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Multiplicative with a(p^e) = A005940(p^e).
12

%I #26 Sep 19 2022 02:02:31

%S 1,2,3,4,5,6,9,8,7,10,15,12,25,18,15,16,11,14,21,20,27,30,45,24,49,50,

%T 75,36,125,30,81,32,45,22,45,28,55,42,75,40,77,54,105,60,35,90,135,48,

%U 121,98,33,100,245,150,75,72,63,250,375,60,625,162,63,64,125,90,39,44,135,90,99,56,91,110,147,84,135,150,189,80,143,154,231,108,55

%N Multiplicative with a(p^e) = A005940(p^e).

%C Question: are there any other numbers n besides 1 and those in A070776, for which a(n) = A005940(n)? At least not below 2^25. This is probably easy to prove.

%H Antti Karttunen, <a href="/A324106/b324106.txt">Table of n, a(n) for n = 1..16384</a>

%H Michael De Vlieger, <a href="/A324106/a324106.png">Fan style binary tree of a(n)</a>, n = 1..2^12, color coded to show the smallest values in the range r = (2^r - 1)..2^(r+1) in blue and highlighting the largest with red.

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%e For n = 85 = 5*17, a(85) = A005940(5) * A005940(17) = 5*11 = 55. Note that A005940(5) is obtained from the binary expansion of 5-1 = 4, which is "100", and A005940(17) is obtained from the binary expansion of 17-1 = 16, which is "1000".

%t nn = 128; Array[Set[a[#], #] &, 2]; Do[If[EvenQ[n], Set[a[n], 2 a[n/2]], Set[a[n], Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &, FactorInteger[a[(n + 1)/2]]]]], {n, 3, nn}]; Array[Times @@ Map[a, Power @@@ FactorInteger[#]] &, nn] (* _Michael De Vlieger_, Sep 18 2022 *)

%o (PARI)

%o A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940

%o A324106(n) = { my(f=factor(n)); prod(i=1, #f~, A005940(f[i,1]^f[i,2])); };

%Y Cf. A005940, A070776, A324107 (fixed points), A324108, A324109.

%K nonn,mult

%O 1,2

%A _Antti Karttunen_, Feb 15 2019