A097327 Least positive integer m such that m*n has greater decimal digit length than n.

10, 5, 4, 3, 2, 2, 2, 2, 2, 10, 10, 9, 8, 8, 7, 7, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 10, 10, 10

Offset: 1

Author

Rick L. Shepherd, Aug 04 2004

Keywords

• base
• easy
• nonn

Comments

For any positive base B >= 2 the corresponding sequence contains only terms from 2 to B inclusive so the corresponding sequence for binary is all 2s (A007395).

Examples

			a(12) = 9 since 12 has two decimal digits and 9*12 = 108 has three (but 8*12 = 96 has only two).
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A089186 (analog for decimal m+n), A080079 (analog for binary m+n), A097326.

Cf. A055642.

Programs

Mathematica

Table[Ceiling[10^IntegerLength[n]/n], {n, 100}] (* Paolo Xausa, Nov 02 2024 *)

PARI

a(n) = my(m=1, sn=#Str(n)); while (#Str(m*n) <= sn, m++); m; \\ Michel Marcus, Oct 05 2021

Python

def a(n): return (10**len(str(n))-1)//n + 1
print([a(n) for n in range(1, 103)]) # Michael S. Branicky, Oct 05 2021

Formula

a(n) = A097326(n) + 1.

a(n) = ceiling(10^A055642(n)/n). - Michael S. Branicky, Oct 05 2021