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A031217
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Number of terms in longest arithmetic progression of consecutive primes starting at n-th prime (conjectured to be unbounded).
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11
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2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2
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OFFSET
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1,1
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COMMENTS
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The first instance of 4 consecutive primes in an arithmetic progression is (251, 257, 263, 269), which starts with the 54th prime. The first instance of 5 consecutive primes in an arithmetic progression is (9843019, 9843049, 9843079, 9843109, 9843139), which starts with the 654926th prime. [Harvey P. Dale, Jul 13 2011]
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, A6.
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LINKS
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EXAMPLE
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At 47 there are 3 consecutive primes in A.P., 47 53 59.
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MATHEMATICA
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max = 5; a[n_] := Catch[pp = NestList[ NextPrime, Prime[n], max-1]; Do[ If[ Length[ Union[ Differences[pp[[1 ;; -k]] ] ] ] == 1, Throw[max-k+1]], {k, 1, max-1}]]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Jul 17 2012 *)
Length[Split[Differences[#]][[1]]]&/@Partition[Prime[Range[120]], 10, 1]+1 (* Harvey P. Dale, Mar 17 2024 *)
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PROG
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(PARI) a(n)=my(p=prime(n), q=nextprime(p+1), g=q-p, k=2); while(nextprime(q+1)==q+g, q+=g; k++); k \\ Charles R Greathouse IV, Jun 20 2013
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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