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

1,1

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 }.

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

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

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).

Sequence in context: A206415 A162317 A162318 * A108910 A308193 A179274

Adjacent sequences: A350590 A350591 A350592 * A350594 A350595 A350596

Keyword

nonn

Author

Jon E. Schoenfield, Jan 08 2022

Status

approved