A359556 Number of ways to represent the average of the n-th twin prime pair as arithmetic mean of the averages of two other twin prime pairs.

0, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 6, 1, 2, 0, 1, 3, 4, 2, 3, 0, 7, 2, 3, 1, 4, 4, 1, 3, 6, 5, 1, 1, 3, 4, 6, 1, 11, 6, 7, 3, 6, 2, 10, 5, 4, 4, 6, 4, 2, 1, 7, 1, 4, 5, 4, 4, 4, 8, 7, 2, 3, 4, 3, 3, 10, 9, 3, 5, 21, 17, 5, 12, 5, 2, 3, 3, 18, 13, 4, 19, 11, 15, 5

Offset

1,7

Links

Amiram Eldar, Table of n, a(n) for n = 1..1200

Example

a(4) = 1 because 18, the average of the 4th twin prime pair (17, 19), can be expressed in one way only as the arithmetic mean of the averages of two other twin prime pairs. These are (5, 7) and (29, 31) with their averages 6 and 30: (6 + 30)/2 = 36/2 = 18.
a(7) = 2 because 60, the average of the 7th twin prime pair (59, 61), can be expressed in two ways as the arithmetic mean of the averages of two other twin prime pairs. Firstly, by the averages 12 and 108 of the twin prime pairs (11, 13) and (107, 109), since (12 + 108)/2 = 120/2 = 60. Secondly, by the averages 18 and 102 of the twin prime pairs (17, 19) and (101, 103), as (18 + 102)/2 = 120/2 = 60 also.
a(15) = 0 because 198, the average of the 15th twin prime pair (197, 199), cannot be expressed as the arithmetic mean of the averages of any other two twin prime pairs.

Mathematica

means = Select[2*Range[3500], PrimeQ[# - 1] && PrimeQ[# + 1] &]; Count[(Plus @@@ Subsets[means, {2}])/2, #] & /@ Select[means, # < Max[means]/2 &] (* Amiram Eldar, Jan 06 2023 *)

Crossrefs

Cf. A014574, A071681.

Sequence in context: A348768 A210673 A129320 * A320844 A212119 A096831

Adjacent sequences: A359553 A359554 A359555 * A359557 A359558 A359559

Keyword

nonn

Author

Tamas Sandor Nagy, Jan 05 2023

Extensions

More terms from Amiram Eldar, Jan 06 2023

Status

approved