cp's OEIS Frontend

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.

%I A381678 #31 Apr 24 2025 13:32:01
%S A381678 0,1,5,11,21,26,31,41,51,55,58,54,72,73,91,113,116,107,150,147,103,
%T A381678 152,158,199,181,202,165,186,215,218,238,231,273,255,266,232,302,317,
%U A381678 297,294,327,320,293,398,339,340,350,356,406,361,380,421,391,330,401,429,438,474,388
%N 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.
%C A381678 If instead of the least exponent, one looks at the greatest exponent, then b(n) ~ n/0.1
%C A381678 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}.
%C A381678 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.
%e A381678 a(1) = 0 since 11^0 = 1 has one occurrence of the decimal digit 1.
%e A381678 a(2) = 1 since 11^1 = 11 which has just two decimal digits of 1;
%e A381678 a(3) = 5 since 11^5 = 161051 which has just three decimal digits of 1;
%e A381678 a(4) = 11 since 11^11 = 285311670611 which has just four decimal digits of 1; etc.
%t A381678 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]
%o A381678 (PARI) a(n) = my(k=0); while (#select(x->(x==1), digits(11^k)) != n, k++); k; \\ _Michel Marcus_, Mar 04 2025
%Y A381678 Cf. A001020, A007377, A195946.
%K A381678 base,easy,sign
%O A381678 1,3
%A A381678 _Robert G. Wilson v_, Mar 03 2025