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
Keywords
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
Programs
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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]; } TreeSetnumResults = 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
Comments