|
%I #4 Oct 02 2022 00:44:15
%S 105263157894736842,
%T 100502512562814070351758793969849246231155778894472361809045226130653266331658291457286432160804020
%N Smallest positive integer that doubles when the n rightmost digits are shifted to the left end.
%C a(n) with n>=3 is too large to be written in data.
%C The following is a method for finding a(n): Let n be the number of digits shifted, and let m be the smallest positive integer such that 10^m = 2 mod 2*10^n-1. We then look for the smallest positive b that is an n+d digit number and satisfies b = c(10^n-2)/(2*10^d-1), where c is a positive integer. Then a(n) = c(10^n-2)/(2*10^d-1)*10^n+c.
%e a(1) = 105263157894736842 because shifting the 1 rightmost digit to the left end gives 210526315789473684 which is double a(1).
%Y Cf. A146088
%K nonn,base
%O 1,1
%A _Joseph C. Y. Wong_, Oct 01 2022
| 