login
Numbers k such that tau(k) + ... + tau(k+7) = 40, where tau is the number of divisors function A000005.
6

%I #16 Jan 21 2022 16:03:25

%S 38,39,41,51,55,67,82,10780552,62198632,884811061,1457032501,

%T 3573315892,7321991041,7391371681,8557865812,11434075381,16893247141,

%U 21599190901,22487905441,28044279892,28273111012,37923188932,50238568801,59635316161,77814456292,86148922852

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

%C 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:

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

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

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

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

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

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

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

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

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

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

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

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

%C The terms of the repeatedly occurring patterns form sequence A071370.

%C 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.

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

%e 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).

%e .

%e . # divisors of factorization of k+j as

%e k+j for j = m*(prime > 8) for j =

%e n a(n)=k 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7 k mod 2520

%e - ---------- - - - - - - - - -- -- -- -- -- -- -- -- ----------

%e 8 10780552 8 4 8 4 6 4 4 2 8p 7q 6r 5s 4t 3u 2v w 2512

%e 10 884811061 2 4 4 8 4 8 4 6 p 2q 3r 8s 5t 6u 7v 4w 1261

%e 12 3573315892 6 4 8 4 8 4 4 2 4p 7q 6r 5s 8t 3u 2v w 1252

%e 13 7321991041 2 4 4 6 4 8 4 8 p 2q 3r 4s 5t 6u 7v 8w 1

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

%o (Python) from sympy import divisor_count as tau

%o taulist = [1, 2, 2, 3, 2, 4, 2, 4]

%o for k in range(1, 10000000):

%o if sum(taulist) == 40: print(k, end=", ")

%o taulist.append(tau(k+8))

%o del taulist[0] # _Karl-Heinz Hofmann_, Jan 21 2022

%Y Cf. A000005, A071370.

%Y 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).

%K nonn

%O 1,1

%A _Jon E. Schoenfield_, Jan 19 2022