%I #19 May 18 2024 19:35:13

%S 1,0,2,3,3,6,4,8,10,11,10,18,18,13,9,14,18,26,24,29,26,27,27,29,32,37,

%T 34,34,40,38,36,39,46,49,38,47,39,49,44,54,60,57,60,64,66,71,52,48,55,

%U 63,71,67,70,59,52,52,71,85,96,96,84,89,87,85,76,74,71,80

%N a(n) = sum of the digits (mod 5) of 5^n.

%H Paolo Xausa, <a href="/A372808/b372808.txt">Table of n, a(n) for n = 0..10000</a>

%H J. M. Borwein and P. B. Borwein, <a href="https://doi.org/10.2307/2324993">Strange Series and High Precision Fraud</a>, The American Mathematical Monthly, Vol. 99, No. 7 (1992), pp. 622-640.

%F Sum_{n >= 1} a(n)/5^n = 1/9. See Example 5.1 (e) in Borwein and Borwein (1992), p. 639.

%e a(7) = 8 since 5^7 = 78125 and (7 mod 5) + (8 mod 5) + (1 mod 5) + (2 mod 5) + (5 mod 5) = 2 + 3 + 1 + 2 + 0 = 8.

%t Array[Total[Mod[IntegerDigits[5^#], 5]] &, 100, 0]

%o (PARI) a(n) = my(d=digits(5^n)); vecsum(apply(x->(x % 5), d)); \\ _Michel Marcus_, May 17 2024

%Y Cf. A000351, A010874, A066001.

%K nonn,base,easy

%O 0,3

%A _Paolo Xausa_, May 13 2024