A122535 Smallest prime of a triple of successive primes, where the middle one is the arithmetic mean of the other two.

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

Offset: 1

Author

Miklos Kristof, Sep 18 2006

Keywords

• nonn

Comments

Subsets are A047948, A052188, A052189, A052190, A052195, A052197, A052198, etc. - R. J. Mathar, Apr 11 2008

Could be generated by searching for cases A001223(i) = A001223(i+1), writing down A000040(i). - R. J. Mathar, Dec 20 2008

a(n) = A006562(n) - A117217(n). - Zak Seidov, Feb 12 2013

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

Examples

			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.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to primes in arithmetic progressions

Crossrefs

Cf. A006562, A062839, A102552, A117217, A181424.

Programs

Haskell

a122535 = a000040 . a064113  -- Reinhard Zumkeller, Jan 20 2012

Mathematica

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 *)

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

Formula

{A000040(i): A000040(i+1)= (A000040(i)+A000040(i+2))/2 }. - R. J. Mathar, Dec 20 2008

a(n) = A000040(A064113(n)). - Reinhard Zumkeller, Jan 20 2012

Extensions

More terms from Roger L. Bagula, Nov 13 2008

Definition rephrased by R. J. Mathar, Dec 20 2008