OFFSET
1,1
COMMENTS
Since tau(k) + tau(k+1) = 6, (tau(k), tau(k+1)) must be (1,5), (2,4), (3,3), (4,2), or (5,1); of these, (1,5) and (5,1) are impossible (tau(m) = 1 only for m=1, but then neither m+1 nor m-1 would have 5 divisors), and (3,3) is also impossible (both k and k+1 would have to be squares of primes), so (tau(k), tau(k+1)) must be either (2,4) or (4,2).
For every prime p, tau(p) = 2. For every semiprime s, tau(s) = 4, with the exception of the squares of primes; for p prime, tau(p^2) = 3, since the divisors of p^2 are 1, p, and p^2.
The only numbers that have exactly 4 divisors but are not semiprimes are the cubes of primes; for prime p, the divisors of p^3 are 1, p, p^2, and p^3.
As a result, this sequence consists of:
(1) the primes p such that (p+1)/2 is prime (A005383), with the exception of p=3 (since p+1 = 4 has 3 divisors, not 4),
(2) semiprimes of the form prime - 1 (A077065), with the exception of the semiprime 4 (since it does not have 4 divisors), and
(3) the special case k = 7, since it is the unique prime p such that p+1 has 4 divisors but is not a semiprime.
For all k > 4, tau(k) + tau(k+1) >= 6; for k = 1..4, tau(k) + tau(k+1) = 3, 4, 5, 5.
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
FORMULA
{ k : tau(k) + tau(k+1) = 6 }.
a(n) = A164977(n+1) for n>=4. - Hugo Pfoertner, Jan 08 2022
EXAMPLE
k tau(k) tau(k+1) tau(k) + tau(k+1)
-- ------ -------- -----------------
1 1 2 1 + 2 = 3
2 2 2 2 + 2 = 4
3 2 3 2 + 3 = 5
4 3 2 3 + 2 = 5
5 2 4 2 + 4 = 6 so 5 = a(1)
6 4 2 4 + 2 = 6 so 6 = a(2)
7 2 4 2 + 4 = 6 so 7 = a(3)
8 4 3 4 + 3 = 7
9 3 4 3 + 4 = 7
10 4 2 4 + 2 = 6 so 10 = a(4)
11 2 6 2 + 6 = 8
12 6 2 6 + 2 = 8
13 2 4 2 + 4 = 6 so 13 = a(5)
MATHEMATICA
Select[Range[1300], Plus @@ DivisorSigma[0, # + {0, 1}] == 6 &] (* Amiram Eldar, Jan 08 2022 *)
Position[Total/@Partition[DivisorSigma[0, Range[1300]], 2, 1], 6]//Flatten (* Harvey P. Dale, Sep 03 2022 *)
PROG
(PARI) isok(k) = numdiv(k) + numdiv(k+1) == 6; \\ Michel Marcus, Jan 08 2022
(Python)
from itertools import count, islice
from sympy import divisor_count
def A350093_gen(): # generator of terms
a, b = divisor_count(1), divisor_count(2)
for k in count(1):
if a + b == 6:
yield k
a, b = b, divisor_count(k+2)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 08 2022
STATUS
approved