A248835 a(n) = n + A033677(n), where A033677(n) is the smallest divisor of n >= sqrt(n).

2, 4, 6, 6, 10, 9, 14, 12, 12, 15, 22, 16, 26, 21, 20, 20, 34, 24, 38, 25, 28, 33, 46, 30, 30, 39, 36, 35, 58, 36, 62, 40, 44, 51, 42, 42, 74, 57, 52, 48, 82, 49, 86, 55, 54, 69, 94, 56, 56, 60, 68, 65, 106, 63, 66, 64, 76, 87, 118, 70, 122, 93

Offset: 1

Author

Bob Selcoe, Oct 15 2014

Keywords

• nonn

Comments

Conjecture: For n > 1 there is at least one prime in [n, a(n)] exclusive.

a(n) = 2*n when n is prime.

When n = A002620(m), then a(n) = A002620(m+1), i.e., quarter-squares. Oppermann's conjecture states that there is at least one prime in [A002620(m), A002620(m+1)] exclusive.

When n is square, repeated values for a(n) occur at n-1 and n. These values are A002378(sqrt(n)), i.e., oblong numbers.

When n = A002378(m), then a(n) = (m+1)^2.

Examples

			When n = 40, the smallest divisor of 40 that is greater than or equal to sqrt(40) is 8 so a(40)=48.
		

Links

• Wikipedia, Oppermann's conjecture

Crossrefs

Cf. A002378, A002620, A033677.

Programs

Mathematica

a248835[n_Integer] := n + Min[Select[Divisors[n], # >= Sqrt[n] &]]; a248835 /@ Range[120] (* Michael De Vlieger, Nov 10 2014 *)

PARI

a(n)=fordiv(n,d,if(d^2>=n,return(n+d))) \\ Charles R Greathouse IV, Oct 21 2014

Sage

[n+min([x for x in divisors(n) if x>=sqrt(n)]) for n in [1..100]] # Tom Edgar, Oct 15 2014