A350773 Numbers k such that tau(k) + ... + tau(k+6) = 32, where tau is the number of divisors function A000005.
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
Keywords
Examples
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
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], 7, 1], 32] // 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, 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
Formula
{ k : Sum_{j=0..6} tau(k+j) = 32 }.
Comments