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A350773
Numbers k such that tau(k) + ... + tau(k+6) = 32, where tau is the number of divisors function A000005.
7
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
1,1
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.
LINKS
FORMULA
{ k : Sum_{j=0..6} tau(k+j) = 32 }.
EXAMPLE
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
MATHEMATICA
Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 7, 1], 32] // Flatten (* Amiram Eldar, Jan 16 2022 *)
PROG
(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
CROSSREFS
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).
Sequence in context: A362435 A248111 A109769 * A229967 A212049 A212052
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 15 2022
STATUS
approved