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%I #15 Oct 20 2021 12:43:49
%S 1,1,1,1,256,1,3125,1,8916100448256,46656,823543,1,
%T 18446744073709551616,1,387420489,16777216,
%U 1461501637330902918203684832716283019655932542976,1,5842587018385982521381124421,1,1333735776850284124449081472843776,10000000000,302875106592253
%N a(n) = (n')^(n'), where ' is the arithmetic derivative of n.
%C a(p) = 1 for primes p since we have a(p) = (p')^(p') = 1^1 = 1.
%F a(n) = A000312(A003415(n)).
%p a:= n-> (t-> t^t)(n*add(i[2]/i[1], i=ifactors(n)[2])):
%p seq(a(n), n=0..23); # _Alois P. Heinz_, Oct 20 2021
%t Array[#^# &@ If[# < 2, 0, # Total[#2/#1 & @@@ FactorInteger[#]]] &, 19, 2] (* _Michael De Vlieger_, Oct 18 2021 *)
%o (PARI) ad(n) = vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415
%o a(n) = my(d=ad(n)); d^d; \\ _Michel Marcus_, Oct 19 2021
%Y Cf. A000312, A003415, A068327.
%K nonn
%O 0,5
%A _Wesley Ivan Hurt_, Oct 18 2021