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Search: id:A094010
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| A094010 |
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Least number k such that k! in binary representation has n consecutive nontrivial zeros. |
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+0
2
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| 6, 7, 9, 11, 15, 19, 40, 54, 38, 67, 63, 56, 37, 69, 352, 236, 258, 600, 1234, 979, 901, 3384, 2268, 4675, 5087, 5820, 3184, 12294, 41082, 25557
(list; graph; listen)
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OFFSET
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1,1
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FORMULA
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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.
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EXAMPLE
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a(4)=11 because 11!_d = 10011000010001010100000000_b. The last zeros are trivial.
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MATHEMATICA
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helper[b_][a : {b_, ___}] := Length[a]; helper[b_][a_List] := 0; maxConsecutiveCount[m_List, x_] := Max[helper[x] /@ Split[m]] (from Dr. Bob drbob@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
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CROSSREFS
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Cf. A094009, A011371.
Sequence in context: A022892 A120164 A074898 this_sequence A100348 A094698 A095908
Adjacent sequences: A094007 A094008 A094009 this_sequence A094011 A094012 A094013
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KEYWORD
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nonn
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AUTHOR
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Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 20 2004
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