A350769 Numbers k such that tau(k) + ... + tau(k+5) = 28, where tau is the number of divisors function A000005.
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
Offset: 1
Keywords
Examples
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
Links
- Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
Position[Plus @@@ Partition[Array[DivisorSigma[0, #] &, 10^7], 6, 1], 28] // Flatten (* Amiram Eldar, Jan 16 2022 *)
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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)) del taulist[0] # Karl-Heinz Hofmann, Jan 18 2022
Formula
{ k : Sum_{j=0..5} tau(k+j) = 28 }.
Comments