cp's OEIS Frontend

Local minima of Euler's phi function. - Walter Nissen

Number of potential primes in a modulus primorial(n+1) sieve. - Robert G. Wilson v, Nov 20 2000

Let p=prime(n) and let p# be the primorial (A002110), then it can be shown that any p# consecutive numbers have exactly a(n-1) numbers whose lowest prime factor is p. For a proof, see the "Proofs Regarding Primorial Patterns" link. For example, if we let p=7 and consider the interval [101,310] containing 210 numbers, we find the 8 numbers 119, 133, 161, 203, 217, 259, 287, 301. - Dennis Martin (dennis.martin(AT)dptechnology.com), Jul 16 2006

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

Equals (-1)^n * (1, 1, 1, 2, 8, 48, ...) dot (-1, 2, -3, 5, -7, 11, ...).

a(6) = 480 = (1, 1, 1, 2, 8, 48) dot (-1, 2, -3, 5, -7, 11) = (-1, 2, -3, 10, -56, 528). (End)

It can be proved that there are at least T prime numbers less than N, where the recursive function T is: T = N - N*Sum_{i=0..T(sqrt(N))} A005867(i)/A002110(i). This can show for example that at least 0.16*N numbers are primes less than N for 29^2 > N > 23^2. - Ben Paul Thurston, Aug 23 2010

First column of A096294. - Eric Desbiaux, Jun 20 2013

Conjecture: The g.f. for the prime(n+1)-rough numbers (A000027, A005408, A007310, A007775, A008364, A008365, A008366, A166061, A166063) is x*P(x)/(1-x-x^a(n)+x^(a(n)+1)), where P(x) is an order a(n) polynomial with symmetric coefficients (i.e., c(0)=c(n), c(1)=c(n-1), ...). - Benedict W. J. Irwin, Mar 18 2016

a(n)/A002110(n+1) (primorial(n+1)) is the ratio of natural numbers whose smallest prime factor is prime(n+1); i.e., prime(n+1) coprime to A002110(n). So the ratio of even numbers to natural numbers = 1/2; odd multiples of 3 = 1/6; multiples of 5 coprime to 6 (A084967) = 2/30 = 1/15; multiples of 7 coprime to 30 (A084968) = 8/210 = 4/105; etc. - Bob Selcoe, Aug 11 2016

The 2-adic valuation of a(n) is A057773(n), being sum of the 2-adic valuations of the product terms here. - Kevin Ryde, Jan 03 2023

For n > 1, a(n) is the number of prime(n+1)-rough numbers in [1, primorial(prime(n))]. - Alexandre Herrera, Aug 29 2023

Also numbers n such that the sum of the 4th powers of the first n positive integers is divisible by n, or A000538(n) = n*(n+1)(2*n+1)(3*n^2+3*n-1)/30 is divisible by n. - Alexander Adamchuk, Jan 04 2007

Also the 7-rough numbers: positive integers that have no prime factors less than 7. - Michael B. Porter, Oct 09 2009

a(n) mod 3 has period 8, repeating [1,1,2,1,2,1,2,2] = (n mod 2) + floor(((n-1) mod 8)/7) - floor(((n-2) mod 8)/7) + 1. floor(a(n)/3) is the set of numbers k such that k is congruent to {0,2,3,4,5,6,7,9} mod 10 = floor((5*n-2)/4)-floor((n mod 8)/6). - Gary Detlefs, Jan 08 2012

Numbers k such that C(k+3,3)==1 (mod k) and C(k+5,5)==1 (mod k). - Gary Detlefs, Sep 15 2013

a(n) mod 30 has period 8 repeating [1, 7, 11, 13, 17, 19, 23, 29]. The mean of these 8 numbers is 120/8 = 15. (a(n)-15) mod 30 has period 8 repeating [-14, -8, -4, -2, 2, 4, 8, 14]. One half of the absolute value produces the symmetric sequence [7, 4, 2, 1, 1, 2, 4, 7] = A061501(((n-1) + 16) mod 8). - Gary Detlefs, Sep 24 2013

a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)*(n + 3*m)(n + 4*m)/120 is integral. Cf. A007310. - Peter Bala, Nov 13 2015

The asymptotic density of this sequence is 4/15. - Amiram Eldar, Sep 30 2020

If a(n) + a(n+1) = 0 (mod 30), then a(n-j) + a(n+j+1) = a(n) + a(n+1) for each j in [1, n-1]. - Alexandre Herrera, Jun 27 2023

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)

For any number n > 0, let f(n) be the polynomial in a single indeterminate x where the coefficient of x^e is the prime(1+e)-adic valuation of n (where prime(k) denotes the k-th prime); f establishes a bijection between the positive numbers and the polynomials in a single indeterminate x with nonnegative integer coefficients; let g be the inverse of f; T(n, k) = g(f(n) * f(k)).

This table has many similarities with A248601.

For any n > 0 and m > 0, f(n * m) = f(n) + f(m).

Also, f(1) = 0 and f(2) = 1.

The function f can be naturally extended to the set of positive rational numbers: if r = u/v (not necessarily in reduced form), then f(r) = f(u) - f(v); as such, f is a homomorphism from the multiplicative group of positive rational numbers to the additive group of polynomials of a single indeterminate x with integer coefficients.

See A297473 for the main diagonal of T.

As a binary operation, T(.,.) is related to A306697(.,.) and A329329(.,.). When their operands are terms of A050376 (sometimes called Fermi-Dirac primes) the three operations give the same result. However the rest of the multiplication table for T(.,.) can be derived from these results because T(.,.) distributes over integer multiplication (A003991), whereas for A306697 and A329329, the equivalent derivation uses distribution over A059896(.,.) and A059897(.,.) respectively. - Peter Munn, Mar 25 2020

From Peter Munn, Jun 16 2021: (Start)

The operation defined by this sequence can be extended to be the multiplicative operator of a ring over the positive rationals that is isomorphic to the polynomial ring Z[x]. The extended function f (described in the author's original comments) is the isomorphism we use, and it has the same relationship with the extended operation that exists between their unextended equivalents.

Denoting this extension of T(.,.) as t_Q(.,.), we get t_Q(n, 1/k) = t_Q(1/n, k) = 1/T(n, k) and t_Q(1/n, 1/k) = T(n, k) for positive integers n and k. The result for other rationals is derived from the distributive property: t_Q(q, r*s) = t_Q(q, r) * t_Q(q, s); t_Q(q*r, s) = t_Q(q, s) * t_Q(r, s). This may look unusual because standard multiplication of rational numbers takes on the role of the ring's additive group.

There are many OEIS sequences that can be shown to be a list of the integers in an ideal of this ring. See the cross-references.

There are some completely additive sequences that similarly define by extension completely additive functions on the positive rationals that can be shown to be homomorphisms from this ring onto the integer ring Z, and these functions relate to some of the ideals. For example, the extended function of A048675, denoted A048675_Q, maps i/j to A048675(i) - A048675(j) for positive integers i and j. For any positive integer k, the set {r rational > 0 : k divides A048675_Q(r)} forms an ideal of the ring; for k=2 and k=3 the integers in this ideal are listed in A003159 and A332820 respectively.

(End)

Also the 17-rough numbers: positive integers that have no prime factors less than 17. - Michael B. Porter, Oct 10 2009

a(n) - (1001/192) n is periodic with period 5760. - Robert Israel, Mar 18 2016

From Peter Bala, May 12 2018: (Start)

The product of two 17-rough numbers is a 17-rough number and the prime factors of a 17-rough number are 17-rough numbers.

Let k equal either 13, 14, 15 or 16. Then the product of k numbers n*(n + a)*(n + 2*a)*...*(n + (k-1)*a) in arithmetical progression is divisible by k! for all integer n if and only if a is a 17-rough number.

The sequence terms satisfy the congruence x^60 = 1 (mod 30030), where 30030 = 2*3*5*7*11*13. (End)

The asymptotic density of this sequence is 192/1001. - Amiram Eldar, Sep 30 2020

Or, positive integers relatively prime to 510510 = 2*3*5*7*11*13*17.

Or, positive integers relatively prime to 9699690 = 2*3*5*7*11*13*17*19.

First composite term is 529 = 23^2.

See sequence A322269 for further explanations. This sequence is related to A322269(5).

The sequence is periodic, repeating itself after phi(9240) terms. Its largest term is 1873, which is A322269(5). In order to satisfy the conditions, both f and b must be coprime to 9240. Otherwise, the product would be zero mod a prime <= 11.

The b(n) corresponding to each a(n) is A008365(n).

The first 28 entries are trivial: f=b, and then the product b*f naturally is a square modulo everything.

Equivalent: Numbers such that the LCM of their palindromic divisors (A087999) is 1, or,

Numbers such that the number of palindromic divisors (A087990) is 1.

All terms are odd.

The 1st family consists of non-palindromic primes that form the subsequence A334321.

The 2nd family consists of {p^k, p prime, k >= 2} such that p^j for 1 <= j <= k is not a palindrome {169 = 13^2, 289 = 17^2, 361 = 19^2, ..., 2197 = 13^3, ...} (see examples).

The 3rd family consists of products p_1^q_1 * ... * p_k^q_k with k >= 2, all of whose divisors are nonpalindromic {221 = 13 * 27, 247 = 13 * 19, 299 = 13 * 23, 377 = 13 * 29, 391 = 17 * 23, 403 = 13 * 31, 481 = 13 * 37, ...}.

Also, equivalent: numbers all of whose divisors > 1 are nonpalindromic (A029742). - Bernard Schott, Jul 14 2022