A385971 Smallest m such that 5^m begins with n 9's after the first digit.

0, 8, 195, 799, 28737, 167821, 325146, 6432162, 543157237, 1807789217, 3731189547, 3731189547

Offset: 0

Author

Giulio Bonfissuto, Jul 13 2025

Keywords

• base
• nonn
• hard
• more

Comments

a(n) is also the smallest m such that 1/2^m begins with n 9's after the first nonzero digit.

a(n) is equal to A152561(n)-1 for n=2, 6, 7 and possibly for many other terms.

When summing a series with dominant term 1/2^m (such as the Riemann zeta function), the n 9's here show how small further terms must be to avoid changing the initial decimal digit from 1/2^m.

Examples

			5^a(0) = 5^0      = 1
5^a(1) = 5^8      = 390625
5^a(2) = 5^195    = 1991364888915565346...
5^a(3) = 5^799    = 2999393627791261909...
5^a(4) = 5^28737  = 1999929120817815105...
5^a(5) = 5^167821 = 6999994116858573262...
		

Links

• Marco Ripà, Is the leading digit of the decimal expansion of the prime zeta function at n equal to the first digit of 5^n, for all integers n≥10?, Stack Exchange (2025)

Crossrefs

Cf. A152561, A000351, A111395, A011557.

Formula

a(n) = Min_{d=1..9} S(d*10^(n+1)-1) where 5^S(k) is the smallest power of 5 beginning with k.