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A280660 Least k such that at least half of the last n digits of 2^k are 9. 1
12, 53, 53, 232, 93, 1862, 93, 3244, 93, 93, 93, 55754, 12864, 55756, 23353, 361353, 16441, 361353, 304362, 361353, 361353, 361353, 361353, 3748854, 3748854, 78055893, 66290232, 119133355, 119133355, 379371432, 20958353, 130883333, 20958353, 130883333 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,1
COMMENTS
See the Mathematical Reflections link for a proof that a(n) exists for all n>1.
LINKS
Jon E. Schoenfield and Chai Wah Wu, Table of n, a(n) for n = 2..44 (terms for n = 2..42 from Jon E. Schoenfield)
Mathematical Reflections, Solution to Problem O316, Issue 6, 2014, p 26.
Jon E. Schoenfield, Magma program
EXAMPLE
For n=2, k=12 with 2^k = 4096.
MAPLE
a:= proc(n) local m, t, k, c, h; m, t:= 10^n, 2048;
for k from 12 do t:= 2*t mod m; h, c:= t, 0;
while h>0 do if irem(h, 10, 'h')=9 then c:= c+2 fi od;
if c >= n then return k fi
od
end:
seq(a(n), n=2..16); # Alois P. Heinz, Jan 07 2017
PROG
(PARI) a(n) = my(k = 1, ok = 0); until (ok, vd = Vecrev(digits(2^k)); nb = sum(j=1, min(n, #vd), vd[j]==9); ok = (nb >= n/2); if (! ok, k++); ); k;
(Python)
def A280660(n):
m, k, l = 10**n, 1, 2
while True:
if 2*str(l).count('9') >= n:
return k
k += 1
l = (l*2) % m # Chai Wah Wu, Jan 07 2017
CROSSREFS
Sequence in context: A195544 A248135 A307916 * A268186 A253130 A213549
KEYWORD
nonn,base
AUTHOR
Michel Marcus, Jan 07 2017
EXTENSIONS
a(17)-a(30) from Alois P. Heinz, Jan 07 2017
a(31)-a(35) from Jon E. Schoenfield, Jan 07 2017
STATUS
approved

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Last modified April 24 12:22 EDT 2024. Contains 371937 sequences. (Running on oeis4.)