A249824 Permutation of natural numbers: a(n) = A078898(A003961(A003961(2*n))).

1, 2, 3, 9, 4, 12, 5, 42, 17, 19, 6, 59, 7, 22, 26, 209, 8, 82, 10, 92, 31, 29, 11, 292, 41, 32, 115, 109, 13, 129, 14, 1042, 40, 39, 48, 409, 15, 49, 45, 459, 16, 152, 18, 142, 180, 52, 20, 1459, 57, 202, 54, 159, 21, 572, 63, 542, 68, 62, 23, 642, 24, 69, 213

Offset: 1

Author

Antti Karttunen, Nov 06 2014

Keywords

• nonn

Examples

			a(4) = 9 because of the following. 2n = 2*4 = 8 = 2^3. We replace the prime factor 2 of 8 with the next prime 3 to get 3^3, then replace 3 with 5 to get 5^3 = 125. The smallest prime factor of 125 is 5. 125 is the 9th term of A084967: 5, 25, 35, 55, 65, 85, 95, 115, 125, ..., thus a(4) = 9.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..255
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Inverse: A249823.

Row 3 of A249822.

Cf. A003961, A048673, A078898, A084967, A243071, A246278, A249734, A249746, A249826, A250475, A275716.

Cf. also A273664, A273669.

Programs

Mathematica

t = PositionIndex[FactorInteger[#][[1, 1]] & /@ Range[10^4]]; f[n_] := Times @@ Power[If[# == 1, 1, NextPrime@ #] & /@ First@ #, Last@ #] &@ Transpose@ FactorInteger@ n; Flatten@ Table[Position[Lookup[t, FactorInteger[#][[1, 1]] ], #] &[f@ f[2 n]], {n, 120}] (* Michael De Vlieger, Jul 25 2016, Version 10 *)

Scheme

(define (A249824 n) (A078898 (A003961 (A003961 (* 2 n)))))

Formula

a(n) = A078898(A003961(A003961(2*n))) = A078898(A003961(A249734(n))).

a(n) = A078898(A246278(3,n)).

As a composition of other permutations:

a(n) = A249746(A048673(n)).

a(n) = A250475(A249826(n)).

a(n) = A275716(A243071(n)).

Other identities. For all n >= 1:

a(2n) = A273669(a(n)) and a(A003961(n)) = A273664(a(n)). -- Antti Karttunen, Aug 07 2016