A337111 Number of length four 1..n vectors that contain their geometric mean.

1, 2, 3, 16, 17, 18, 19, 56, 105, 106, 107, 144, 145, 146, 147, 208, 209, 306, 307, 320, 321, 322, 323, 432, 529, 530, 723, 736, 737, 738, 739, 968, 969, 970, 971, 1176, 1177, 1178, 1179, 1288, 1289, 1290, 1291, 1304, 1401, 1402, 1403, 1608, 1777, 2018, 2019, 2032

Offset: 1

Author

Hywel Normington, Aug 16 2020

Keywords

• nonn
• easy

Comments

From David A. Corneth, Aug 27 2020: (Start)

Let (a, b, g, n) be a tuple where g is the geometric mean of a*b*g*n and a, b, g <= n. Then there are two cases: g < n and g = n.

If g = n then (a, b, g, n) = (n, n, n, n) adding 1 to the number of permutations.

If g < n then a*b = g^4 / (g*n) = g^3 / n. Furthermore let s be the squarefree part of n (Cf. A007947). Then s | g and so candidates for g are (s*i) where 1 <= i < floor(n/s), depending on whether g^3 / n = (s*i)^3 / n is an integer.

It follows that for suitable values of i, (a, b) are a pair of divisors of (s*i)^3 / n where a*b = (s*i)^3 / n and max(a, b) <= n. (End)

Bounds: n + 12*(floor(n/4) + 2*floor(n/8) + 4*floor(n/9) + 2*floor(n/12) + 2*floor(n/16) + 4*floor(n/18)) <= a(n) <= n^4 - 14*(n-1). - Hywel Normington, Jan 25 2021

Examples

			For n = 2 the a(2) = 2 solutions are: (1,1,1,1) and (2,2,2,2).
For n = 4 the a(4) = 16 solutions are:
  (1, 1, 1, 1), (2, 2, 2, 2), (3, 3, 3, 3), (4, 4, 4, 4),
  and the 12 permutations of (1, 2, 2, 4).
For n = 40, the a(40)-a(39) = 109 new solutions are:
  (40,40,40,40),
  the 24 permutations of (1, 10, 25, 40),
  the 12 permutations of (5, 5, 10, 40),
  the 12 permutations of (5, 20, 40, 40),
  the 24 permutations of (8, 20, 25, 40),
  the 12 permutations of (10, 20, 20, 40),
  and the 24 permutations of (25, 27, 30, 40).
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000
• Hywel Normington, Python code, 2020.
• Hywel Normington, Julia code, 2023.

Crossrefs

Cf. A007947, A000189, A248435, A337110.

Programs

PARI

first(n) = { my(res = vector(n)); s = 0; for(i = 1, n, s += b(i);  res[i] = s; ); res }
b(n) = {my(resa = 1); my(s = factorback(factor(n)[, 1])); for(i = 1, n \ s - 1, s4 = (s*i)^3; if(s4 % n == 0, c = tuples((s*i)^3/n, s*i, n); for(i = 1, #c, resa+=qperms(c[i]) ) ) ); resa }
qperms(v) = {my(r=1,t); v = vecsort(v); for(i=1,#v-1,if(v[i]==v[i+1],t++,r*=binomial(i,t+1);t=0));r*=binomial(#v,t+1)}
tuples(n, s, u) = {my(res = List(), u4n, d, i); d = divisors(n); i = (#d + 1) \ 2; while(i > 0 && d[#d - i + 1] <= u, c = vecsort([d[i], d[#d - i + 1], s, u]); listput(res, c); i--); res} \\ David A. Corneth, Aug 28 2020

Formula

Empirical: if A000189(n) = 1 then a(n) = a(n-1) + 1.

From David A. Corneth, Aug 25 2020: (Start)

The above holds. That is: if x^3 == 0 (mod n) has only one solution then a(n) = a(n-1) + 1. Proof:

Let (a, b, c, n) be such a tuple. Let without loss of generality c be the geometric mean of the tuple. Then a*b*c*n = c^4 and as c is not 0 we have c^3 = a*b*n. So then c^3 == 0 (mod n). If c^3 == 0 (mod n) has only 1 solution then c = n. This gives the tuple (n, n, n, n) which has 1 permutation. So giving a(n) = a(n-1) + 1. (End)

a(n) - a(n-1) == 1 (mod 12). - Hywel Normington, Sep 28 2020