A381678 a(n) is the least exponent k such that there are exactly n 1's in the decimal expansion of 11^k, or -1 if no such k exists.

0, 1, 5, 11, 21, 26, 31, 41, 51, 55, 58, 54, 72, 73, 91, 113, 116, 107, 150, 147, 103, 152, 158, 199, 181, 202, 165, 186, 215, 218, 238, 231, 273, 255, 266, 232, 302, 317, 297, 294, 327, 320, 293, 398, 339, 340, 350, 356, 406, 361, 380, 421, 391, 330, 401, 429, 438, 474, 388

Offset: 1

Author

Robert G. Wilson v, Mar 03 2025

Keywords

• base
• easy
• sign

Comments

If instead of the least exponent, one looks at the greatest exponent, then b(n) ~ n/0.1

Define f(x) to be the number of exponents k which have x ones in the decimal expansion of 11^k. Then f(x) = 1 for x = {0, 9, 10, 14, 17, 20, 23}.

It is provable that a(0) = -1 using the expansion of (x+y)^n. Conjecture: a(n) = -1 for n = 18622, 62206, 74453, 133125, etc. Search limit was k=1.5*10^6.

Examples

			a(1) = 0 since 11^0 = 1 has one occurrence of the decimal digit 1.
a(2) = 1 since 11^1 = 11 which has just two decimal digits of 1;
a(3) = 5 since 11^5 = 161051 which has just three decimal digits of 1;
a(4) = 11 since 11^11 = 285311670611 which has just four decimal digits of 1; etc.
		

Crossrefs

Cf. A001020, A007377, A195946.

Programs

Mathematica

t[_] := -1; k = 0; While[k < 1000, a = DigitCount[11^k, 10, 1]; If[t[a] == -1, t[a] = k]; k++]; t /@ Range[0, 100]

PARI

a(n) = my(k=0); while (#select(x->(x==1), digits(11^k)) != n, k++); k; \\ Michel Marcus, Mar 04 2025