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A350675
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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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8
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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{ k : tau(k) + tau(k+1) + tau(k+2) = 10 }.
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EXAMPLE
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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.
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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
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MATHEMATICA
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Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^4], 3, 1], 10] // Flatten (* Amiram Eldar, Jan 11 2022 *)
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PROG
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(PARI) isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) == 10; \\ Michel Marcus, Jan 16 2022
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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