|
%I #4 Mar 30 2012 17:31:00
%S 6,7,9,11,15,19,40,54,38,67,63,56,37,69,352,236,258,600,1234,979,901,
%T 3384,2268,4675,5087,5820,3184,12294,41082,25557
%N Least number k such that k! in binary representation has n consecutive nontrivial zeros.
%F Eliminate the trailing (least significant) trivial zeros by dividing n! by the Sum_{e=1..max) Floor(n/2^e), max is the first exponent where 2^e >= n. See A011371.
%e a(4)=11 because 11!_d = 10011000010001010100000000_b. The last zeros are trivial.
%t helper[b_][a : {b_, ___}] := Length[a]; helper[b_][a_List] := 0; maxConsecutiveCount[m_List, x_] := Max[helper[x] /@ Split[m]] (from Dr. Bob drbob(AT)bigfoot.com Apr 20 2004)
%t a = Table[0, {30}]; Do[ b = maxConsecutiveCount[ IntegerDigits[ n! / 2^IntegerExponent[n!, 2], 2], 0]; If[ a[[b]] == 0, a[[b]] = n; Print[b, " = ", n]], {n, 16500}]; a
%Y Cf. A094009, A011371.
%K nonn
%O 1,1
%A _Robert G. Wilson v_, Apr 20 2004
|
