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

Author

Lekraj Beedassy, Dec 01 2003

Keywords

• base
• nonn

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

Examples

			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.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000
• T. Sillke, Powers of 2 and 5 Puzzle

Crossrefs

Cf. A088995.

Programs

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 ] ] & ]

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

Extensions

Edited by Robert G. Wilson v, Dec 02 2003