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)

Arises in Hardy-Littlewood k-tuple conjecture. Also a(n) is the exact number of d=2 and also d=4 differences in dRRS[modulus=n-th primorial]; see A049296 (dRRS[m]=set of first differences of reduced residue system modulo m).

For n>1 this is the determinant of the (n-1) X (n-1) matrix whose diagonal is A006093(n) = {1, 2, 4, 6, 10, 12, 16, 18..} = the first primes minus 1 and all other elements are 1's. The determinant begins: / (2-1) 1 1 1 1 1 1 ... / 1 (3-1) 1 1 1 1 1 ... / 1 1 (5-1) 1 1 1 1 ... / 1 1 1 (7-1) 1 1 1 ... / 1 1 1 1 (11-1) 1 1 ... / 1 1 1 1 1 (13-1) 1 ... - Alexander Adamchuk, May 21 2006

From Gary W. Adamson, Apr 21 2009: (Start)

Equals (-1)^n * (1, 1, 1, 3, 15, ...) dot (1, -2, 4, -6, 10, ...).

a(6) = 135 = (1, 1, 1, 3, 15) dot (1, -2, 4, -6, 10) = (1, -2, 4, -18, 150). (End)

Arises in Hardy-Littlewood prime k-tuplet conjectural formulas. Also the sequence gives the exact numbers of X42424Y difference-pattern in dRRS[m], where m=modulus=A002110(n). See A049296 (=dRRS[210]=list of first differences of reduced residue system modulo 210=4th primorial). A pattern X42424Y corresponds to a residue-sextuple or it is their difference-quintuple, X,Y > 4. Analogous pattern for primes is in A022008.

a(352) has 1001 decimal digits. - Michael De Vlieger, Mar 06 2017

Technically, the formula is undefined modulo 2# or 3#, but their values are listed as "0", since there are no 6's in the first differences of their reduced residue systems. For our purposes, by "6's", we mean n such that n,n+6 are relatively prime to the primorial modulus, while n+1,n+2,n+3,n+4,n+5 all share a factor (or factors) with p#. The values of this sequence are tied to actual distribution of sexy primes over N (conjecture).

Technically, the formula is undefined modulo 2# or 3#, but I have listed their values as "0", since there are no 8's in the first differences of their reduced residue systems. For our purposes, by "8's", we mean n such that n,n+8 are relatively prime to the primorial modulus, while n+1,n+2,n+3,n+4,n+5,n+6,n+7 all share a factor (or factors) with p#.

Such products arise in Hardy-Littlewood prime k-tuplet conjectural formulas.

P(x) is a function which represents a prime number at a particular ordinal x. This pattern, dp(x), describes the difference between consecutive prime numbers as described by p(x) (see A236175) and therefore the length of dp(x) is len(p(x)) - 1 and each value in dp(x) times P(x) is the difference between values determined not primed when running one pass of a reductive sieve, starting at P(x)^2. See A236185.