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A374911 a(n) = a(2^n mod n) + a(3^n mod n), with a(0) = 1.

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%I A374911 #26 Nov 04 2024 18:12:33
%S A374911 1,2,3,4,3,7,7,7,3,4,7,7,7,7,7,10,3,7,11,7,5,10,7,7,7,18,7,8,21,7,7,7,
%T A374911 3,11,7,18,25,7,7,11,5,7,17,7,10,18,7,7,14,14,21,11,10,7,29,14,7,11,7,
%U A374911 7,13,7,7,11,3,17,7,7,10,11,21,7,7,7,7,21,10,32,11,7,5,6,7,7,14,10,7,11,19
%N A374911 a(n) = a(2^n mod n) + a(3^n mod n), with a(0) = 1.
%C A374911 Conjectured to contain all positive integers. Here are the indexes where each of the first few positive integers appear:
%C A374911 1: 0
%C A374911 2: 1
%C A374911 3: 2, 4, 8, 16, 32, ... (2^k, k > 0)
%C A374911 4: 3, 9, ...
%C A374911 5: 20, 40, 80, 272, 320, 328, ...
%C A374911 6: 81, 66469, 144937, ...
%C A374911 7: 5, 6, 7, 10, 11, 12, 13,... (all primes appear except 2 and 3)
%C A374911 8: 27, 301, 729, 1099, 2107, 2187, 85085, 1594323, ...
%C A374911 Most solutions to a(n) = 5 seem to be divisible by 5 and all of them seem to be even. Why?
%C A374911 Are 3 and 9 the only solutions to a(n) = 4?
%H A374911 John Tyler Rascoe, <a href="/A374911/b374911.txt">Table of n, a(n) for n = 0..10000</a>
%F A374911 a(p) = 7 for primes p except 2 and 3.
%F A374911 a(2^n) = 3 for n > 0.
%t A374911 a[0]=1; a[n_]:=a[PowerMod[2,n,n]]+a[PowerMod[3,n,n]]; Array[a,89,0] (* _Stefano Spezia_, Jul 23 2024 *)
%o A374911 (Python)
%o A374911 def a(n):
%o A374911   return 1 if n == 0 else a(pow(2, n, n)) + a(pow(3, n, n))
%o A374911 (PARI) a(n) = if (n==0, 1, a(lift(Mod(2,n)^n)) + a(lift(Mod(3,n)^n))); \\ _Michel Marcus_, Jul 25 2024
%Y A374911 Cf. A000079, A015910, A066601.
%K A374911 nonn,easy
%O A374911 0,2
%A A374911 _Bryle Morga_, Jul 23 2024