A350854 Numbers k such that tau(k) + ... + tau(k+7) = 40, where tau is the number of divisors function A000005.

38, 39, 41, 51, 55, 67, 82, 10780552, 62198632, 884811061, 1457032501, 3573315892, 7321991041, 7391371681, 8557865812, 11434075381, 16893247141, 21599190901, 22487905441, 28044279892, 28273111012, 37923188932, 50238568801, 59635316161, 77814456292, 86148922852

Offset: 1

Author

Jon E. Schoenfield, Jan 19 2022

Keywords

• nonn

Comments

It can be shown that if tau(k) + ... + tau(k+7) = 40, the octuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5), tau(k+6), tau(k+7)) must be one of the following, each of which might plausibly occur infinitely often:

(2, 4, 4, 6, 4, 8, 4, 8), which first occurs at k = 7321991041, 7391371681, 22487905441, ...;

(2, 4, 4, 8, 4, 8, 4, 6), which first occurs at k = 884811061, 1457032501, 11434075381, ...;

(6, 4, 8, 4, 8, 4, 4, 2), which first occurs at k = 3573315892, 8557865812, 28044279892, ...;

(8, 4, 8, 4, 6, 4, 4, 2), which first occurs at k = 10780552, 62198632, 139738178152, ...;

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

(4, 4, 8, 2, 8, 2, 6, 6), which occurs only at k = 38;

(4, 8, 2, 8, 2, 6, 6, 4), which occurs only at k = 39;

(2, 8, 2, 6, 6, 4, 2, 10), which occurs only at k = 41;

(4, 6, 2, 8, 4, 8, 4, 4), which occurs only at k = 51;

(4, 8, 4, 4, 2, 12, 2, 4), which occurs only at k = 55;

(2, 6, 4, 8, 2, 12, 2, 4), which occurs only at k = 67;

(4, 2, 12, 4, 4, 4, 8, 2), which occurs only at k = 82.

The terms of the repeatedly occurring patterns form sequence A071370.

Tau(k) + ... + tau(k+7) >= 40 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+7) < 40 are 1..34, 36, 37, 40, 43, 46, 52, and 61.

Examples

			The table below lists each term k that is the smallest one having a pattern (tau(k), ..., tau(k+7)) that appears repeatedly for large k. Each such pattern corresponds to one of the 4 possible orders in which the multipliers m=1..8 can appear among 8 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=884811061 begins a run of 8 consecutive integers having the form (p, 2*q, 3*r, 8*s, 5*t, 6*u, 7*v, 4*w), where p, q, r, s, t, u, v, and w are distinct primes > 8, and all such runs satisfy k == 1261 (mod 2520).
.
.                   # divisors of    factorization of k+j as
                     k+j for j =      m*(prime > 8) for j =
   n      a(n)=k   0 1 2 3 4 5 6 7    0  1  2  3  4  5  6  7   k mod 2520
   -  ----------   - - - - - - - -   -- -- -- -- -- -- -- --   ----------
   8    10780552   8 4 8 4 6 4 4 2   8p 7q 6r 5s 4t 3u 2v  w      2512
  10   884811061   2 4 4 8 4 8 4 6    p 2q 3r 8s 5t 6u 7v 4w      1261
  12  3573315892   6 4 8 4 8 4 4 2   4p 7q 6r 5s 8t 3u 2v  w      1252
  13  7321991041   2 4 4 6 4 8 4 8    p 2q 3r 4s 5t 6u 7v 8w         1
		

Crossrefs

Cf. A000005, A071370.

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), A350773 (N=7), (this sequence) (N=8).

Programs

Mathematica

Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 100], 8, 1], 40] // Flatten (* Amiram Eldar, Jan 19 2022 *)

Python

from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4, 2, 4]
for k in range(1, 10000000):
    if sum(taulist) == 40: print(k, end=", ")
    taulist.append(tau(k+8))
    del taulist[0] # Karl-Heinz Hofmann, Jan 21 2022

Formula

{ k : Sum_{j=0..7} tau(k+j) = 40 }.