OFFSET
1,1
FORMULA
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.
EXAMPLE
a(4)=11 because 11!_d = 10011000010001010100000000_b. The last zeros are trivial.
MATHEMATICA
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)
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Apr 20 2004
STATUS
approved