A064113 Indices k such that (1/3)*(prime(k)+prime(k+1)+prime(k+2)) is a prime.

2, 15, 36, 39, 46, 54, 55, 73, 102, 107, 110, 118, 129, 160, 164, 184, 187, 194, 199, 218, 239, 271, 272, 291, 339, 358, 387, 419, 426, 464, 465, 508, 520, 553, 599, 605, 621, 629, 633, 667, 682, 683, 702, 709, 710, 733, 761, 791, 813, 821, 822, 829, 830

Offset: 1

Author

Jason Earls, Sep 08 2001

Keywords

• easy
• nonn

Comments

n such that d(n) = d(n+1), where d(n) = prime(n+1) - prime(n) = A001223(n).

Of interest because when I generalize it to d(n) = d(n+2), d(n) = d(n+3), etc. I am unable to find any positive number k such that d(n) = d(n+k) has no solution.

From Lei Zhou, Dec 06 2005: (Start)

When (1/3)*(prime(k) + prime(k+1) + prime(k+2)) is prime, then it is equal to prime(k+1).

Also, indices k such that (prime(k)+prime(k+2))/2 = prime(k+1).

The Mathematica program is based on the alternative definition. (End)

Inflection and undulation points of the primes, i.e., positions of zeros in A036263, the second differences of the primes. - Gus Wiseman, Mar 24 2020

Examples

			a(2) = 15 because (p(15)+p(16)+p(17)) = 1/3(47 + 53 + 59) = 53 (prime average of three successive primes).
Splitting the prime gaps into anti-runs gives: (1,2), (2,4,2,4,2,4,6,2,6,4,2,4,6), (6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,10,2,6), (6,4,6), ... Then a(n) is the n-th partial sum of the lengths of these anti-runs. - _Gus Wiseman_, Mar 24 2020
		

Links

• Harry J. Smith, Table of n, a(n) for n=1..1000
• Wikipedia, Inflection point

Crossrefs

Indices of zeros in A036263 (second differences of primes).

Indices (A000720 = primepi) of balanced primes A006562, minus 1.

Cf. A001223, A024675, A075540, A075541.

Cf. A262138.

Complement of A333214.

First differences are A333216.

The version for strict ascents is A258025.

The version for strict descents is A258026.

The version for weak ascents is A333230.

The version for weak descents is A333231.

A triangle for anti-runs of compositions is A106356.

Lengths of maximal runs of prime gaps are A333254.

Anti-runs of compositions in standard order are A333381.

Cf. A000040, A084758, A124762, A124767, A238279.

Programs

Haskell

import Data.List (elemIndices)
a064113 n = a064113_list !! (n-1)
a064113_list = map (+ 1) $ elemIndices 0 a036263_list
-- Reinhard Zumkeller, Jan 20 2012

Mathematica

ct = 0; Do[If[(Prime[k] + Prime[k + 2] - 2*Prime[k + 1]) == 0, ct++; n[ct] = k], {k, 1, 2000}]; Table[n[k], {k, 1, ct}] (* Lei Zhou, Dec 06 2005 *)
Join@@Position[Differences[Array[Prime,100],2],0] (* Gus Wiseman, Mar 24 2020 *)

PARI

d(n) = prime(n+1)-prime(n); j=[]; for(n=1,1500, if(d(n)==d(n+1), j=concat(j,n))); j

PARI

{ n=0; for (m=1, 10^9, if (d(m)==d(m+1), write("b064113.txt", n++, " ", m); if (n==1000, break)) ) } \\ Using d(n) above. - Harry J. Smith, Sep 07 2009

PARI

[n | n<-[1..888], !A036263(n)] \\ M. F. Hasler, Oct 15 2024

PARI

\\ More efficient for larges range of n:
A064113_upto(N, n=1, L=List(), q=prime(n+1), d=q-prime(n))={forprime(p=1+q,, if(d==d=p-q, listput(L,n); #LM. F. Hasler, Oct 15 2024

Python

from itertools import count, islice
from sympy import prime, nextprime
def A064113_gen(startvalue=1): # generator of terms >= startvalue
    c = max(startvalue,1)
    p = prime(c)
    q = nextprime(p)
    r = nextprime(q)
    for k in count(c):
        if p+r==(q<<1):
            yield k
        p, q, r = q, r, nextprime(r)
A064113_list = list(islice(A064113_gen(),20)) # Chai Wah Wu, Feb 27 2024

Formula

A036263(a(n)) = 0; A122535(n) = A000040(a(n)); A006562(n) = A000040(a(n) + 1); A181424(n) = A000040(a(n) + 2). - Reinhard Zumkeller, Jan 20 2012

A262138(2*a(n)) = 0. - Reinhard Zumkeller, Sep 12 2015

a(n) = A000720(A006562(n)) - 1, where A000720 = (prime)pi, A006562 = balanced primes. - M. F. Hasler, Oct 15 2024