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A098726
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Take three consecutive primes started with the n-th prime:{p(n),p(n+1),p(n+2)}. Calculate d(i,j)=Abs[p(i)-p(j)], for all {i,j}, i.e. all possible differences. a[n] is the number of distinct differences which are either 3 or 2.
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1
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3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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EXAMPLE
| a[n]=2 iff the consecutive prime-differences are equal.
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MATHEMATICA
| k=3; t=Table[Abs[Prime[n+i]-Prime[n+j]], {i, 0, k-1}, {j, 0, k-1}]; u=Delete[Union[Flatten[t]], 1]; a(n)=Length[u]
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CROSSREFS
| Cf. A080370, A054643.
Sequence in context: A080428 A120004 A079790 * A065801 A205237 A086920
Adjacent sequences: A098723 A098724 A098725 * A098727 A098728 A098729
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KEYWORD
| nonn
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AUTHOR
| Labos E. (labos(AT)ana.sote.hu), Oct 05 2004
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