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 1-4 of 4 results.

A251428 T(n,k)=Number of length n+2 0..k arrays with the sum of the maximum minus twice the median plus the minimum of adjacent triples multiplied by some arrangement of +-1 equal to zero.

Original entry on oeis.org

2, 9, 12, 16, 41, 12, 29, 116, 97, 40, 42, 237, 380, 341, 56, 61, 432, 1113, 1888, 1003, 144, 80, 725, 2532, 6589, 7458, 3129, 240, 105, 1128, 5097, 18952, 34893, 31980, 9439, 544, 130, 1641, 9120, 44465, 122452, 183341, 127566, 28717, 992, 161, 2316, 15449
Offset: 1

Views

Author

R. H. Hardin, Dec 02 2014

Keywords

Comments

Table starts
....2......9......16........29.........42.........61..........80..........105
...12.....41.....116.......237........432........725........1128.........1641
...12.....97.....380......1113.......2532.......5097........9120........15449
...40....341....1888......6589......18952......44465.......94452.......180625
...56...1003....7458.....34893.....122452.....349257......863840......1913597
..144...3129...31980....183341.....791668....2647405.....7658288.....19250197
..240...9439..127566....938895....4868362...19285211....64105154....183394583
..544..28717..520568...4771117...29782800..137949185...526064608...1698285517
..992..86695.2080650..24063611..179832500..976032575..4251906158..15488167451
.2112.261789.8370976.120999825.1084104408.6871688649.34199112972.140282803741

Examples

			Some solutions for n=5 k=4
..3....3....1....0....3....1....4....3....3....1....3....1....0....4....0....3
..2....1....4....4....1....0....4....0....0....0....1....0....2....2....1....2
..3....2....1....4....1....0....1....3....1....4....2....3....4....1....3....3
..0....3....0....0....0....4....2....2....3....3....1....1....0....3....1....3
..4....1....3....1....3....3....4....3....3....0....3....3....2....4....3....1
..0....3....4....3....0....1....4....4....0....3....2....4....0....1....1....3
..3....3....0....0....3....1....1....4....0....0....2....4....2....2....4....3

Links

Crossrefs

Row 1 is A248434

Formula

Empirical for column k:
k=1: a(n) = 2*a(n-1) +2*a(n-2) -4*a(n-3)
k=2: [order 14]
k=3: [order 29]
Empirical for row n:
n=1: a(n) = 2*a(n-1) -2*a(n-3) +a(n-4); also a polynomial of degree 2 plus a quasipolynomial of degree 0 with period 2
n=2: a(n) = a(n-1) +2*a(n-3) -a(n-4) -a(n-5) -a(n-6) -a(n-7) +2*a(n-8) +a(n-10) -a(n-11); also a polynomial of degree 3 plus a quasipolynomial of degree 1 with period 12
n=3: [order 38]

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

Original entry on oeis.org

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

Views

Author

Hywel Normington, Aug 16 2020

Keywords

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

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).

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 1-4 of 4 results.

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