A243909 Least number k > 0 such that 2^k contains exactly n different digits.

1, 4, 7, 10, 14, 21, 20, 37, 29, 68

Offset

1,2

Links

Table of n, a(n) for n=1..10.

Example

2^7 = 128 is the first power of 2 with 3 different digits. Thus a(3) = 7.

Mathematica

dd[n_]:=Count[DigitCount[2^n], _?(#>0&)]; Table[SelectFirst[Table[{n, dd[n]}, {n, 70}], #[[2]] == k&], {k, 10}][[All, 1]] (* Harvey P. Dale, Jan 22 2023 *)

Prog

(PARI) a(n)=for(k=1, 10^3, st=2^k; if(#digits(st)>=n, c=0; for(i=0, 9, for(j=1, #digits(st), if(digits(st)[j]==i, c++; break))); if(c==n, return(k))))
n=1; while(n<11, print1(a(n), ", "); n++)

Crossrefs

Sequence in context: A310696 A310697 A133497 * A187334 A310698 A310699

Adjacent sequences: A243906 A243907 A243908 * A243910 A243911 A243912

Keyword

nonn,base,full,fini

Author

Derek Orr, Jun 14 2014

Status

approved