A088935 Numbers n such that leading digits of 2^n and 5^n are equal.

0, 5, 15, 78, 88, 98, 108, 118, 181, 191, 201, 211, 274, 284, 294, 304, 367, 377, 387, 397, 407, 470, 480, 490, 500, 563, 573, 583, 593, 603, 666, 676, 686, 696, 759, 769, 779, 789, 852, 862, 872, 882, 892, 955, 965, 975, 985, 1048, 1058, 1068, 1078, 1088

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)}, so we have {k > 0 : 2^k, 5^k, 8^k all start with a} = {k: {k*lg(2)} is in I_a}, where I_a = (lg(a), lg(a+1)) intersect (1-lg(a+1), 1-lg(a)). Note that I_3 = (lg(3), 1-lg(3)) and I_a is empty otherwise. As a result, k > 0 is a term if and only if lg(3) < {k*lg(2)} < 1-lg(3). - Jianing Song, Dec 26 2022

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

T. Sillke, Powers of 2 and 5 Puzzle

Example

78 is in the sequence since 2^78 = 302231454903657293676544 and 5^78 = 3308722450212110699485634768279851414263248443603515625
98 is in the sequence since 2^98 = 316912650057057350374175801344 and 5^98 = 315544362088404722164691426113114491869282574043609201908111572265625.

Maple

filter:= n -> convert(2^n, base, 10)[-1]=convert(5^n, base, 10)[-1]:
select(filter, [$0..1000]); # Robert Israel, Aug 09 2018

Mathematica

Select[ Range[ 1000 ], IntegerDigits[ 2^# ][ [ 1 ] ] == IntegerDigits[ 5^# ][ [ 1 ] ] & ]

Prog

(PARI) is(n)=(digits(2^n)[1]==digits(5^n)[1]);
for(n=0, 10^3, if(is(n), print1(n, ", "))); \\ Joerg Arndt, Aug 10 2018
(Python)
def ok(n): return str(2**n)[0] == str(5**n)[0]
print([k for k in range(1100) if ok(k)]) # Michael S. Branicky, Nov 03 2022

Crossrefs

Cf. A088995.

Sequence in context: A032122 A220818 A064678 * A183937 A275971 A030487

Adjacent sequences: A088932 A088933 A088934 * A088936 A088937 A088938

Keyword

base,nonn

Author

Lekraj Beedassy, Dec 01 2003

Extensions

Edited by Robert G. Wilson v, Dec 02 2003

Status

approved