|
|
A075541
|
|
Indices of primes p(i) such that (1/3) (p(i)+p(i+1)+p(i+2)) is an integer.
|
|
5
|
|
|
2, 15, 36, 39, 46, 54, 55, 73, 96, 99, 102, 107, 110, 118, 129, 160, 164, 167, 179, 184, 187, 194, 199, 202, 218, 231, 238, 239, 242, 271, 272, 273, 274, 290, 291, 292, 311, 326, 339, 356, 357, 358, 362, 387, 419, 426, 437, 438, 449, 452, 464, 465, 489, 508
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Not every three successive primes have an integer average. The integer averages are in A075540.
Not all of these 3-averages are prime: the prime 3-averages are in A006562 (balanced primes). There are surprisingly many prime 3-averages: among first 117 3-averages, there are 59 primes. Indices i(n) of first prime in sequence of three primes with integer average are in sequence A064113. Interprimes (s-averages with s=2) are all composite, see A024675.
|
|
LINKS
|
|
|
FORMULA
|
i(n)-> 1/3 (p(i)+p(i+1)+p(i+2)) is integer.
|
|
EXAMPLE
|
i(2) = 15 because (p(15)+p(16)+p(17)) = 1/3(47 + 53 + 59)=53 (integer average of three successive primes).
|
|
MATHEMATICA
|
A075541= {}; Do[If[IntegerQ[s3 = (Prime[i] + Prime[i + 1] + Prime[i + 2])/3], A075541 = Append[A075541, i]], {i, 1000}]; (* 119 terms*)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|