A349897 Start with b(0)=4. a(n) is the smallest integer such that b(n)=a(n)*b(n-1) has twice as many divisors as b(n-1) has.

3, 5, 6, 7, 11, 12, 13, 17, 19, 20, 23, 29, 31, 35, 37, 41, 43, 47, 53, 59, 61, 63, 67, 71, 73, 79, 83, 89, 97, 99, 101, 103, 107, 109, 113, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 208, 211, 223, 227, 229, 233, 239

Offset

1,1

Comments

If b(0)=4 is replaced by b(0)=1 we get numbers of the form p^(2^k) where p is prime; that is, A050376. The number of divisors of 4*the n-th partial product of sequence terms is 3*2^n.

Conjecture: a composite number is in this sequence if and only if it is the product of 2 consecutive terms of A050376.

Links

Table of n, a(n) for n=1..59.

Example

4 has 3 divisors. a(1) must be a number by which you can multiply 4 to get a number with 3*2=6 divisors. It can't be 2 because 4*2=8 has 4 divisors. It can be 3 because 4*3=12 has 6 divisors.

Prog

(PARI) lista(nn) = {my(b = 4); for (n=1, nn, my(a=1, d=numdiv(b)); while(numdiv(a*b) != 2*d, a++); print1(a, ", "); b = b*a; ); } \\ Michel Marcus, Dec 05 2021

Crossrefs

Cf. A050376.

Sequence in context: A360008 A092559 A242076 * A064728 A046839 A003601

Adjacent sequences: A349894 A349895 A349896 * A349898 A349899 A349900

Keyword

nonn

Author

J. Lowell, Dec 04 2021

Extensions

More terms from Jon E. Schoenfield, Dec 04 2021

Status

approved