A384835 The exponents (j, k) of the numbers 2^j*3^k that are averages of twin primes, with both j and k > 0, in the order of their sum, and then by j.

1, 1, 1, 2, 2, 1, 2, 3, 3, 2, 4, 3, 6, 1, 5, 4, 7, 2, 3, 10, 6, 7, 2, 15, 12, 5, 18, 1, 18, 5, 21, 4, 24, 5, 27, 4, 11, 24, 30, 7, 32, 9, 33, 8, 31, 12, 36, 7, 43, 2, 32, 15, 43, 8, 50, 9, 63, 2, 66, 25, 79, 20, 99, 10, 57, 64, 82, 63, 63, 88, 56, 99, 148, 27

Offset: 1

Author

Ken Clements, Jun 10 2025

Keywords

• nonn
• tabf

Comments

These are the (j,k) exponents of the numbers 2^j*3^k that are averages of twin primes, ordered by j+k, j. They are remarkable in structure because except for the first pair (2^1*3^1), j+k is always an odd number. I have proof of this, and the reason it is not the case for the first pair is that 6-1=5 is the only number divisible by 5 that is prime.

Examples

			2^a(1) * 3^a(2) = 6.
2^a(3) * 3^a(4) = 18.
2^a(5) * 3^a(6) = 12.
2^a(7) * 3^a(8) = 108.
2^a(9) * 3^a(10) = 72.
		

Links

• Ken Clements, Table of n, a(n) for n = 1..162

Crossrefs

Cf. A027856, A384639 (ordered by value of 2^j*3^k).

Programs

Mathematica

seq[max_] := Flatten@ Transpose[IntegerExponent[Select[Flatten[Table[2^j*3^(m-j), {m, 2, max}, {j, 1, m-1}]], And @@ PrimeQ[# + {-1, 1}] &], #] & /@ {2, 3}]; seq[200] (* Amiram Eldar, Jun 26 2025 *)

Python

from sympy import isprime
def is_TP_pi_2(j, k):
    N = 2**j * 3**k
    return isprime(N-1) and isprime(N+1)
def aupto(limit):
    result = [1, 1]
    for exponent_sum in range(3, limit+1, 2):
        for j in range(1, exponent_sum):
             k = exponent_sum - j
             if is_TP_pi_2(j, k):
                  result.append(j)
                  result.append(k)
    return result
print(aupto(10_000))

Python

import heapq
from gmpy2 import is_prime
from itertools import islice
def agen(): # generator of terms
    v, oldv, h = 1, 0, [(2, 1, 1, 6)]
    while True:
        s, e2, e3, v = heapq.heappop(h)
        if v != oldv:
            if is_prime(v-1) and is_prime(v+1):
                yield from (e2, e3)
            oldv = v
            heapq.heappush(h, (s+1, e2+1, e3, 2*v))
            heapq.heappush(h, (s+1, e2, e3+1, 3*v))
print(list(islice(agen(), 70))) # Michael S. Branicky, Jun 26 2025