A307120 -id:A307120 - cp's OEIS frontend

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

Jon E. Schoenfield, Jan 10 2022

nonn

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.

			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

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

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).

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

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

{ 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.

A307118 a(1) = 0; for n>1, a(n) = dr(n-1) + dr(n) + dr(n+1), where dr(n) is the number of nontrivial divisors of n (A070824).

Original entry on oeis.org

0, 0, 1, 1, 3, 2, 4, 3, 5, 3, 6, 4, 6, 4, 7, 5, 7, 4, 8, 6, 8, 4, 8, 7, 9, 5, 8, 6, 10, 6, 10, 6, 8, 6, 11, 9, 9, 4, 10, 8, 12, 6, 10, 8, 10, 6, 10, 9, 13, 7, 10, 6, 10, 8, 14, 10, 10, 4, 12, 10, 12, 6, 11, 11, 13, 8, 10, 6, 12, 8, 16, 10, 12, 6, 10, 10, 12, 8, 14, 11, 13, 5, 12, 12, 14, 6, 10, 8, 16, 12, 16

Offset: 1

Todor Szimeonov, Mar 25 2019

nonn

Real divisibility of n's one-area (or 1-area). This is the first step to examine the divisibility of n's k-area. n's k-area is the set of m for which |n-m| is less than or equal to k (where n, k, m are natural numbers). 1's 1-area is {1,2}, 5's 1-area {4,5,6}, 3's 2-area {1,2,3,4,5}. We could call this natural area, and still talk about nonnegative or integer areas, etc.

Cf. A070824, A307119, A307120.

{0}~Join~MapAt[# + 1 &, Total /@ Partition[DivisorSigma[0, Range@ 82] - 2, 3, 1], 1] (* Michael De Vlieger, Jun 06 2019 *)

dr(n) = if (n<2, 0, numdiv(n)-2);
a(n) = if (n==1, 0, dr(n-1) + dr(n) + dr(n+1)); \\ Michel Marcus, Apr 11 2019

A307119 a(1) = 1, for n>1, a(n) = dp(n-1) + dp(n) + dp(n+1), where dp(n) is the number of divisors of n less than n (A032741).

Original entry on oeis.org

1, 2, 4, 4, 6, 5, 7, 6, 8, 6, 9, 7, 9, 7, 10, 8, 10, 7, 11, 9, 11, 7, 11, 10, 12, 8, 11, 9, 13, 9, 13, 9, 11, 9, 14, 12, 12, 7, 13, 11, 15, 9, 13, 11, 13, 9, 13, 12, 16, 10, 13, 9, 13, 11, 17, 13, 13, 7, 15, 13, 15, 9, 14, 14, 16, 11, 13, 9, 15, 11, 19, 13, 15, 9, 13, 13, 15, 11, 17, 14, 16, 8

Offset: 1

Todor Szimeonov, Mar 25 2019

nonn

Proper divisibility of n's 1-area. This is the first step to examine the divisibility of n's k-area. n's k-area is the set of m for which |n-m| is less than or equal to k (n, k, m are natural numbers). 1's 1-area is {1,2}, 5's 1-area {4,5,6}, 3's 2-area {1,2,3,4,5}. We could call this natural area, and still talk about nonnegative or integer areas etc.

Cf. A032741, A307118, A307120.

{1}~Join~MapAt[# + 1 &, Total /@ Partition[DivisorSigma[0, Range@82] - 1, 3, 1], 1] (* Michael De Vlieger, Jun 06 2019 *)

dp(n) = if (n < 1, 0, numdiv(n) - 1);
a(n) = dp(n-1) + dp(n) + dp(n+1); \\ Michel Marcus, Jun 11 2019

cp's OEIS Frontend

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

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A307118 a(1) = 0; for n>1, a(n) = dr(n-1) + dr(n) + dr(n+1), where dr(n) is the number of nontrivial divisors of n (A070824).

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A307119 a(1) = 1, for n>1, a(n) = dp(n-1) + dp(n) + dp(n+1), where dp(n) is the number of divisors of n less than n (A032741).

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