A350699 Numbers k such that tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20, where tau is the number of divisors function A000005.
17, 31, 37, 43, 211, 2305, 2731, 19441, 116131, 174595, 222931, 229945, 232051, 243091, 266401, 334315, 350785, 423481, 495265, 523945, 530545, 535915, 539401, 556705, 600601, 663601, 671035, 689131, 721891, 907195, 908041, 1105105, 1113961, 1289731, 1338241
Offset: 1
Keywords
Examples
The table below lists each term k with a pattern (tau(k), ..., tau(k+4)) that appears only once (these appear at n = 1, 3, and 4), as well as each term k that is the smallest one having a pattern that appears repeatedly for large k (these are at n = 2, 6, 8, and 10). It also includes k = a(5) = 211, which is the smallest k that not only has a pattern that appears repeatedly for large k but also has each of k, ..., k+4 divisible by a prime > 5. (k = a(2) = 31 is a special case in that, while it and k = 211 share the same pattern of tau values, i.e., (2, 6, 4, 4, 4), their prime signatures differ at k+1: both 31+1=32 and 211+1=212 have 6 divisors, but 32 is a 5th power.)
Each of the repeatedly occurring patterns corresponds to one of the four possible orders in which the multipliers m=1..5 can appear among 5 consecutive integers of the form m*prime, and thus to a single residue of k modulo 120; e.g., k=2305 begins a run of 5 consecutive integers having the form (5*p, 2*q, 3*r, 4*s, t), where p, q, r, s, and t are distinct primes > 5, and all such runs satisfy k == 25 (mod 120).
For each of the patterns of tau values that does not occur repeatedly, and also for the special case k = 31, one or more of the five consecutive integers in k..k+4 has no prime factor > 5; each such integer appears in parentheses in the "factorization" columns.
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factorization as
# divisors of m*(prime > 5)
n a(n)=k k k+1 k+2 k+3 k+4 k k+1 k+2 k+3 k+4 k mod 120
- ------ --- --- --- --- --- --- --- --- --- --- ---------
1 17 2 6 2 6 4 p (18) r (20) 3t 17
2 31 2 6 4 4 4 p (32) 3r 2s 5t 31
3 37 2 4 4 8 2 p 2q 3r (40) t 37
4 43 2 6 6 4 2 p 4q (45) 2s t 43
5 211 2 6 4 4 4 p 4q 3r 2s 5t 91
6 2305 4 4 4 6 2 5p 2q 3r 4s t 25
8 19441 2 4 4 6 4 p 2q 3r 4s 5t 1
10 174595 4 6 4 4 2 5p 4q 3r 2s t 115
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, #] &, 10^6], 5, 1], 20] // Flatten (* Amiram Eldar, Jan 13 2022 *)
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PARI
isok(k) = numdiv(k) + numdiv(k+1) + numdiv(k+2) + numdiv(k+3) + numdiv(k+4) == 20; \\ Michel Marcus, Jan 13 2022
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Python
from labmath import divcount print([k for k in range(1, 1338242) if divcount(k) + divcount(k+1) + divcount(k+2) + divcount(k+3) + divcount(k+4) == 20]) # Karl-Heinz Hofmann, Jan 13 2022
Formula
{ k : tau(k) + tau(k+1) + tau(k+2) + tau(k+3) + tau(k+4) = 20 }.
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