A086139 Let p = A046133(n), that is, let p run through the list of primes such that p+12 is also prime (A046133); a(n) = number of primes in the interval p + 1 through p + 11 inclusive.

3, 3, 3, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 3, 3, 2, 1, 1, 1, 1, 1, 1, 0, 0, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 0, 1, 1, 1, 0, 1, 1, 1, 2, 0, 1, 2, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 0, 2, 2, 2, 2, 0, 1, 2, 1, 2, 0, 1, 3, 2, 0, 0, 0, 1, 1, 1, 2, 2

Offset

1,1

Comments

From Michael De Vlieger, Jul 30 2017: (Start)

a(n) = 0 for n = {24, 25, 44, 48, 53, 57, 62, 70, 82, 84, 89, 94, ...}.

a(n) = 1 for n = {9, 14, 18, 19, 20, 21, 22, 23, 28, 29, 30, 33, ...}.

a(n) = 2 for n = {4, 5, 6, 7, 8, 10, 11, 12, 13, 17, 26, 27, 31, ...}.

a(n) = 3 for n = {1, 2, 3, 15, 16, 96, 118, 183, 266, 570, 581, ...}.

(End)

Links

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Example

For n=1, we have p=5, the primes between 5 and 5+12=17 are 7,11,13, so a(1)=3.

Maple

a:=[]; b:=[];
for n from 1 to 200 do if isprime(ithprime(n)+12) then
   a:=[op(a), ithprime(n)];
c:=0;
for i from 1 to 11 do if isprime(ithprime(n)+i) then c:=c+1; fi; od;
b:=[op(b), c];
fi;
od:
a; # A046133b; # this sequence

Mathematica

cp[x_, y_] := Count[Table[PrimeQ[i], {i, x, y}], True]; d = 12; Do[s = Prime[n]; If[PrimeQ[s+d], Print[cp[s+1, s+d-1]]], {n, 1, 1000}]
(* Second program: *)
With[{d = 12}, DeleteCases[#, -1] &@ Table[Function[p, If[PrimeQ[p + d],
Count[Range[p + 1, p + d - 1], _?PrimeQ], -1] ]@ Prime@ n, {n, 252}]]
PrimePi[#+11]-PrimePi[#+1]&/@Select[Prime[Range[400]], PrimeQ[#+12]&] (* Harvey P. Dale, Jul 30 2022 *)

Crossrefs

Cf. A031930, A031931, A046133.

Sequence in context: A136218 A112106 A010608 * A237879 A074804 A242465

Adjacent sequences: A086136 A086137 A086138 * A086140 A086141 A086142

Keyword

nonn

Author

Labos Elemer, Jul 29 2003

Extensions

Definition edited by N. J. A. Sloane, Aug 05 2017 following analysis by Michael De Vlieger, Jul 30 2017

Status

approved