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A350854 Numbers k such that tau(k) + ... + tau(k+7) = 40, where tau is the number of divisors function A000005. 6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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.

LINKS

Table of n, a(n) for n=1..26.

FORMULA

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

EXAMPLE

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

MATHEMATICA

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

PROG

(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

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).

Sequence in context: A341708 A031958 A288036 * A272035 A056027 A072585

Adjacent sequences:  A350851 A350852 A350853 * A350855 A350856 A350857

KEYWORD

nonn

AUTHOR

Jon E. Schoenfield, Jan 19 2022

STATUS

approved

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Last modified September 27 07:56 EDT 2022. Contains 357052 sequences. (Running on oeis4.)