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

1,1

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

Links

Harry J. Smith, Table of n, a(n) for n=1..1000

Wikipedia, Inflection point

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

Example

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

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 *)

Prog

(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); #L<N||break); n++; q=p); L} \\ M. F. Hasler, Oct 15 2024
(Haskell)
import Data.List (elemIndices)
a064113 n = a064113_list !! (n-1)
a064113_list = map (+ 1) $ elemIndices 0 a036263_list
-- Reinhard Zumkeller, Jan 20 2012
(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

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.

Sequence in context: A346641 A075541 A075542 * A007217 A295367 A214541

Adjacent sequences: A064110 A064111 A064112 * A064114 A064115 A064116

Keyword

easy,nonn

Author

Jason Earls, Sep 08 2001

Status

approved