A350593 Numbers k such that tau(k) + tau(k+1) = 6, where tau is the number of divisors function A000005.

5, 6, 7, 10, 13, 22, 37, 46, 58, 61, 73, 82, 106, 157, 166, 178, 193, 226, 262, 277, 313, 346, 358, 382, 397, 421, 457, 466, 478, 502, 541, 562, 586, 613, 661, 673, 718, 733, 757, 838, 862, 877, 886, 982, 997, 1018, 1093, 1153, 1186, 1201, 1213, 1237, 1282

Offset: 1

Author

Jon E. Schoenfield, Jan 08 2022

Keywords

• nonn

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.

Examples

			   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)
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A005383, A077065, A092405, A164977.

Numbers k such that Sum_{j=0..N-1} tau(k+j) = 2*Sum_{k=1..N} tau(k): A000040 (N=1), (this sequence) (N=2), A350675 (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

Programs

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 *)

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)
A350093_list = list(islice(A350093_gen(),12)) # Chai Wah Wu, Jan 11 2022

Formula

{ k : tau(k) + tau(k+1) = 6 }.

UNION(A005383 \ {3}, A077065 \ {4}, {7}).

a(n) = A164977(n+1) for n>=4. - Hugo Pfoertner, Jan 08 2022