A324222 - OEIS

A324222

a(n) is defined by the condition that the decimal expansion of Sum_{n>0} 1/a(n)^n = 1/a(1)^1 + 1/a(2)^2 + 1/a(3)^3 + ... begins with the concatenation of these numbers; also a(1) = 3 and a(n) > a(n-1).

3, 6, 14, 75, 574, 2029, 4589, 7927, 78325, 681667, 720945

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

a(11) is the last term because the sequence cannot be extended. At any step a(n) is chosen as the least number greater than a(n-1) that meets the requirement. Up to 720945 the sum is 0.3 6 14 75 574 2029 4589 7927 78325 681667 720945 0664... and the zero after 720945 cannot be removed.

If the limitation a(n) > a(n-1) were removed then the sequence would be 3, 6, 14, 75, 57, 58, 91, 197, 53, 423, 613, 102, 956 and 956 would be the last term because after it the sum presents 0316... and the zero cannot be removed. - Giovanni Resta, Feb 20 2019

LINKS

Table of n, a(n) for n=1..11.

EXAMPLE

1/3^1 = 0.3333...

1/3^1 + 1/6^2 = 0.36111...

1/3^1 + 1/6^2 + 1/14^3 = 0.3614755...

The sum is 0.3 6 14 75 574 ...

MAPLE

P:=proc(q, h) local a, b, d, n, t; a:=1/h; b:=ilog10(h)+1;

d:=h; print(d); t:=2; for n from 1 to q do

if trunc(evalf(a+1/n^t, 100)*10^(b+ilog10(n)+1))=d*10^(ilog10(n)+1)+n

then b:=b+ilog10(n)+1; d:=d*10^(ilog10(n)+1)+n; a:=a+1/n^t; t:=t+1;