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A122535
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Smallest prime of a triple of successive primes, where the middle one is the arithmetic mean of the other two.
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14
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3, 47, 151, 167, 199, 251, 257, 367, 557, 587, 601, 647, 727, 941, 971, 1097, 1117, 1181, 1217, 1361, 1499, 1741, 1747, 1901, 2281, 2411, 2671, 2897, 2957, 3301, 3307, 3631, 3727, 4007, 4397, 4451, 4591, 4651, 4679, 4987, 5101, 5107, 5297, 5381, 5387
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OFFSET
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1,1
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COMMENTS
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These are primes for which the subsequent prime gaps are equal, so (p(k+2)-p(k+1))/(p(k+1)-p(k)) = 1. It is conjectured that prime gaps ratios equal to one are less frequent than those equal to 1/2, 2, 3/2, 2/3, 1/3 and 3. - Andres Cicuttin, Nov 07 2016
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LINKS
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FORMULA
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EXAMPLE
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The prime 7 is not in the list, because in the triple (7, 11, 13) of successive primes, 11 is not equal (7 + 13)/2 = 10.
The second term, 47, is the first prime in the triple (47, 53, 59) of primes, where 53 is the mean of 47 and 59.
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MATHEMATICA
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Clear[d2, d1, k]; d2[n_] = Prime[n + 2] - 2*Prime[n + 1] + Prime[n]; d1[n_] = Prime[n + 1] - Prime[n]; k[n_] = -d2[n]/(1 + d1[n])^(3/2); Flatten[Table[If[k[n] == 0, Prime[n], {}], {n, 1, 1000}]] (* Roger L. Bagula, Nov 13 2008 *)
Transpose[Select[Partition[Prime[Range[750]], 3, 1], #[[2]] == (#[[1]] + #[[3]])/2 &]][[1]] (* Harvey P. Dale, Jan 09 2011 *)
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PROG
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(Haskell)
(PARI) A122535()={n=3; ctr=0; while(ctr<50, avgg=( prime(n-2)+prime(n) )/2;
if( prime(n-1) ==avgg, ctr+=1; print( ctr, " ", prime(n-2) ) ); n+=1); } \\ Bill McEachen, Jan 19 2015
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CROSSREFS
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KEYWORD
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nonn,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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