A350686 Numbers k such that tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16, where tau is the number of divisors function A000005.

12, 17, 19, 20, 26, 31, 211, 716, 1226, 1436, 2306, 2731, 2971, 5636, 8011, 12146, 12721, 16921, 18266, 19441, 24481, 24691, 25796, 28316, 30026, 34651, 35876, 37171, 45986, 49681, 51691, 56036, 58676, 61561, 67531, 77276, 98731, 98996, 104161, 104756, 108571

Offset: 1

Author

Jon E. Schoenfield, Jan 11 2022

Keywords

• nonn

Comments

It can be shown that if tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16, the quadruple (tau(k), tau(k+1), tau(k+2), tau(k+3)) must be one of the following, each of which might plausibly occur infinitely often:

(2, 4, 4, 6), which first occurs at k = 12721, 16921, 19441, 24481, ... (A163573);

(2, 6, 4, 4), which first occurs at k = 19, 31, 211, 2731, ...;

(4, 4, 6, 2), which first occurs at k = 26, 1226, 2306, 12146, ...;

(6, 4, 4, 2), which first occurs at k = 20, 716, 1436, 5636, ...; ({A247347(n)-3}, other than its first term)

or one of the following, each of which occurs only once:

(2, 6, 2, 6), which occurs only at k = 17; and

(6, 2, 4, 4), which occurs only at k = 12.

Tau(k) + tau(k+1) + tau(k+2) + tau(k+3) >= 16 for all sufficiently large k; the only numbers k for which tau(k) + tau(k+1) + tau(k+2) + tau(k+3) < 16 are 1..11, 13, 14, and 16.

Examples

			The table below includes all terms k such that at least one of the four numbers k, k+1, k+2, k+3 has no prime factor > 5; each such number appears in parentheses in the columns under "factorization".
The table also includes, for each of the patterns (tau(k), tau(k+1), tau(k+2), tau(k+3)) that continues to appear for large k, the smallest such k for which each of the four numbers k, k+1, k+2, k+3 has a prime factor > 5. For each such quadruple, each of the four numbers is the product of a distinct multiplier m from 1..4 and a prime > 5, and each pattern corresponds to a distinct value of k mod 120: the tau patterns (2, 4, 4, 6), (2, 6, 4, 4), (4, 4, 6, 2), and (6, 4, 4, 2) correspond to k mod 120 = 1, 91, 26, and 116, respectively.
.
                                factorization as
                # divisors of     m*(prime > 5)
   n  a(n)=k    k  k+1 k+2 k+3    k  k+1 k+2 k+3   k mod 120
   -  ------   --- --- --- ---   --- --- --- ---   ---------
   1      12    6   2   4   4    (12)  q  2r  3s       12
   2      17    2   6   2   6      p (18)  r  4s       17
   3      19    2   6   4   4      p (20) 3r  2s       19
   4      20    6   4   4   2    (20) 3q  2r   s       20
   5      26    4   4   6   2     2p (27) 4r   s       26
   6      31    2   6   4   4      p (32) 3r  2s       31
   7     211    2   6   4   4      p  4q  3r  2s       91
   8     716    6   4   2   2     4p  3q  2r   s      116
   9    1226    4   4   6   2     2p  3q  4r   s       26
  17   12721    2   4   4   6      p  2q  3r  4s        1
		

Links

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

Crossrefs

Cf. A000005, A163573, A247347.

Numbers k such that Sum_{j=0..N-1} tau(k+j) = 2*Sum_{k=1..N} tau(k): A000040 (N=1), A350593 (N=2), A350675 (N=3), (this sequence) (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

Programs

Mathematica

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^5], 4, 1], 16] // Flatten (* Amiram Eldar, Jan 12 2022 *)

PARI

isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) == 16; \\ Michel Marcus, Jan 12 2022

Python

from sympy import divisor_count as tau
print([k for k in range( 1, 108572) if tau(k) + tau(k+1) + tau(k+2) + tau(k+3) == 16]) # Karl-Heinz Hofmann, Jan 12 2022

Formula

{ k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16 }.