A104321 Smallest number m such that A104320(m)=n.
0, 5, 8, 18, 13, 26, 27, 23, 42, 25, 37, 58, 47, 46, 61, 67, 54, 71, 77, 73, 88, 99, 141, 100, 115, 114, 119, 117, 113, 112, 109, 135, 110, 127, 133, 136, 164, 162, 177, 186, 193, 195, 163, 189, 201, 196, 191, 199, 206, 188, 208, 200, 221, 266, 235, 234, 238, 280
Offset: 0
Keywords
A102483 Numbers k such that 2^k contains no zeros in base 3.
0, 1, 2, 3, 4, 15
Offset: 1
Comments
I conjectured in 1973 that there are no further terms. This question is still open.
A104320(a(n)) = 0. - Reinhard Zumkeller, Mar 01 2005
No other terms less than 200000. - Robert G. Wilson v, Dec 06 2005
a(7) > 10^7. - Martin Ehrenstein, Jul 27 2021
If it exists, a(7) > 10^21. - Robert Saye, Mar 23 2022
Links
- Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, J. Integer Seq. 25 (2022) Article 22.3.4.
- N. J. A. Sloane, The persistence of a number, J. Recreational Math., 6 (1973), 97-98.
- Eric Weisstein's World of Mathematics, Ternary
Programs
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Mathematica
Select[ Range@1000, FreeQ[ IntegerDigits[2^#, 3], 0] &] (* Robert G. Wilson v, Dec 06 2005 *)
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PARI
for (n=0, 100, if (vecmin(digits(2^n, 3)), print1(n, ", "))) \\ Michel Marcus, Mar 25 2015
A117970 Position of first 0 counting from the least significant digit in the ternary expansion of 2^n.
0, 0, 0, 0, 0, 3, 2, 2, 4, 4, 5, 4, 2, 2, 4, 0, 3, 4, 2, 2, 3, 3, 8, 3, 2, 2, 5, 5, 6, 5, 2, 2, 11, 4, 3, 5, 2, 2, 3, 3, 11, 3, 2, 2, 5, 8, 4, 5, 2, 2, 13, 5, 3, 5, 2, 2, 3, 3, 4, 3, 2, 2, 4, 4, 7, 4, 2, 2, 4, 6, 3, 4, 2, 2, 3, 3, 7, 3, 2, 2, 6, 6, 7, 6, 2, 2, 10, 4, 3, 7, 2, 2, 3, 3, 5, 3, 2, 2, 11, 5, 4, 17
Offset: 0
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
- Eric Weisstein's World of Mathematics, Ternary
Programs
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PARI
a(n) = { my (p=2^n); for (k=1, oo, if (p==0, return (0), p%3==0, return (k), p\=3)) } \\ Rémy Sigrist, Dec 20 2019
Extensions
Edited by Charles R Greathouse IV, Aug 05 2010
a(0) = 0 prepended by Rémy Sigrist, Dec 20 2019
A260683 Number of 2's in the expansion of 2^n in base 3.
0, 1, 0, 2, 1, 1, 1, 2, 0, 4, 2, 4, 3, 3, 2, 6, 5, 5, 3, 7, 4, 7, 5, 4, 1, 5, 2, 8, 8, 7, 9, 9, 8, 7, 7, 8, 4, 6, 8, 9, 11, 11, 7, 11, 10, 8, 9, 8, 8, 10, 11, 16, 13, 10, 9, 12, 13, 16, 12, 13, 15, 15, 11, 15, 16, 14, 14, 12, 14, 15, 14, 16, 11, 18, 11, 17, 10
Offset: 0
Comments
Erdős conjectures that a(n) > 0 for n > 8.
Examples
For n=5, the expansion of 2^n in number base 3 is 1012, thus: a(n)=1 For n=10, the expansion of 2^n in number base 3 is 1101221, thus: a(n)=2
References
- R. K. Guy, Unsolved Problems in Number Theory, B33. [Does not seem to be in section B33.]
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
- Paul Erdős, Some unconventional problems in number theory, Mathematics Magazine, Vol. 52, No. 2 (1979), pp. 67-70.
Crossrefs
Programs
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Maple
seq(numboccur(2, convert(2^n,base,3)),n=0..100); # Robert Israel, Nov 15 2015
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Mathematica
S={};n=-1;While[n<150,n++;A=IntegerDigits[2^n,3];k=Count[A,2];AppendTo[S, k]];S -
PARI
c(k, d, b) = {my(c=0, f); while (k>b-1, f=k-b*(k\b); if (f==d, c++); k\=b); if (k==d, c++); return(c)} for(n=0, 300, print1(c(2^n, 2, 3)", ")) \\ Altug Alkan, Nov 15 2015 -
PARI
a(n) = #select(x->(x==2), digits(2^n, 3)); \\ Michel Marcus, Nov 28 2018
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PARI
a(n) = hammingweight(digits(2^n, 3)\2); \\ Ruud H.G. van Tol, May 09 2024
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Perl
use ntheory ":all"; sub a260683 { scalar grep { $==2 } todigits(vecprod((2) x shift), 3) } # _Dana Jacobsen, Aug 16 2016
Formula
A346497 List of powers of 2 written in base 3 which contain no zero digits.
1, 2, 11, 22, 121, 1122221122
Offset: 1
Comments
The listed terms are the base-3 expansions of 1, 2, 4, 8, 16, and 32768.
The program shows that there are no other terms less than 2^1000.
a(7) > 2^(10^7). - Martin Ehrenstein, Jul 27 2021
If it exists, a(7) > 2^(10^21). - Robert Saye, Mar 23 2022
References
- David Wells, "The Penguin Dictionary of Curious and Interesting Numbers" (1997), p. 123.
Links
- R. C. Couto, Number of nonzero digits in 2^n base 3
- Robert I. Saye, On two conjectures concerning the ternary digits of powers of two, arXiv:2202.13256 [math.NT], 2022; J. Integer Seq. 25 (2022) Article 22.3.4.
Crossrefs
Programs
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Mathematica
pwr = 1; Do[pwr = Mod[2*pwr, 3^100]; d = Union[IntegerDigits[pwr, 3]]; If[Intersection[d, {0}] == {}, Print[IntegerString[pwr, 3]]], {n, 10000000}] (* Ricardo Bittencourt, Jul 07 2021 *) Select[Table[FromDigits[IntegerDigits[2^n,3]],{n,0,100}],DigitCount[#,10,0]==0&] (* Harvey P. Dale, Feb 18 2025 *)
Formula
A036461 Number of 1 digits in base 3 representation of 2^n.
1, 0, 2, 0, 2, 2, 2, 2, 4, 0, 4, 2, 4, 2, 6, 4, 2, 4, 6, 2, 6, 4, 6, 4, 8, 2, 10, 4, 4, 8, 6, 8, 8, 8, 8, 6, 10, 8, 10, 10, 6, 6, 12, 8, 10, 14, 8, 10, 10, 12, 16, 8, 12, 18, 10, 10, 14, 10, 14, 14, 16, 10, 16, 12, 16, 16, 14, 16, 14, 18, 20, 12, 20, 10, 22, 12, 26, 8, 20, 12, 22, 14, 16
Offset: 0
Comments
The number of 1's in the base 3 representation of any even(odd) number is even(odd).
Links
- Robert Israel, Table of n, a(n) for n = 0..10000
Programs
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Maple
seq(numboccur(1,convert(2^n,base,3)),n=0..100); # Robert Israel, Apr 04 2018
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Mathematica
Table[DigitCount[2^n,3,1],{n,0,120}] (* Harvey P. Dale, Mar 14 2011 *) -
PARI
a(n) = #select(x->(x==1), digits(2^n, 3)); \\ Michel Marcus, Apr 04 2018
A036462 Conjecturally, a power of 2 written in base 3 cannot have this many 0's.
115, 124, 139, 243, 367, 445, 783, 914, 958, 1095, 1112, 1200, 1239, 1312, 1487, 1752, 1902, 2013, 2504, 2583, 2620, 2697, 2725, 2754, 2881, 3015, 3365, 3443, 3612, 3673, 3980, 3984, 4002, 4105, 4184, 4212, 4315, 4343, 4394, 4477, 4516, 4862, 4918, 5100
Offset: 1
Comments
Conjecture 1: Sequence A104320 never obtains the values in this sequence, so A104321(a(n)) is undefined. Conjecture 2: This sequence is infinite. - David W. Wilson, Oct 21 2016, edited by Antti Karttunen, Oct 29 2016
Comments
Links
Crossrefs
Programs
PARI
a(n) = {my(k = 0); while (#select(x->(x==0), digits(2^k, 3)) != n, k++); k;} \\ Michel Marcus, Oct 19 2016