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A330024
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a(n) = floor(n/z) where z is the number of zeros in the decimal expansion of 2^n, and a(n)=0 when z=0.
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1
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0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 11, 12, 0, 0, 0, 0, 17, 0, 0, 20, 21, 22, 23, 0, 0, 26, 0, 0, 29, 30, 0, 0, 0, 0, 0, 0, 0, 38, 0, 40, 41, 21, 14, 44, 45, 46, 47, 48, 0, 50, 0, 26, 17, 27, 27, 28, 57, 58, 29, 30, 20, 31, 31, 32, 65, 66, 0, 68, 23, 23, 71, 0
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OFFSET
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0,11
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COMMENTS
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Is a(229)=229 the largest term?
a(8949)=41; is 8949 the largest n such that a(n) >= 41?
Is 79391 the largest n such that a(n) <= 30?
Is 30 <= a(n) <= 36 true for all n >= 713789?
Conjecture: For every sequence which can be named as "digit k appears m times in the decimal expansion of 2^n", the sequences are finite for 0 <= k <= 9 and any given m >= 0. Every digit from 0 to 9 are inclined to appear an equal number of times in the decimal expansion of 2^n as n increases.
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LINKS
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FORMULA
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Conjecture: a(n) = 33 (= floor(10/log_10(2))) for all sufficiently large n. - Pontus von Brömssen, Jul 23 2021
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EXAMPLE
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a(11) = 11 because 2^11 = 2048, there is 1 zero in 2048 and the integer part of 11/1 is 11.
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MAPLE
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f:= proc(n) local z;
z:= numboccur(0, convert(2^n, base, 10));
if z = 0 then 0 else floor(n/z) fi
end proc:
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MATHEMATICA
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Do[z=DigitCount[2^n, 10, 0]; an=IntegerPart[n/z]; If[z==0, Print[0], Print[an]], {n, 0, 8000}]
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PROG
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(Magma) a:=[0]; for n in [1..72] do z:=Multiplicity(Intseq(2^n), 0); if z ne 0 then Append(~a, Floor(n/z)); else Append(~a, 0); end if; end for; a; // Marius A. Burtea, Nov 27 2019
(PARI) a(n) = my(z=#select(d->!d, digits(2^n))); if (z, n\z, 0); \\ Michel Marcus, Jan 07 2020
(Python)
z=str(2**n).count('0')
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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