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A073490
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Number of prime gaps in factorization of n.
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14
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0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,110
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COMMENTS
| a(A000040(n))=0; a(A000961(n))=0; a(A006094(n))=0; a(A002110(n))=0; a(A073485(n))=0; a(A073486(n))>0; a(A073487(n)) = 1; a(A073488(n))=2; a(A073489(n))=3;
a(n)=0 iff A073483(n) = 1.
a(A097889(n)) = 0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 20 2004
A137723(n) is the smallest number of the first occurring set of exactly n consecutive numbers with at least one prime gap in their factorization: a(A137723(n)+k)>0 for 0<=k<n and a(A137723(n)-1)=a(A137723(n)+n)=0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 09 2008
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LINKS
| R. Zumkeller, Table of n, a(n) for n = 1..50000
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FORMULA
| a(n) = A073484(A007947(n)).
0 <= a(m*n) <= a(m) + a(n) + 1. A137794(n) = 0^a(n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 11 2008
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EXAMPLE
| 84=2*2*3*7 with one gap between 3 and 7, therefore a(84) = 1;
110=2*5*11 with two gaps: between 3 and 5 and between 5 and 11, therefore a(110)=2.
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MATHEMATICA
| gaps[n_Integer/; n>0]:=If[n===1, 0, Complement[Prime[PrimePi[Rest[ # ]]-1], # ]&[First/@FactorInteger[n]]]; Table[Length[gaps[n]], {n, 1, 105}] (Wouter Meeussen (wouter.meeussen(AT)pandora.be), Oct 30 2004)
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CROSSREFS
| Cf. A073491, A073492, A073493, A073494, A073495.
Cf. A137721, A137722.
Sequence in context: A085737 A191904 A005090 * A194285 A135341 A033665
Adjacent sequences: A073487 A073488 A073489 * A073491 A073492 A073493
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KEYWORD
| nonn,nice
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AUTHOR
| Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 03 2002
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EXTENSIONS
| More terms from Frank Adams-Watters (FrankTAW(AT)Netscape.net), May 19 2006
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