A293756 a(n) = smallest number k with n prime factors such that d + k/d is prime for every d | k.

1, 2, 6, 30, 210, 186162

Offset: 0

Author

Thomas Ordowski, Nov 11 2017

Keywords

• nonn
• more

Comments

For n > 0, a(n) is even and squarefree.

For n > 0, a(n) gives 2^(n-1) distinct primes.

If the k-tuple conjecture is true, then this sequence is infinite. - Carl Pomerance, Nov 12 2017

a(n) is the least integer k with n prime divisors such that A282849(k) = A000005(k). - Michel Marcus, Nov 13 2017

a(n) is the smallest k with n prime factors such that A282849(k) = 2^n. - Thomas Ordowski, Nov 13 2017

a(6), if it exists, has a prime divisor greater than 10^3. - Arkadiusz Wesolowski, Nov 14 2017

Examples

			a(2) = 2*3 = 6 because k = 6 is the smallest number with 2 prime factors such that for d = {1, 2, 3, 6} we have 1 + 6/1 = 6 + 6/6 = 7 is prime and 2 + 6/2 = 3 + 6/3 = 5 is prime.
From _Michael De Vlieger_, Nov 14 2017: (Start)
First differences of prime indices of a(n):
n       a(n)   A287352(a(n))
-----------------------------
1         2    1
2         6    1, 1
3        30    1, 1, 1
4       210    1, 1, 1, 1
5    186162    1, 1, 6, 1, 11
(End)
		

Links

• Eric Weisstein's World of Mathematics, k-Tuple Conjecture

Crossrefs

Subsequence of A080715 (d + k/d is prime for every d|k).

Cf. A103787, A282849, A294925, A295124, A295169.

Programs

Maple

with(numtheory): P:=proc(q) local a,b,j,k,n,ok; print(1);for n from 1 to q do for k from 2 to q do a:=ifactors(k)[2]; a:=add(a[j][2],j=1..nops(a)); if a=n then b:=divisors(k); ok:=1;
for j from 1 to nops(b) do if not isprime(b[j]+k/b[j]) then ok:=0; break; fi; od; if ok=1 then print(k); break; fi; fi; od; od; end: P(10^8); # Paolo P. Lava, Nov 16 2017

PARI

isok(k, n)=if (!issquarefree(k), return (0)); if (omega(k) != n, return (0)); fordiv(k, d, if (!isprime(d+k/d), return(0))); 1;
a(n) = {my(k=1); while( !isok(k, n), k++); k;} \\ Michel Marcus, Nov 11 2017

Formula

a(n) = 2*A295124(n-1) for n > 0. - Thomas Ordowski, Nov 15 2017

Extensions

a(5) from Michel Marcus, Nov 11 2017