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a(n) is the period of the decimal expansion of 1/(100^n - 10^n - 1).
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%I #26 Jan 06 2022 08:15:50

%S 44,468,496620,16090340,2499916380,499999499999,47368416315780,

%T 71942445323740,71428571357142857,2413792682560590100,

%U 661025543433488700,998035336547180189380,9826562531691739684620,3086415088517393302531635,33093525179856082014388489204

%N a(n) is the period of the decimal expansion of 1/(100^n - 10^n - 1).

%C It appears that when Fibonacci numbers are written in base 10 diagonally (from top left to bottom right) such that each lower number is n digits farther to the right than its neighbor above, and the columns of digits are summed, the resulting total digit string recurs after a(n) digits.

%e For n = 1: the Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, when written in a diagonal, with each number 1 digit farther to the right than its predecessor, and the columns summed, gives the digit string 112359... . After this is extended to 44 numbers, the digit string has another occurrence of 112359. I conjecture that this is because the reciprocal of 89 has a period of 44 digits. It also demonstrates an amazing property of Fibonacci numbers.

%t Array[MultiplicativeOrder[10, (100^# - 10^# - 1)] &, 15] (* _Michael De Vlieger_, Jun 29 2018 *)

%o (PARI) a(n) = znorder(Mod(10, 100^n-10^n-1)) \\ _Felix Fröhlich_, Jun 18 2018

%Y Cf. A007732, A021093 (1/89), A086695.

%K base,nonn

%O 1,1

%A _Stephen Tucker_, Jun 17 2018

%E More terms from _Jon E. Schoenfield_ and _Felix Fröhlich_, Jun 18 2018