A350773 Numbers k such that tau(k) + ... + tau(k+6) = 32, where tau is the number of divisors function A000005.

18, 26, 27, 28, 53, 73, 2914913, 5516281, 6618241, 9018353, 10780553, 18164161, 20239913, 45652313, 51755761, 62198633, 81235441, 91986833, 158764313, 175472641, 191010953, 197375753, 215206201, 322030801, 322461713, 362007353, 513284401, 668745001, 757892513

Offset: 1

Author

Jon E. Schoenfield, Jan 15 2022

Keywords

• nonn

Comments

It can be shown that if tau(k) + ... + tau(k+6) = 32, the septuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5), tau(k+6)) must be one of the following, each of which might plausibly occur infinitely often:

(2, 4, 4, 6, 4, 8, 4), which first occurs at k = 5516281, 18164161, 51755761, ... (A207825);

(2, 8, 4, 6, 4, 4, 4), which first occurs at k = 2914913, 9018353, 20239913, ...;

(4, 4, 4, 6, 4, 8, 2), which first occurs at k = 6618241, 81235441, 215206201, ...;

(4, 8, 4, 6, 4, 4, 2), which first occurs at k = 10780553, 45652313, 62198633, ...;

or one of the following, each of which occurs only once:

(6, 2, 6, 4, 4, 2, 8), which occurs only at k = 18;

(4, 4, 6, 2, 8, 2, 6), which occurs only at k = 26;

(4, 6, 2, 8, 2, 6, 4), which occurs only at k = 27;

(6, 2, 8, 2, 6, 4, 4), which occurs only at k = 28;

(2, 8, 4, 8, 4, 4, 2), which occurs only at k = 53;

(2, 4, 6, 6, 4, 8, 2), which occurs only at k = 73.

The terms of the repeatedly occurring patterns form sequence A071369.

Tau(k) + ... + tau(k+6) >= 32 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+6) < 32 are 1..17, 19..23, 25, 29, 31, 33, 37, and 41.

Examples

			The table below lists each term k with a pattern (tau(k), ..., tau(k+6)) that appears only once (these appear at n = 1..6) as well as each term k that is the smallest one having a pattern that appears repeatedly for large k. (a(10)=9018353 is omitted from the table because it has the same pattern as a(7)=2914913.)
Each of the repeatedly occurring patterns corresponds to one of the 4 possible orders in which the multipliers m=1..7 can appear among 7 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=2914913 begins a run of 7 consecutive integers having the form (1*p, 6*q, 5*r, 4*s, 3*t, 2*u, 7*v), where p, q, r, s, t, u, and v are distinct primes > 7, and all such runs satisfy k == 1793 (mod 2520).
For each of the patterns that does not occur repeatedly, one or more of the seven consecutive integers in k..k+6 has no prime factor > 7; each such integer appears in parentheses in the columns under "factorization".
.
.                # divisors of          factorization
                  k+j for j =          as m*(prime > 7)
   n    a(n)=k   0 1 2 3 4 5 6    k  k+1 k+2 k+3 k+4 k+5 k+6   k mod 2520
   -  --------   - - - - - - -   --- --- --- --- --- --- ---   ----------
   1        18   6 2 6 4 4 2 8   (18)  q (20)(21) 2t   u (24)       18
   2        26   4 4 6 2 8 2 6    2p (27)(28)  s (30)  u (32)       26
   3        27   4 6 2 8 2 6 4   (27)(28)  r (30)  t (32) 3v        27
   4        28   6 2 8 2 6 4 4   (28)  q (30)  s (32) 3u  2v        28
   5        53   2 8 4 8 4 4 2     p (54) 5r (56) 3t  2u   v        53
   6        73   2 4 6 6 4 8 2     p  2q (75) 4s  7t  6u   v        73
   7   2914913   2 8 4 6 4 4 4     p  6q  5r  4s  3t  2u  7v      1793
   8   5516281   2 4 4 6 4 8 4     p  2q  3r  4s  5t  6u  7v         1
   9   6618241   4 4 4 6 4 8 2    7p  2q  3r  4s  5t  6u   v       721
  11  10780553   4 8 4 6 4 4 2    7p  6q  5r  4s  3t  2u   v      2513
		

Links

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

Crossrefs

Cf. A000005, A071369, A207825.

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), A350675 (N=3), A350686 (N=4), A350699 (N=5), A350769 (N=6), (this sequence) (N=7), A350854 (N=8).

Programs

Mathematica

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 7, 1], 32] // Flatten (* Amiram Eldar, Jan 16 2022 *)

Python

from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4, 2]
for k in range(2, 10000000):
    taulist.append(tau(k+6))
    del taulist[0]
    if sum(taulist) == 32: print(k, end=", ") # Karl-Heinz Hofmann, Jan 15 2022

Formula

{ k : Sum_{j=0..6} tau(k+j) = 32 }.