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%I #25 Aug 07 2016 20:43:50
%S 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,
%T 32,115,109,13,129,14,1042,40,39,48,409,15,49,45,459,16,152,18,142,
%U 180,52,20,1459,57,202,54,159,21,572,63,542,68,62,23,642,24,69,213
%N Permutation of natural numbers: a(n) = A078898(A003961(A003961(2*n))).
%H Antti Karttunen, <a href="/A249824/b249824.txt">Table of n, a(n) for n = 1..255</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F a(n) = A078898(A003961(A003961(2*n))) = A078898(A003961(A249734(n))).
%F a(n) = A078898(A246278(3,n)).
%F As a composition of other permutations:
%F a(n) = A249746(A048673(n)).
%F a(n) = A250475(A249826(n)).
%F a(n) = A275716(A243071(n)).
%F Other identities. For all n >= 1:
%F a(2n) = A273669(a(n)) and a(A003961(n)) = A273664(a(n)). -- _Antti Karttunen_, Aug 07 2016
%e 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.
%t 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 *)
%o (Scheme) (define (A249824 n) (A078898 (A003961 (A003961 (* 2 n)))))
%Y Inverse: A249823.
%Y Row 3 of A249822.
%Y Cf. A003961, A048673, A078898, A084967, A243071, A246278, A249734, A249746, A249826, A250475, A275716.
%Y Cf. also A273664, A273669.
%K nonn
%O 1,2
%A _Antti Karttunen_, Nov 06 2014
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