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 1-7 of 7 results.

A183897 Number of nondecreasing arrangements of n+3 numbers in 0..2 with each number being the sum mod 3 of three others.

Original entry on oeis.org

1, 7, 17, 25, 34, 44, 55, 67, 80, 94, 109, 125, 142, 160, 179, 199, 220, 242, 265, 289, 314, 340, 367, 395, 424, 454, 485, 517, 550, 584, 619, 655, 692, 730, 769, 809, 850, 892, 935, 979, 1024, 1070, 1117, 1165, 1214, 1264, 1315, 1367, 1420, 1474, 1529, 1585, 1642
Offset: 1

Views

Author

R. H. Hardin, Jan 07 2011

Keywords

Comments

Column 2 of A183904.

Examples

			Some solutions for n=4:
..0....0....0....0....1....0....0....0....1....0....0....0....0....0....0....0
..1....1....0....0....1....0....0....0....1....0....0....0....0....0....1....0
..1....1....2....0....1....0....1....0....1....0....1....0....0....1....1....1
..2....1....2....0....2....1....1....0....1....1....1....0....0....1....1....1
..2....1....2....1....2....1....1....2....2....2....2....0....1....1....1....1
..2....2....2....1....2....1....1....2....2....2....2....1....2....1....1....2
..2....2....2....1....2....1....1....2....2....2....2....1....2....2....2....2

Links

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/2)*n^2 + (9/2)*n - 1 for n>2.
Conjectures from Colin Barker, Apr 05 2018: (Start)
G.f.: x*(1 + x - x^2)*(1 + 3*x - 3*x^2) / (1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>5.
(End)

A183898 Number of nondecreasing arrangements of n+3 numbers in 0..3 with each number being the sum mod 4 of three others.

Original entry on oeis.org

5, 14, 38, 70, 111, 162, 224, 298, 385, 486, 602, 734, 883, 1050, 1236, 1442, 1669, 1918, 2190, 2486, 2807, 3154, 3528, 3930, 4361, 4822, 5314, 5838, 6395, 6986, 7612, 8274, 8973, 9710, 10486, 11302, 12159, 13058, 14000, 14986, 16017, 17094, 18218, 19390
Offset: 1

Views

Author

R. H. Hardin, Jan 07 2011

Keywords

Comments

Column 3 of A183904.

Examples

			All solutions for n=2:
..0....2....0....1....0....0....1....1....0....0....0....1....0....0
..2....2....0....3....1....0....1....1....0....0....1....1....1....0
..2....2....2....3....1....0....3....1....2....0....2....1....1....1
..3....2....3....3....2....0....3....3....2....2....3....1....2....1
..3....2....3....3....3....0....3....3....2....2....3....3....2....2

Links

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/6)*n^3 + (5/2)*n^2 + (25/3)*n - 14 for n>1.
Conjectures from Colin Barker, Apr 05 2018: (Start)
G.f.: x*(5 - 6*x + 12*x^2 - 18*x^3 + 8*x^4) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>5.
(End)

A183899 Number of nondecreasing arrangements of n+3 numbers in 0..4 with each number being the sum mod 5 of three others.

Original entry on oeis.org

1, 5, 66, 174, 329, 539, 815, 1169, 1614, 2164, 2834, 3640, 4599, 5729, 7049, 8579, 10340, 12354, 14644, 17234, 20149, 23415, 27059, 31109, 35594, 40544, 45990, 51964, 58499, 65629, 73389, 81815, 90944, 100814, 111464, 122934, 135265, 148499, 162679
Offset: 1

Views

Author

R. H. Hardin, Jan 07 2011

Keywords

Comments

Column 4 of A183904.

Examples

			All solutions for n=2:
..0....0....0....0....0
..1....1....2....1....0
..3....2....3....1....0
..3....2....4....2....0
..4....4....4....3....0

Links

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/24)*n^4 + (11/12)*n^3 + (179/24)*n^2 + (199/12)*n - 81 for n>3.
Conjectures from Colin Barker, Apr 05 2018: (Start)
G.f.: x*(1 + 51*x^2 - 116*x^3 + 74*x^4 - 2*x^5 - 5*x^6 - 2*x^7) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>8.
(End)

A183900 Number of nondecreasing arrangements of n+3 numbers in 0..5 with each number being the sum mod 6 of three others.

Original entry on oeis.org

3, 24, 130, 364, 771, 1386, 2281, 3534, 5236, 7492, 10422, 14162, 18865, 24702, 31863, 40558, 51018, 63496, 78268, 95634, 115919, 139474, 166677, 197934, 233680, 274380, 320530, 372658, 431325, 497126, 570691, 652686, 743814, 844816, 956472, 1079602
Offset: 1

Views

Author

R. H. Hardin, Jan 07 2011

Keywords

Comments

Column 5 of A183904.

Examples

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

Links

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/120)*n^5 + (1/4)*n^4 + (71/24)*n^3 + (57/4)*n^2 - (262/15)*n - 50 for n>4.
Conjectures from Colin Barker, Apr 05 2018: (Start)
G.f.: x*(3 + 6*x + 31*x^2 - 116*x^3 + 102*x^4 - 38*x^5 + 59*x^6 - 78*x^7 + 38*x^8 - 6*x^9) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>10.
(End)

A183901 Number of nondecreasing arrangements of n+3 numbers in 0..6 with each number being the sum mod 7 of three others.

Original entry on oeis.org

1, 1, 136, 664, 1720, 3491, 6263, 10400, 16357, 24694, 36091, 51364, 71482, 97585, 131003, 173276, 226175, 291724, 372223, 470272, 588796, 731071, 900751, 1101896, 1339001, 1617026, 1941427, 2318188, 2753854, 3255565, 3831091, 4488868, 5238035
Offset: 1

Views

Author

R. H. Hardin, Jan 07 2011

Keywords

Comments

Column 6 of A183904.

Examples

			Some solutions for n=3:
..0....0....3....0....0....0....0....0....0....0....0....1....0....0....2....1
..1....0....3....1....0....0....0....3....2....1....1....1....3....0....4....2
..1....1....3....1....1....3....1....4....3....2....4....2....3....1....4....3
..2....1....4....4....2....3....1....5....4....3....5....3....4....2....5....3
..4....2....4....4....2....5....3....6....4....4....5....4....5....4....5....5
..4....4....4....6....4....6....3....6....6....4....6....5....5....4....6....6

Links

Crossrefs

Cf. A183904.

Formula

Empirical: a(n) = (1/720)*n^6 + (13/240)*n^5 + (125/144)*n^4 + (117/16)*n^3 + (12287/360)*n^2 - (4421/30)*n - 44 for n>3.
Conjectures from Colin Barker, Apr 05 2018: (Start)
G.f.: x*(1 - 6*x + 150*x^2 - 302*x^3 - 72*x^4 + 649*x^5 - 548*x^6 + 60*x^7 + 102*x^8 - 33*x^9) / (1 - x)^7.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>10.
(End)

A183902 Number of nondecreasing arrangements of n+3 numbers in 0..7 with each number being the sum mod 8 of three others.

Original entry on oeis.org

5, 22, 243, 1160, 3347, 7416, 14376, 25584, 42852, 68552, 105736, 158272, 230997, 329888, 462252, 636936, 864558, 1157760, 1531484, 2003272, 2593591, 3326184, 4228448, 5331840, 6672312, 8290776, 10233600, 12553136, 15308281
Offset: 1

Views

Author

R. H. Hardin Jan 07 2011

Keywords

Comments

Column 7 of A183904

Examples

			Some solutions for n=2
..2....1....1....0....1....0....0....2....0....4....1....2....3....0....1....3
..6....5....1....0....3....0....0....2....2....4....3....2....3....0....1....5
..6....5....3....2....3....0....0....2....2....4....3....6....5....4....5....7
..6....5....5....2....3....4....0....6....4....4....7....6....7....6....5....7
..6....7....5....4....7....4....0....6....4....4....7....6....7....6....7....7

Formula

Empirical: a(n) = (1/5040)*n^7 + (7/720)*n^6 + (29/144)*n^5 + (329/144)*n^4 + (1259/90)*n^3 + (5167/180)*n^2 - (6955/21)*n + 312 for n>4

A183903 Number of nondecreasing arrangements of n+3 numbers in 0..8 with each number being the sum mod 9 of three others.

Original entry on oeis.org

1, 9, 242, 1866, 6503, 15973, 33236, 62678, 110487, 185231, 298538, 465912, 707702, 1050242, 1527181, 2181023, 3064898, 4244586, 5800817, 7831871, 10456503
Offset: 1

Views

Author

R. H. Hardin Jan 07 2011

Keywords

Comments

Column 8 of A183904

Examples

			All solutions for n=2
..0....0....0....0....0....0....0....1....2
..0....0....3....0....0....3....0....3....5
..3....3....6....3....6....3....0....3....6
..6....3....6....3....6....3....0....4....6
..6....3....6....6....6....6....0....7....8

Formula

Empirical: a(n) = (1/40320)*n^8 + (1/672)*n^7 + (37/960)*n^6 + (9/16)*n^5 + (9709/1920)*n^4 + (917/32)*n^3 + (14159/10080)*n^2 - (119065/168)*n + 801 for n>5
Showing 1-7 of 7 results.

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