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

1, 2, 3, 10, 11, 12, 13, 20, 33, 34, 35, 42, 43, 44, 45, 64, 65, 78, 79, 86, 87, 88, 89, 96, 121, 122, 135, 142, 143, 144, 145, 164, 165, 166, 167, 198, 199, 200, 201, 208, 209, 210, 211, 218, 231, 232, 233, 252, 289, 314, 315, 322, 323, 336, 337, 344, 345, 346

Offset: 1

Author

Hywel Normington, Aug 16 2020

Keywords

• nonn
• easy

Comments

From David A. Corneth, Aug 26 2020: (Start)

If x^2 == 0 (mod n) has only 1 solution then a(n) = a(n-1) + 1. Proof:

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)

Examples

			For n = 2, the a(2) = 2 solutions are: (1,1,1) and (2,2,2).
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).
		

Links

• Hywel Normington, Python code, 2020.
• Hywel Normington, Julia code, 2023.

Crossrefs

Cf. A120486, A248434, A337111, A013929, A057918, A000188.

Programs

PARI

first(n) = {my(s = 0, res = vector(n)); for(i = 1, n, s+=b(i); res[i] = s ); res }
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

Formula

a(n) = a(n-1) + 1 + 6*A057918(n).