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

1,2

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)

Links

Table of n, a(n) for n=1..58.

Hywel Normington, Python code, 2020.

Hywel Normington, Julia code, 2023.

Formula

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

Example

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).

Prog

(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

Crossrefs

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

Sequence in context: A069967 A061909 A345358 * A007961 A212067 A060811

Adjacent sequences: A337107 A337108 A337109 * A337111 A337112 A337113

Keyword

nonn,easy

Author

Hywel Normington, Aug 16 2020

Status

approved