A060298 Number of powers x^y (x,y > 1) with n digits.

3, 12, 34, 94, 263, 768, 2333, 7167, 22291, 69751, 219081, 689736, 2174856, 6864354, 21679391, 68497906, 216485583, 684323923, 2163459803, 6840258025, 21628220224, 68388917596, 216252901472, 683826283482, 2162393925204, 6837972506895, 21623315009817

Offset: 1

Author

Michel ten Voorde, Apr 10 2001

Keywords

• nonn
• base

Comments

Conjectures from Robert G. Wilson v, Aug 29 2012: (Start)

Limit_{n->oo} a(2n)/10^n = 1 - 1/sqrt(10).

Limit_{n->oo} a(2n-1)/10^n = 1/sqrt(10) - 1/10. (End)

These follow from the Formula. - Robert Israel, Apr 29 2020

Limit_{n->oo} a(n)/a(n-1) = sqrt(10). - Bernard Schott, Jan 21 2023

Examples

			a(1) = 3 because there are 3 powers with 1 digit: 2^2, 2^3 and 3^2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000
• Karl-Heinz Hofmann, Python program

Crossrefs

Cf. A001597, A089580 (partial sums).

Programs

Maple

f:= proc(n) local y;
  add(ceil(10^(n/y))-ceil(10^((n-1)/y)), y=2..floor(n*log[2](10)))
end proc:
f(1):= 3:
map(f, [$1..20]); # Robert Israel, Apr 29 2020

Python

# see link

Python

from sympy import integer_nthroot, integer_log
def A060298(n):
    if n == 1: return 3
    c, y, a, b, t = 0, 2, 10**n-1, 10**(n-1)-1, (10**n).bit_length()
    while yChai Wah Wu, Oct 16 2023

Formula

a(n) = Sum_{y=2..floor(n*log_2(10))} (ceiling(10^(n/y)) - ceiling(10^((n-1)/y))) for n >= 2. - Robert Israel, Apr 29 2020

a(n) = A089580(n+1) - A089580(n) for n > 1. - Karl-Heinz Hofmann, Sep 18 2023

Extensions

a(10)-a(18) from Donovan Johnson, Dec 14 2009

a(19)-a(27) from Donovan Johnson, Aug 29 2012