A100322 - OEIS

%I #16 Jun 18 2022 14:06:06

%S 1,7,6,4,8,23,25,2,15,91,51,307,49,1,102,315,112,12,76,26,115,208,77,

%T 276,161,40,13,41,7,99,174,169,86,453,110,204,53,6,67,4,228,123,37,

%U 134,158,192,33,45,61,200,31,324,8,56,34,105,148,17,19,92,23,38,27,39,32,82

%N a(n) is the smallest positive integer k such that the digits of the fractional part of Pi^k begin with n.

%e Pi^1 = 3.14159..., whose digits after the decimal point begin with 1, so a(1)=1.

%e Pi^2 = 9.869..., whose digits after the decimal point begin with 8, so a(8)=2.

%e a(14)=1 because Pi^1 = 3.14....

%o (PARI) a(n) = my(k=1); while (floor(frac(Pi^k)*10^(1+logint(n, 10))) != n, k++); k; \\ _Michel Marcus_, Jun 18 2022

%Y Cf. A000796, A002388, A091925, A092425, A092731, A092732, A092735, A092736.

%K base,nonn

%O 1,2

%A Mark Hudson (mrmarkhudson(AT)hotmail.com), Nov 16 2004