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Least positive integer such that a(n)! starts with n, both written in base 3.
3

%I #2 Mar 31 2012 13:48:25

%S 1,2,12,5,13,3,19,4,31,18,12,21,5,33,17,13,34,8,38,7,19,92,24,4,11,6,

%T 46,39,31,22,18,77,12,55,32,75,21,52,74,5,15,33,61,122,166,17,47,13,

%U 65,54,49,148,34,71,8,117,107,60,38,51,10,7,111,25,19,92,123,128,29,24,56,99

%N Least positive integer such that a(n)! starts with n, both written in base 3.

%e a(4) = 5 since 4 = 11[3] (i.e. written in base 2) and 5! = 11110[3] is the least factorial to have 11 as most significant digits in base 3.

%o (PARI) A155892(n)={ local( F=1,k=1 ); while( F\1!=n, F*=k++; while( F>=n+1, F/=3 )); k}

%Y Cf. A155891, A018799, A076219.

%K base,nonn

%O 1,2

%A _M. F. Hasler_, Feb 01 2009