A278311 Numbers n such that n-1 and n+1 have the same number of prime factors as n (with multiplicity).

34, 86, 94, 122, 142, 171, 202, 214, 218, 245, 285, 302, 394, 429, 435, 446, 507, 603, 604, 605, 634, 638, 698, 842, 922, 963, 1042, 1075, 1084, 1085, 1131, 1138, 1245, 1262, 1275, 1310, 1346, 1402, 1413, 1431, 1435, 1449, 1491, 1533, 1557, 1587, 1605, 1635, 1642, 1676, 1762, 1772, 1838, 1886, 1894, 1925, 1942

Offset: 1

Author

Ely Golden, Nov 17 2016

Keywords

• nonn

Examples

			a(1) = 34, as 33, 34, and 35 all have 2 prime factors.
a(2) = 86, as 85, 86, and 87 all have 2 prime factors.
		

Links

• Ely Golden, Table of n, a(n) for n = 1..10000

Crossrefs

Intersection of A045920 and A278291.

a(n) = A045939(n) + 1.

Programs

Java

public class A278311{
public static void main(String[] args)throws Exception{
    long dim0=numberOfPrimeFactors(2);//note that this method must be manually implemented by the user
    long dim1=numberOfPrimeFactors(3);
    long dim2;
    long counter=4;
    long index=1;
    while(index<=10000){
      dim2=numberOfPrimeFactors(counter);
      if(dim2==dim1&&dim1==dim0){System.out.println(index+" "+(counter-1));index++;}
      dim0=dim1;
      dim1=dim2;
      counter++;
    }
  }
}

PARI

isok(n) = (bigomega(n-1) == bigomega(n)) && (bigomega(n) == bigomega(n+1)); \\ Michel Marcus, Nov 17 2016

SageMath

def bigomega(x):
    s=0;
    f=list(factor(x));
    for c in range(len(f)):
        s+=f[c][1]
    return s;
dim0=bigomega(2);
dim1=bigomega(3);
counter=4
index=1
while(index<=10000):
    dim2=bigomega(counter);
    if(dim2==dim1&dim1==dim0):
        print(str(index)+" "+str(counter-1))
        index+=1;
    dim0=dim1;
    dim1=dim2;
    counter+=1;