A350675 Numbers k such that tau(k) + tau(k+1) + tau(k+2) = 10, where tau is the number of divisors function A000005.

6, 11, 13, 17, 21, 37, 57, 157, 177, 381, 501, 541, 717, 877, 1201, 1317, 1381, 1437, 1621, 1821, 2017, 2557, 2577, 2857, 2901, 3061, 3117, 3777, 4281, 4357, 4441, 4677, 4701, 5077, 5097, 5581, 5637, 5701, 5937, 6337, 6637, 6661, 6717, 6997, 7417, 8221, 8781

Offset

1,1

Comments

Since tau(k) + tau(k+1) + tau(k+2) = 10 and no three consecutive integers include more than one square, the triple (tau(k), tau(k+1), tau(k+2)) must consist of three even numbers, so it must be one of (2, 2, 6), (2, 4, 4), (2, 6, 2), (4, 2, 4), (4, 4, 2), and (6, 2, 2). Of these, (2, 2, 6) and (6, 2, 2) are impossible. Of the remaining patterns:

(2, 4, 4) requires that k be an odd prime other than 3, followed by two semiprimes, so k is a prime p such that (p+1)/2 and (p+2)/3 are also prime, and such primes are 13, 37, 157, 541, ... (A036570);

(2, 6, 2) requires that (k, k+2) be a twin prime pair whose average has exactly 6 divisors, and is thus either 12 or 18, so k is 11 or 17;

(4, 2, 4) requires that k+1 be an odd prime, with both k and k+2 having exactly 4 divisors, even though one of them is a multiple of 4, so that one is k+2 = 2^3 = 8, so k = 6;

(4, 4, 2) requires that k+2 be an odd prime > 3, preceded by two semiprimes, so k+2 is a prime p such that (p-1)/2 and (p-2)/3 are also prime, so k+2 is in {23, 59, 179, 383, ...} (which is A181841, after its first two terms, 7 and 11), so k is in {A181841(n) - 2} \ {5, 9}, i.e., k is in {21, 57, 177, 381, ...}.

Tau(k) + tau(k+1) + tau(k+2) >= 10 for all sufficiently large k; the only numbers k for which tau(k) + tau(k+1) + tau(k+2) < 10 are 1..5, 7, and 9.

Links

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

Formula

{ k : tau(k) + tau(k+1) + tau(k+2) = 10 }.

UNION({6}, {11, 17}, A036570, {A181841(n) - 2} \ {5, 9}).

a(n) = A317670(n) - 1.

Example

Each of the patterns (tau(k), ..., tau(k+2)) that appears repeatedly for large k corresponds to one of the two possible orders in which the multipliers m=1..3 can appear among 3 consecutive integers of the form m*prime. E.g., k=37 begins a run of 3 consecutive integers having the form (p, 2*q, 3*r), where p, q, and r are distinct primes > 3; k=57 begins a similar run, but there the 3 consecutive integers have the form (3*p, 2*q, r).
For each of the patterns of tau values that does not occur repeatedly for large k, one or more of the three consecutive integers in k..k+2 has no prime factor > 3; in the table below, each such integer appears in parentheses in the columns on the right.
.
                              factorization as
              # divisors of    m*(prime > 3)
  n  a(n)=k     k  k+1 k+2      k   k+1  k+2
  -  ------    --- --- ---    ---- ---- ----
  1      6      4   2   4      (6)   q   (8)
  2     11      2   6   2       p  (12)   r
  3     13      2   4   4       p   2q   3r
  4     17      2   6   2       p  (18)   r
  5     21      4   4   2      3p   2q    r
  6     37      2   4   4       p   2q   3r
  7     57      4   4   2      3p   2q    r

Mathematica

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^4], 3, 1], 10] // Flatten (* Amiram Eldar, Jan 11 2022 *)

Prog

(PARI) isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) == 10; \\ Michel Marcus, Jan 16 2022

Crossrefs

Cf. A000005, A036570, A181841, A307120, A317670.

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), (this sequence) (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), A350773 (N=7), A350854 (N=8).

Sequence in context: A121765 A315368 A315369 * A102974 A315370 A315371

Adjacent sequences: A350672 A350673 A350674 * A350676 A350677 A350678

Keyword

nonn

Author

Jon E. Schoenfield, Jan 10 2022

Status

approved