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.

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A297834 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 4, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

Original entry on oeis.org

1, 2, 5, 8, 12, 17, 19, 22, 27, 29, 32, 35, 40, 44, 46, 51, 53, 56, 59, 64, 68, 70, 75, 77, 82, 84, 87, 90, 95, 97, 100, 105, 109, 111, 114, 117, 122, 126, 128, 133, 135, 140, 142, 145, 148, 153, 155, 158, 163, 167, 169, 172, 175, 180, 184, 186, 189, 192
Offset: 0

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A297830 for a guide to related sequences.
Conjecture: -3 < a(n) - (2 +sqrt(2))*n <= 1 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 5.
Complement: (b(n)) = (3,4,6,7,9,10,11,13,14,15,16,18,20,...)

Links

Crossrefs

Cf. A297826, A297830.

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 2 n - 4;
j = 1; While[j < 100, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A297834 *)

A298001 Solution of the complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + 3*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

Original entry on oeis.org

1, 2, 12, 16, 20, 24, 28, 32, 36, 42, 45, 49, 55, 58, 62, 68, 71, 75, 81, 84, 88, 94, 97, 101, 107, 110, 114, 120, 123, 127, 131, 135, 141, 144, 150, 153, 157, 163, 166, 170, 174, 178, 184, 187, 193, 196, 200, 206, 209, 213, 217, 221, 227, 230, 236, 239, 243
Offset: 0

Views

Author

Clark Kimberling, Feb 08 2018

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A298000 for a guide to related sequences.
Conjecture: a(n) - n*L < 4 for n >= 1, where L = (5 + sqrt(13))/2.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, so that a(2) = 12.
Complement: (b(n)) = (3,4,5,6,8,9,10,11,14,15,17,18,19,21...)

Links

Crossrefs

Cf. A297800, A297826, A297836, A297837.

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5;
a[n_] := a[n] = a[1]*b[n] - a[0]*b[n - 1] + 3 n;
j = 1; While[j < 100, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A298001 *)

A298002 Solution of the complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + 4*n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

Original entry on oeis.org

1, 2, 14, 19, 24, 29, 34, 39, 44, 49, 54, 61, 65, 70, 75, 82, 86, 91, 96, 103, 107, 112, 117, 124, 128, 133, 138, 145, 149, 154, 159, 166, 170, 175, 180, 187, 191, 196, 201, 208, 212, 217, 222, 229, 233, 238, 243, 248, 253, 260, 264, 269, 276, 280, 285, 290
Offset: 0

Views

Author

Clark Kimberling, Feb 08 2018

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A298000 for a guide to related sequences.
Conjecture: a(n) - n*L < 4 for n >= 1, where L = 3 + sqrt(5).

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, so that a(2) = 14.
Complement: (b(n)) = (3,4,5,6,7,8,9,10,11,13,15,17,18,20...)

Links

Crossrefs

Cf. A297800, A297826, A297836, A297837.

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5;
a[n_] := a[n] = a[1]*b[n] - a[0]*b[n - 1] + 4 n;
j = 1; While[j < 100, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A298002 *)

A297828 Difference sequence of A297997.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 2, 1, 3, 2, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 3, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 3, 2, 1, 1, 3, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1
Offset: 1

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

Conjectures:
(1) 2 <= a(k) <= 4 for k>=1;
(2) if d is in {1,2,3}, then a(k) = d for infinitely many k.

Links

Crossrefs

Cf. A297826, A297827.

Programs

  • Mathematica
    mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
tbl = {}; a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + n;
b[n_] := b[n] = mex[tbl = Join[{a[n], a[n - 1], b[n - 1]}, tbl], b[n - 1]];
u = Table[a[n], {n, 0, 300}](* A297826 *)
v = Table[b[n], {n, 0, 300}](* A297997 *)
Differences[u];  (* A297827 *)
Differences[v];  (* A297828 *)
(* Peter J. C. Moses, Jan 03 2017 *)

A297835 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n + 1, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

Original entry on oeis.org

1, 2, 10, 13, 16, 19, 22, 25, 30, 32, 37, 39, 44, 46, 51, 53, 58, 60, 65, 67, 70, 73, 78, 82, 84, 87, 90, 95, 99, 101, 104, 107, 112, 116, 118, 121, 124, 129, 133, 135, 138, 141, 146, 150, 152, 155, 158, 163, 167, 169, 174, 176, 181, 183, 186, 189, 194, 196
Offset: 0

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A297830 for a guide to related sequences.
Conjecture: a(n) - (2 +sqrt(2))*n < 7 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 10.
Complement: (b(n)) = (3,4,6,7,8,9,11,12,14,15,17,18,20,...)

Links

Crossrefs

Cf. A297826, A297830.

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + 2 n + 1;
j = 1; While[j < 100, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A297835 *)

A297998 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + floor(5*n/2), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.

Original entry on oeis.org

1, 2, 10, 13, 17, 20, 24, 27, 33, 35, 41, 43, 47, 52, 55, 60, 63, 66, 72, 74, 80, 82, 86, 89, 93, 98, 103, 105, 109, 112, 116, 121, 126, 128, 132, 137, 140, 143, 147, 152, 155, 160, 163, 166, 170, 175, 178, 183, 186, 191, 194, 197, 201, 204, 210, 214, 217
Offset: 0

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 10.
Complement: (b(n)) = (3,4,5,6,7,8,9,11,12,14,15,16,18,19,21,...)

Links

Crossrefs

Cf. A297826, A297830, A297836.

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n - 1] - a[0]*b[n - 2] + Floor[5/2];
j = 1; While[j < 100, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++]; k
Table[a[n], {n, 0, k}]  (* A297998 *)
Previous Showing 11-16 of 16 results.

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