A246541 Take the squares of all P_(n+2)-rough numbers less than the (n+1)-th primorial and mod each by the (n+1)-th primorial. There will be a(n) different results.

1, 2, 6, 30, 180, 1440, 12960, 142560, 1995840, 29937600, 538876800, 10777536000, 226328256000, 5205549888000, 135344297088000, 3924984615552000, 117749538466560000, 3885734769396480000, 136000716928876800000, 4896025809439564800000, 190945006568143027200000

Offset: 1

Author

John B. Yin, Aug 29 2014

Keywords

• nonn

Comments

The P_(n+2)-rough numbers less than the (n+1)-th primorial also comprise the reduced residue system of the (n+1)-th primorial.

The conjectured formula from Jon E. Schoenfield is true. This can be seen by considering that each odd prime p has exactly (p+1)/2 quadratic residues (mod p), of which (p-1)/2 are nonzero. The P_(n+2)-rough numbers less than the (n+1)-th primorial comprise all combinations of nonzero residues modulo the first n+1 primes. So for each odd prime p, the p-1 nonzero residues map to (p-1)/2 (nonzero) residues after squaring. - Bert Dobbelaere, Aug 09 2023

Examples

			For n=2, P_(n+2) = 7.
The 7-rough numbers less than 2*3*5 are 1,7,11,13,17,19,23,29.
The squares of those numbers mod 2*3*5 are 1,19,1,19,19,1,19,1.
There are 2 different results: 1 and 19; so a(2) = 2.
		

Links

• Eric Weisstein's World of Mathematics, Primorial
• Eric Weisstein's World of Mathematics, Rough Number

Crossrefs

Cf. A002110 (primorial).

Cf. k-rough numbers A007310 (k=5), A007775 (k=7), A008364 (k=11), A008365 (k=13), A008366 (k=17), A166061 (k=19), A166063 (k=23).

Cf. A323739.

Programs

Java

import java.util.TreeSet;
for(int z = 1; z < 10 ; z++) {
int n = z;
int numNumPerLine = 210;
int[] primes = {2,3,5,7,11,13,17,19,23,29,31,37,41,43};
int numRepeats = 1;
int numSpaces = 1;
for(int i = 0; i < n + 1; i++) {
numSpaces *= (primes[i] - 1);
}
int counter = 0;
long integerLength = 1;
for(int i = 0; i < n + 1; i++) {
integerLength *= primes[i];
}
TreeSet numResults = new TreeSet();
numSpaces/=2;
for(int i = 1; i < integerLength / 2; i+=2) {
boolean isInList = true;
for(int j = 1; j < n + 1; j++) {
if(i % primes[j] == 0) {
isInList = false;
}
}
if(isInList) {
long k = i % integerLength;
if(k != 0) {
long l = (k * k) % integerLength;
if(!numResults.contains(l)) {
numResults.add(l);
}
}
}
}
System.out.println(numResults.size());
}

PARI

a(n) = {hp = prod(k=1, n+1, prime(k)); rp = prod(k=1, n+2, prime(k)); v = []; for (i=1, hp, if (gcd(i, rp) == 1, nv = i^2 % hp; if (! vecsearch(v, nv), v = vecsort(concat(v, nv))););); #v;} \\ Michel Marcus, Sep 06 2014

Formula

Conjecture: a(n) = (1/2^n)*Product_{j=1..n} (prime(j+1)-1) = A005867(n+1)/2^n. - Jon E. Schoenfield, Feb 20 2019

a(n) = A323739(n+1). - Bert Dobbelaere, Aug 09 2023