - 10*x)*(1 + 10*x) / ((1 - x)*(1 - 20*x)*(1 - 100*x)).
a(n) = (-80 - 3*4^n*5^(1+n) + 19*100^n) / 760.
a(n) = 121*a(n-1) - 2120*a(n-2) + 2000*a(n-3) for n>3.
(End)
EXAMPLE
For n = 4:
2000000 8 - 2 = 6
400000
80000
16000 4 - 1 = 3
800
40
+ 2
-------
2496842
For n = 12:
2*10^(24-2) + 4*10^(24-3) + 8*10^(24-4) + ... + 4096*10^11 + ... + 8*10^2 + 4*10^1 + 2
20000000000000000000000 24 - 2 = 22
4000000000000000000000
800000000000000000000
160000000000000000000
32000000000000000000
6400000000000000000
1280000000000000000
256000000000000000
51200000000000000
10240000000000000
2048000000000000
409600000000000 12 - 1 = 11
20480000000000
1024000000000
51200000000
2560000000
128000000
6400000
320000
16000
800
40
+ 2
-----------------------
24999999919157894736842
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Eliora Ben-Gurion, Jul 08 2019
STATUS
approved