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A350769
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Numbers k such that tau(k) + ... + tau(k+5) = 28, where tau is the number of divisors function A000005.
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7
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27, 28, 30, 37, 38, 41, 42, 57, 18362, 2914913, 5516281, 6618242, 7224834, 9018353, 9339114, 10780554, 16831081, 17800553, 18164161, 18646202, 20239913, 29743561, 32464433, 32915513, 42464514, 43502033, 45652314, 51755761, 53464314, 62198634, 69899754
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OFFSET
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1,1
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COMMENTS
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It can be shown that if tau(k) + ... + tau(k+5) = 28, the sextuple (tau(k), tau(k+1), tau(k+2), tau(k+3), tau(k+4), tau(k+5)) must be one of the following, each of which might plausibly occur infinitely often:
(2, 4, 4, 6, 4, 8), which first occurs at k = 5516281, 16831081, 18164161, ... (A208455);
(2, 8, 4, 6, 4, 4), which first occurs at k = 2914913, 9018353, 17800553, ...;
(4, 4, 6, 4, 8, 2), which first occurs at k = 18362, 6618242, 18646202, ...;
(8, 4, 6, 4, 4, 2), which first occurs at k = 7224834, 9339114, 10780554, ...;
or one of the following, each of which occurs only once:
(4, 6, 2, 8, 2, 6), which occurs only at k = 27;
(6, 2, 8, 2, 6, 4), which occurs only at k = 28;
(8, 2, 6, 4, 4, 4), which occurs only at k = 30;
(2, 4, 4, 8, 2, 8), which occurs only at k = 37;
(4, 4, 8, 2, 8, 2), which occurs only at k = 38;
(2, 8, 2, 6, 6, 4), which occurs only at k = 41;
(8, 2, 6, 6, 4, 2), which occurs only at k = 42;
(4, 4, 2, 12, 2, 4), which occurs only at k = 57.
The terms of the repeatedly occurring patterns form sequence A071368.
Tau(k) + ... + tau(k+5) >= 28 for all sufficiently large k; the only numbers k for which tau(k) + ... + tau(k+5) < 28 are 1..26, 29, 33, and 34.
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LINKS
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FORMULA
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{ k : Sum_{j=0..5} tau(k+j) = 28 }.
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EXAMPLE
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The table below lists each term k with a pattern (tau(k), ..., tau(k+5)) that appears only once (these appear at n = 1..8) as well as each term k that is the smallest one having a pattern that appears repeatedly for large k. (a(12)=6618242 is omitted from the table because it has the same pattern as a(9)=18362.)
Each of the repeatedly occurring patterns corresponds to one of the four possible orders in which the multipliers m=1..6 can appear among 6 consecutive integers of the form m*prime, and thus to a single residue of k modulo 2520; e.g., k=18362 begins a run of 6 consecutive integers having the form (2*p, 3*q, 4*r, 5*s, 6*t, 1*u), where p, q, r, s, t, and u are distinct primes > 6, and all such runs satisfy k == 722 (mod 2520).
For each of the patterns that does not occur repeatedly, one or more of the six consecutive integers in k..k+5 has no prime factor > 6; each such integer appears in parentheses in the "factorization" columns.
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. factorization as k
# divisors of m*(prime > 6) mod
n a(n)=k k k+1 k+2 k+3 k+4 k+5 k k+1 k+2 k+3 k+4 k+5 2520
- -------- --- --- --- --- --- --- --- --- --- --- --- --- ----
1 27 4 6 2 8 2 6 (27) 4q r (30) t (32) 27
2 28 6 2 8 2 6 4 4p q (30) s (32) 3u 28
3 30 8 2 6 4 4 4 (30) q (32) 3s 2t 5u 30
4 37 2 4 4 8 2 8 p 2q 3r (40) t 6u 37
5 38 4 4 8 2 8 2 2p 3q (40) s 6t u 38
6 41 2 8 2 6 6 4 p 6q r 4s (45) 2u 41
7 42 8 2 6 6 4 2 6p q 4r (45) 2t u 42
8 57 4 4 2 12 2 4 3p 2q r (60) t 2u 57
9 18362 4 4 6 4 8 2 2p 3q 4r 5s 6t u 722
10 2914913 2 8 4 6 4 4 p 6q 5r 4s 3t 2u 1793
11 5516281 2 4 4 6 4 8 p 2q 3r 4s 5t 6u 1
13 7224834 8 4 6 4 4 2 6p 5q 4r 3s 2t u 2514
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MATHEMATICA
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Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 6, 1], 28] // Flatten (* Amiram Eldar, Jan 16 2022 *)
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PROG
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(Python) from sympy import divisor_count as tau
taulist = [1, 2, 2, 3, 2, 4]
for k in range(1, 10000000):
if sum(taulist) == 28: print(k, end=", ")
taulist.append(tau(k+6))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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