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A363060
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Numbers k such that 5 is the first digit of 2^k.
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5
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9, 19, 29, 39, 49, 59, 69, 102, 112, 122, 132, 142, 152, 162, 172, 195, 205, 215, 225, 235, 245, 255, 265, 298, 308, 318, 328, 338, 348, 358, 391, 401, 411, 421, 431, 441, 451, 461, 494, 504, 514, 524, 534, 544, 554, 587, 597, 607, 617, 627, 637, 647, 657, 680, 690
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listen;
history;
text;
internal format)
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OFFSET
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1,1
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COMMENTS
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The asymptotic density of this sequence is log_10(6/5) = 0.0791812... . - Amiram Eldar, May 16 2023
In base B we may consider numbers k such that some integer Y >= 1 forms the first digit(s) of X^k. For such numbers k the following inequality holds: log_B(Y) - floor(log_B(Y)) <= k*log_B(X) - floor(k*log_B(X)) < log_B(Y+1) - floor(log_B(Y+1)). The irrationality of log_B(X) is the necessary condition; see the Links section. Examples in the OEIS: B = 10, X = 2; Y = 1 (A067497), Y = 2 (A067469), Y = 3 (A172404).
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LINKS
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EXAMPLE
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k = 9: the first digit of 2^9 = 512 is 5, thus k = 9 is a term.
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MAPLE
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R:= NULL: count:= 0: t:= 1:
for k from 1 while count < 100 do
t:= 2*t;
if floor(t/10^ilog10(t)) = 5 then R:= R, k; count:= count+1 fi
od:
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MATHEMATICA
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Select[Range[700], IntegerDigits[2^#][[1]] == 5 &] (* Amiram Eldar, May 16 2023 *)
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PROG
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(PARI) isok(k) = digits(2^k)[1] == 5; \\ Michel Marcus, May 16 2023
(Python)
from itertools import count, islice
def A363060_gen(startvalue=1): # generator of terms >= startvalue
m = 1<<(k:=max(startvalue, 1))
for i in count(k):
if str(m)[0]=='5':
yield i
m <<= 1
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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