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%I #16 Nov 16 2016 11:06:58
%S 1,2,3,3,5,6,7,6,5,10,11,9,13,14,15,12,17,10,19,15,21,22,23,18,7,26,
%T 15,21,29,30,31,24,33,34,35,27,37,38,39,30,41,42,43,33,25,46,47,36,11,
%U 14,51,39,53,30,55,42,57,58,59,45,61,62,35,48,65,66,67,51,69,70,71,54,73,74,21,57,77,78,79,60,45
%N If n is squarefree, a(n) = n, else a(n) = A000040(1+A277885(n)) * (n/(A249739(n)^2)).
%C If n has non-unitary prime divisors, then divide it by the square of the smallest of them and multiply by a single instance of the next larger prime.
%C This differs from related A097246 for the first time at n=16. For both sequences A097248 gives the eventual stable points reached when starting iterating from n.
%H Antti Karttunen, <a href="/A277886/b277886.txt">Table of n, a(n) for n = 1..10000</a>
%F If A277885(n) = 0 [when n is squarefree], then a(n) = n, otherwise a(n) = A000040(1+A277885(n)) * (n/(A249739(n)^2)).
%F Other identities. For all n >= 1:
%F A048675(a(n)) = A048675(n).
%e For n = 12 = 2*2*3, the smallest non-unitary prime divisor (and in this case the only one) is 2, thus we divide with 2^2 and multiply with the next larger prime 3, to get ((2^2 * 3)/(2^2))*3 = 3*3, thus a(12) = 9.
%e For n = 16 = 2^4, we divide two instances of 2 out and multiply by a single instance of 3 to get 2*2*3 = 12.
%t Table[If[SquareFreeQ@ n, n, Prime[1 + PrimePi@ Min[Select[FactorInteger[n][[All, 1]], ! CoprimeQ[#, n/#] &] /. {} -> 0]] (n/If[SquareFreeQ@ n, 1, p = 2; While[! Divisible[n, p^2], p = NextPrime@ p]; p]^2)], {n, 81}] (* _Michael De Vlieger_, Nov 15 2016 *)
%o (Scheme) (define (A277886 n) (if (zero? (A277885 n)) n (* (A000040 (+ 1 (A277885 n))) (/ n (expt (A249739 n) 2)))))
%Y Cf. A000040, A048675, A097246, A097248, A249739, A277885, A277887, A277888, A277896.
%K nonn
%O 1,2
%A _Antti Karttunen_, Nov 08 2016