OFFSET
1,1
COMMENTS
It can be shown that if tau(k) + ... + tau(k+4) = 20, the quintuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4)) must be one of the following, each of which might plausibly occur infinitely often:
(2, 4, 4, 6, 4), which first occurs at k = 19441, 266401, 423481, 539401, ... (A204592);
(2, 6, 4, 4, 4), which first occurs at k = 31, 211, 2731, 116131, ...;
(4, 4, 4, 6, 2), which first occurs at k = 2305, 229945, 350785, 495265, ...;
(4, 6, 4, 4, 2), which first occurs at k = 174595, 334315, 535915, 671035, ... ({A247348(n)} - 4);
or one of the following, each of which occurs only once:
(2, 6, 2, 6, 4), which occurs only at k = 17;
(2, 4, 4, 8, 2), which occurs only at k = 37;
(2, 6, 6, 4, 2), which occurs only at k = 43.
Tau(k) + ... + tau(k+4) >= 20 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+4) < 20 are 1..11, 13, 15, 19, and 25.
LINKS
Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
FORMULA
{ k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20 }.
EXAMPLE
The table below lists each term k with a pattern (tau(k), ..., tau(k+4)) that appears only once (these appear at n = 1, 3, and 4), as well as each term k that is the smallest one having a pattern that appears repeatedly for large k (these are at n = 2, 6, 8, and 10). It also includes k = a(5) = 211, which is the smallest k that not only has a pattern that appears repeatedly for large k but also has each of k, ..., k+4 divisible by a prime > 5. (k = a(2) = 31 is a special case in that, while it and k = 211 share the same pattern of tau values, i.e., (2, 6, 4, 4, 4), their prime signatures differ at k+1: both 31+1=32 and 211+1=212 have 6 divisors, but 32 is a 5th power.)
Each of the repeatedly occurring patterns corresponds to one of the four possible orders in which the multipliers m=1..5 can appear among 5 consecutive integers of the form m*prime, and thus to a single residue of k modulo 120; e.g., k=2305 begins a run of 5 consecutive integers having the form (5*p, 2*q, 3*r, 4*s, t), where p, q, r, s, and t are distinct primes > 5, and all such runs satisfy k == 25 (mod 120).
For each of the patterns of tau values that does not occur repeatedly, and also for the special case k = 31, one or more of the five consecutive integers in k..k+4 has no prime factor > 5; each such integer appears in parentheses in the "factorization" columns.
.
factorization as
# divisors of m*(prime > 5)
n a(n)=k k k+1 k+2 k+3 k+4 k k+1 k+2 k+3 k+4 k mod 120
- ------ --- --- --- --- --- --- --- --- --- --- ---------
1 17 2 6 2 6 4 p (18) r (20) 3t 17
2 31 2 6 4 4 4 p (32) 3r 2s 5t 31
3 37 2 4 4 8 2 p 2q 3r (40) t 37
4 43 2 6 6 4 2 p 4q (45) 2s t 43
5 211 2 6 4 4 4 p 4q 3r 2s 5t 91
6 2305 4 4 4 6 2 5p 2q 3r 4s t 25
8 19441 2 4 4 6 4 p 2q 3r 4s 5t 1
10 174595 4 6 4 4 2 5p 4q 3r 2s t 115
MATHEMATICA
Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^6], 5, 1], 20] // Flatten (* Amiram Eldar, Jan 13 2022 *)
PROG
(Python) from labmath import divcount
print([k for k in range(1, 1338242) if divcount(k) + divcount(k+1) + divcount(k+2) + divcount(k+3) + divcount(k+4) == 20]) # Karl-Heinz Hofmann, Jan 13 2022
(PARI) isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) + numdiv(k+4) == 20; \\ Michel Marcus, Jan 13 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 12 2022
STATUS
approved