login
A165712
a(n) = the smallest integer > n that is divisible by exactly the same number of primes (counted with multiplicity) as n is.
4
3, 5, 6, 7, 9, 11, 12, 10, 14, 13, 18, 17, 15, 21, 24, 19, 20, 23, 27, 22, 25, 29, 36, 26, 33, 28, 30, 31, 42, 37, 48, 34, 35, 38, 40, 41, 39, 46, 54, 43, 44, 47, 45, 50, 49, 53, 72, 51, 52, 55, 63, 59, 56, 57, 60, 58, 62, 61, 81, 67, 65, 66, 96, 69, 68, 71, 70, 74, 75, 73, 80
OFFSET
2,1
LINKS
EXAMPLE
8 = 2^3, and so is divisible by exactly 3 primes counted with multiplicity. The next larger number divisible by exactly 3 primes counted with multiplicity is 12, which is 2^2 *3. So a(8) = 12.
MATHEMATICA
a[n_] := For[Om = PrimeOmega[n]; k = n+1, True, k++, If[PrimeOmega[k] == Om, Return[k]]]; Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Jul 21 2017 *)
Module[{nn = 10^2, s, t}, s = PositionIndex@ Array[PrimeOmega, {nn}]; t = ConstantArray[0, nn]; TakeWhile[#, # > 0 &] &@ Rest@ ReplacePart[t, Flatten@ Map[#1 -> #2 & @@ # &, Map[Partition[Lookup[s, #], 2, 1] &, Keys@ s], {2}]]] (* Michael De Vlieger, Jul 21 2017 *)
PROG
(Haskell)
a165712 n = head [x | x <- [n + 1 ..], a001222 x == a001222 n]
-- Reinhard Zumkeller, Aug 29 2013
(PARI) a(n) = {my(bon = bigomega(n)); my(k = n+1); while (bigomega(k) != bon, k++); k; } \\ Michel Marcus, Jul 21 2017
CROSSREFS
Cf. A165713.
Sequence in context: A031948 A247523 A169957 * A325390 A296365 A291166
KEYWORD
nonn
AUTHOR
Leroy Quet, Sep 24 2009
EXTENSIONS
Extended by Ray Chandler, Mar 12 2010
STATUS
approved