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A098726
Take three consecutive primes starting with the n-th prime. Calculate d(i,j) = abs(prime(i) - prime(j)), for all {i,j}, i.e., all possible differences. a(n) is the number of distinct differences (which can be either 3 or 2).
2
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
OFFSET
1,1
COMMENTS
a(n) = 2 iff the consecutive prime differences are equal.
It appears that a(n) = 2 for n in A064113. - Michel Marcus, Jul 27 2017
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]
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Oct 05 2004
EXTENSIONS
Name edited by Michel Marcus, Jul 27 2017
STATUS
approved