A358196 Numbers k such that 5^k and 8^k have the same leading digit.

0, 5, 9, 15, 19, 29, 34, 39, 44, 49, 54, 59, 98, 102, 108, 112, 118, 122, 132, 137, 142, 147, 152, 162, 191, 195, 201, 205, 211, 215, 225, 230, 235, 240, 245, 250, 255, 284, 294, 298, 304, 308, 318, 328, 333, 338, 343, 348, 387, 391, 397, 401, 407, 411, 421, 426, 431, 436, 441, 446, 451, 480, 490, 494, 500

Offset

1,2

Comments

Write lg = log_10, let {x} denote the fractional part of x. Note that {k*lg(5)} = 1 - {k*lg(2)} and that {k*lg(8)} = {k*lg(2)} + 0, 1, or 2, so we have {k > 0 : 5^k and 8^k both start with a} = {k: {k*lg(2)} is in I_a}, where I_a = (1-lg(a+1), 1-lg(a)) intersect (((lg(a))/3, (lg(a+1))/3) U ((lg(a)+1)/3, (lg(a+1)+1)/3) U ((lg(a)+2)/3, (lg(a+1)+2)/3)). Note that I_1 = (1-lg(2), (lg(2)+2)/3), I_3 = ((lg(3)+1)/3, 1-lg(3)), I_5 = (lg(5)/3, lg(6)/3) and that I_a is empty otherwise. As a result, k > 0 is a term if and only if {k*lg(2)} is in (lg(5)/3, lg(6)/3) U ((lg(3)+1)/3, 1-lg(3)) U (1-lg(2), (lg(2)+2)/3). We see that when 5^k and 8^k both start with 1, 3, or 5, 2^k starts with 5, 3, or 1 respectively. - Jianing Song, Dec 26 2022

Links

Table of n, a(n) for n=1..65.

Nicolay Avilov, Problem 2405. 4 and 5 (in Russian).

Example

5 is a term because 5^5 = 3125 and 8^5 = 32768;
9 is a term because 5^9 = 1953125 and 8^9 = 134217728.

Mathematica

Select[Range[0, 500], Equal @@ IntegerDigits[{5, 8}^#][[;; , 1]] &] (* Amiram Eldar, Nov 02 2022 *)

Prog

(PARI) isok(k) = digits(5^k)[1] == digits(8^k)[1]; \\ Michel Marcus, Nov 02 2022
(Python)
def ok(n): return str(5**n)[0] == str(8**n)[0]
print([k for k in range(501) if ok(k)]) # Michael S. Branicky, Nov 03 2022

Crossrefs

Cf. A000351, A001018, A088935, A111395.

Sequence in context: A315038 A315039 A315040 * A315041 A315042 A315043

Adjacent sequences: A358193 A358194 A358195 * A358197 A358198 A358199

Keyword

nonn,base

Author

Nicolay Avilov, Nov 02 2022

Status

approved