cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Hywel Normington, Aug 16 2020

Keywords

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

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

A337559 Number of length three 1..n vectors that contain their harmonic mean.

Original entry on oeis.org

1, 2, 3, 4, 5, 18, 19, 20, 21, 22, 23, 36, 37, 38, 51, 52, 53, 66, 67, 80, 81, 82, 83, 96, 97, 98, 99, 112, 113, 138, 139, 140, 141, 142, 155, 168, 169, 170, 171, 184, 185, 210, 211, 212, 237, 238, 239, 252, 253, 254, 255, 256, 257, 270, 271, 284, 285, 286, 287, 324, 325, 326, 339
Offset: 1

Views

Author

Hywel Normington, Aug 31 2020

Keywords

Examples

			For n = 1, the only solution is (1,1,1).
For n = 6, the a(6) = 18 solutions are (k,k,k) for k=1,..,6, the 6 permutations of (2,3,6) and the 6 permutations of (3,4,6).
For n = 40, the a(40)-a(39) = 13 new solutions are (40,40,40), the 6 permutations of (10,16,40) and the 6 permutations of (24,30,40).

Links

Crossrefs

Cf. A248434, A337110, A337111, A337560.

Formula

Empirical: If A174903(n) = 0, a(n) = a(n-1) + 1.
a(n)-a(n-1) = 1 (mod 6).

A337560 Number of length four 1..n vectors that contain their harmonic mean.

Original entry on oeis.org

1, 2, 3, 28, 29, 78, 79, 104, 105, 154, 155, 252, 253, 254, 351, 376, 377, 426, 427, 620, 717, 718, 719, 912, 913, 914, 915, 1084, 1085, 1374, 1375, 1400, 1425, 1426, 1619, 1812, 1813, 1814, 1839, 2128, 2129, 2442, 2443, 2564, 2829, 2830, 2831, 3192, 3193
Offset: 1

Views

Author

Hywel Normington, Aug 31 2020

Keywords

Examples

			For n = 1, the a(1) = 1 solution is (1,1,1,1).
For n = 4, the a(4) = 28 solutions are (k,k,k,k) for k=1,2,3,4, the 12 permutations of (1,2,4,4) and the 12 permutations of (2,3,4,4).
For n = 40, the a(40) - a(39) = 289 new solutions are:
  the 24 permutations of (4, 8, 10, 40),
  the 12 permutations of (4, 10, 40, 40),
  the 12 permutations of (5, 12, 40, 40),
  the 24 permutations of (8, 10, 12, 40),
  the 24 permutations of (8, 15, 20, 40),
  the 12 permutations of (10, 16, 16, 40),
  the 24 permutations of (10, 18, 24, 40),
  the 12 permutations of (10, 20, 40, 40),
  the 24 permutations of (12, 20, 24, 40),
  the 24 permutations of (14, 24, 35, 40),
  the 24 permutations of (15, 24, 30, 40),
  the 12 permutations of (16, 16, 20, 40),
  the 12 permutations of (20, 20, 24, 40),
  the 12 permutations of (20, 30, 40, 40),
  the 12 permutations of (24, 30, 30, 40),
  the 12 permutations of (28, 35, 40, 40),
  the 12 permutations of (30, 36, 40, 40),
  and (40, 40, 40, 40).

Links

Crossrefs

Cf. A248434, A337110, A337111, A337559.

Formula

a(n) - a(n-1) = 1 (mod 12).

Extensions

a(43)-a(49) from Alois P. Heinz, Feb 04 2023
Showing 1-3 of 3 results.

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