cp's OEIS Frontend

A337110 Number of length three 1..n vectors that contain their geometric mean.

%I A337110 #42 Jan 02 2023 09:00:42
%S A337110 1,2,3,10,11,12,13,20,33,34,35,42,43,44,45,64,65,78,79,86,87,88,89,96,
%T A337110 121,122,135,142,143,144,145,164,165,166,167,198,199,200,201,208,209,
%U A337110 210,211,218,231,232,233,252,289,314,315,322,323,336,337,344,345,346
%N A337110 Number of length three 1..n vectors that contain their geometric mean.
%C A337110 From _David A. Corneth_, Aug 26 2020: (Start)
%C A337110 If x^2 == 0 (mod n) has only 1 solution then a(n) = a(n-1) + 1. Proof:
%C A337110 Let (a, b, n) be such a tuple. Let without loss of generality b be the geometric mean of the tuple. Then a*b*n = b^3 and as b is not 0 we have b^2 = a*n. So then b^2 == 0 (mod n). If b^2 == 0 (mod n) has only 1 solution then b = n. This gives the tuple (n, n, n) which has 1 permutation. So giving a(n) = a(n-1) + 1. (End)
%H A337110 Hywel Normington, <a href="https://github.com/Horep/Number-of-vectors-that-contain-their-average/blob/master/A337110.py">Python code</a>, 2020.
%H A337110 Hywel Normington, <a href="https://github.com/Horep/Number-of-vectors-that-contain-their-average/blob/master/Julia_Edition/A337110.jl">Julia code</a>, 2023.
%F A337110 a(n) = a(n-1) + 1 + 6*A057918(n).
%e A337110 For n = 2, the a(2) = 2 solutions are: (1,1,1) and (2,2,2).
%e A337110 For n = 4, the a(4) = 10 solutions are: (1,1,1),(2,2,2),(3,3,3),(4,4,4) and the 6 permutations of (1,2,4).
%o A337110 (PARI) first(n) = {my(s = 0, res = vector(n)); for(i = 1, n, s+=b(i); res[i] = s ); res }
%o A337110 b(n) = { my(s = factorback(factor(n)[, 1]), res = 1); for(i = 1, n \ s - 1, c = (s*i)^2/n; if(denominator(c) == 1 && c <= n, res+=6; ) ); res } \\ _David A. Corneth_, Aug 26 2020
%Y A337110 Cf. A120486, A248434, A337111, A013929, A057918, A000188.
%K A337110 nonn,easy
%O A337110 1,2
%A A337110 _Hywel Normington_, Aug 16 2020