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%I #12 Jan 15 2015 10:24:15
%S 0,0,0,0,0,0,0,0,1,0,0,0,1,2,1,1,1,0,0,1,1,1,1,2,2,1,1,2,2,2,0,1,2,0,
%T 3,3,2,2,3,3,4,1,1,1,4,7,4,4,5,4,3,4,6,6,3,5,2,2,0,3,4,5,6,7,8,6,6,5,
%U 7,8,8,6,8,4,3,3,6,5,4,4,8,7,4,4,3,1,4,6,4,4,6,5,6,7,6,4,4,4,6,9,12,8,5,9,7,6,4,2,9,8,5,5,3,4,6,6,9,14,12,12,12,12,13
%N Number of zeros in the decimal expansion of 5^n.
%C Probably a(58) is the last 0 term.
%H Zak Seidov, <a href="/A253638/b253638.txt">Table of n, a(n) for n = 0..1000</a>
%F a(n) = A055641(A000351(n)). - _Michel Marcus_, Jan 15 2015
%e 5^57 = 6938893903907228377647697925567626953125, 2 zeros hence a(57) = 2,
%e 5^58 = 34694469519536141888238489627838134765625, no zeros hence a(58) = 0,
%e 5^59 = 173472347597680709441192448139190673828125, 3 zeros hence a(59) = 3.
%t Table[Count[IntegerDigits[5^n],0],{n,0,200}]
%o (PARI) a(n) = my(d = digits(5^n)); sum(i=1, #d, d[i] == 0); \\ _Michel Marcus_, Jan 15 2015
%Y Cf. A008839 (no zeros in 5^n), A055641 (number of zeros for n), A000351 (5^n).
%K nonn,base
%O 0,14
%A _Zak Seidov_, Jan 07 2015
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