cp's OEIS Frontend

The first A005867(4) = 48 terms give the reduced residue system for the 4th primorial number 210 = A002110(4).

This sequence is closed under multiplication: any product of terms is also a term. - Labos Elemer, Feb 26 2003

Conjecture: these are numbers n such that (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0. - Gary Detlefs, Dec 20 2011

From Peter Bala, May 03 2018: (Start)

The above conjecture is true. Let m be even and let the m-th Bernoulli number be written in reduced form as Bernoulli(m) = N(m)/D(m). Apply Ireland and Rosen, Proposition 15.2.2, to show the congruence D(m)*( Sum_{k = 1..n} k^m )/n = N(m) (mod n) holds for all n >= 1. It follows easily from this congruence that ( Sum_{k = 1..n} k^m )/n is integral iff n is coprime to D(m). Now Bernoulli(4) = -1/(2*3*5) and Bernoulli(6) = 1/(2*3*7) so the numbers n such that both (Sum_{k=1..n} k^4) mod n = 0 and (Sum_{k=1..n} k^6) mod n = 0 are exactly those numbers coprime to the primes 2, 3, 5 and 7, that is, the 11-rough numbers. (End)

Conjecture: these are numbers n such that (n^6 mod 210 = 1) or (n^6 mod 210 = 169). - Gary Detlefs, Dec 30 2011

The second Detlefs conjecture above is true and extremely easy to verify with some basic properties of congruences: take the terms of this sequence up to 209 and compute their sixth powers modulo 210: there should only be 1's and 169's there. Then take the complement of this sequence up to 210, where you will see no instances of 1 or 169. - Alonso del Arte, Jan 12 2014

It is well-known that the product of 7 consecutive integers is divisible by 7!. Conjecture: This sequence is exactly the set of positive values of r such that ( Product_{k = 0..6} n + k*r )/7! is an integer for all n. - Peter Bala, Nov 14 2015

From Ruediger Jehn, Nov 05 2020: (Start)

This conjecture is true. The first part of the proof deals with numbers not in A008364, i.e., numbers which are divisible by p (p either 2, 3, 5, 7). Let r = p*s and n = 1, then (Product_{k = 0..6} n + k*r) is not divisible by p, because none of the factors 1 + k*p*s are divisible by p. Hence dividing the product by 7! does not return an integer.

The second part deals with numbers in A008364. If r and q are coprime, then for any i < q there exists k < q with (k*r mod q) = i. From this, it also follows that for any n there exists k < q with ((n + k*r) mod q) = 0. But this means that Product_{k = 0..6} n + k*r is divisible by all numbers from 2 to 7 because there is always a factor that is divisible. We still have to show that the product is also divisible by 2 times 3 times 4 times 6. If the k_1 with ((n + k_1*r) mod 4) = 0 is even, then (n mod 2) = ((n + 2*r) mod 2) = ((n + 4*r) mod 2) = ((n + 6*r) mod 2) = 0. If this k_1 is odd, then ((n + r) mod 2) = ((n + 3*r) mod 2) = ((n + 5*r) mod 2) = 0. In both cases there are at least 2 other factors divisible by 2. If the k_2 with ((n + k_2*r) mod 6) = 0 is smaller than 4, then ((n + (k_2 + 3)*r) mod 3) = 0. Otherwise, ((n + (k_2 - 3)*r) mod 3) = 0. In both cases there is at least 1 other factor divisible by 3. And therefore Product_{k = 0..6} n + k*r is divisible by 7! for any n.

(End)

The number of primes p <= n with p coprime to n. - Enrique Pérez Herrero, Jul 23 2011

Numbers coprime to 30 in that number's reduced residue system. - Alonso del Arte, Oct 03 2017

From Reinhard Zumkeller, Oct 27 2010: (Start)

Conjecture: the sequence is finite and 60 is the largest term, empirically verified up to 10^7;

A139555(a(n)) = A000010(a(n)). (End)

The sequence is indeed finite. Let pi*(x) denote the number of prime powers (including 1) up to x. Dusart's bounds plus finite checking [up to 60184] shows that pi*(x) <= x/(log(x) - 1.1) + sqrt(x) for x >= 4. phi(n) > n/(e^gamma log log n + 3/(log log n)) for n >= 3. Convexity plus finite checking [up to 1096] allows a quick proof that phi(n) > pi*(n) for n > 420. So if n > 420, the reduced residue system mod n must contain at least one number that is neither 1 nor a prime power. Hence 60 is the last term in the sequence. - Charles R Greathouse IV, Jul 14 2011

Number of occurrences of n in A060646 or n-th prime in A076367.

Sequences A060646, A076367, A076368 were used in proving a property of 30. See A048597, A060646 and corresponding References. It is provable [Bonse] that a(n)>=3 if n>3.

For n>1, a(n) is the sum of the digits of prime(n) in the base prime(n-1). - R. J. Cano, Dec 31 2016; corrected by Michel Marcus, Jan 01 2017

30 is the largest number n with the property that if 1 < k < n and k is relatively prime to n, then k is prime. In other words, a(30) = 0 and if m > 30, then a(m) > 0. - Jonathan Sondow, Dec 08 2012

It is conjectured that x is always bounded.

If p and q are primes < sqrt(x) that do not divide x, then p*q is in RRS(x). Thus the number of composites in RRS(x) is at least (pi(sqrt(x)) - log_2(x))^2/2. If x is too large, this must be greater than n. Thus suppose N is large enough that pi(sqrt(N)) > 2*sqrt(2*n) and for all x >= N, pi(sqrt(x)) > 2*log_2(x). Then a(n) <= N. It appears that the condition pi(sqrt(x)) > 2*log_2(x) is true for all x >= 103^2. - Robert Israel, Aug 26 2018, corrected Feb 24 2020

From Giovanni Resta, Feb 25 2020: (Start)

The following bounds (valid for n>1) are known:

primepi(n) < 1.256*n/log(n),

omega(n) > 0,

phi(n) > n/(3/log(log(n)) + exp(g)*log(log(n))), where g = A001620 = 0.5770836... is the Euler-Mascheroni constant.

Combining these bounds we obtain a lower bound for A048864(k) = phi(k) - primepi(k) + omega(k), which allows the establishment of a finite search range when solving A048864(x) = n. (End)

From Michael De Vlieger, May 21 2017: (Start)

Conjecture: all terms are even.

Let b(n) = differences of a(n). Many of the terms in b(n) are in A060735(n), i.e., b(n) >= A002110(n) and divisible by A002110(n). The smallest b(n) that is not in A060735 is b(450) = 66; b(450) > A002110(n) but not divisible by it; instead it is divisible by A002110(n-1). b(n) seems to "prefer" values of A002110(n) as n increases. There is a run of 22 values of 2 starting at b(2), of 22 values of 6 starting at b(178), of 7 values of 210 starting at b(543), and 3 values of 2310 starting at b(577). In the case of A002110(n) with n < 3, the length of runs of A002110(n) is greater than that of runs of any other number of comparable magnitude. The last observed position of A002110(n) in b(n) is {201, 442, (557), ...}.

Questions: are there increasingly more terms such that b(n) is not in A060735 as n increases? Are terms such that b(n) is in A060735 prevalent? Why does b(n) seem to "prefer" values in A002110 at certain magnitudes of n as n increases? Are there differences of a(n) = A002110(k) at positions greater than those observed, and if not, why? (End)