|
| |
|
|
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.
|
|
7
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
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.
|
|
|
LINKS
|
|
|
|
FORMULA
|
{ k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) = 16 }.
|
|
|
EXAMPLE
|
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
|
|
|
MATHEMATICA
|
Position[Plus @@@ Partition[Array[DivisorSigma[0, #] & , 10^5], 4, 1], 16] // Flatten (* Amiram Eldar, Jan 12 2022 *)
|
|
|
PROG
|
(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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
