A309037 Exponential Demlo sequence, like 12345...54321, but for powers of 2 instead.

2, 242, 24842, 2496842, 249936842, 24998736842, 2499974736842, 249999494736842, 24999989894736842, 2499999797894736842, 249999995957894736842, 24999999919157894736842, 2499999998383157894736842

Offset: 1

Author

Eliora Ben-Gurion, Jul 08 2019

Keywords

• base
• nonn

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.

I propose that the Demlo function should be generalized, so that the function A002477(A000079(n)) produces this sequence. As another example, A002477(A000040(n)) should produce 2, 232, 23532, 2357532, 235817532, 23582417532, etc.

Examples

			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
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..500
• Index entries for linear recurrences with constant coefficients, signature (121, -2120, 2000).

Crossrefs

Cf. A002477, A000079. Numbers produced from A000079 using A002477 algorithm.

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)