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 1-10 of 16 results. Next

A297827 Difference sequence of A297826.

Original entry on oeis.org

1, 5, 2, 2, 4, 3, 3, 1, 2, 4, 1, 4, 1, 6, 2, 1, 2, 6, 0, 2, 6, 0, 2, 2, 2, 4, 5, 2, 1, 2, 2, 2, 4, 3, 1, 2, 2, 2, 4, 3, 3, 3, 1, 2, 4, 1, 2, 2, 4, 5, 2, 3, 3, 1, 2, 4, 1, 6, 2, 3, 3, 1, 2, 4, 1, 4, 1, 4, 1, 6, 2, 1, 2, 6, 2, 3, 1, 2, 4, 1, 2, 2, 4, 3, 1, 4
Offset: 1

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

Conjectures:
(1) 0 <= a(k) <= 6 for k>=1;
(2) if d is in {0,1,2,3,4,5,6}, then a(k) = d for infinitely many k; for d = 0, see A297829.

Links

Crossrefs

Cf. A297826, A297828.

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 *)

A297829 Duplicates in A297826.

Original entry on oeis.org

51, 59, 243, 279, 287, 295, 363, 371, 422, 430, 538, 587, 684, 872, 934, 1075, 1083, 1091, 1232, 1268, 1304, 1312, 1320, 1388, 1396, 1515, 1598, 1634, 1642, 1650, 1718, 1726, 1855, 1891, 1899, 1907, 1975, 1983, 2034, 2042, 2093, 2101, 2209, 2258, 2355, 2363
Offset: 1

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Crossrefs

Cf. A297826.

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]];
du = Differences[Table[a[n], {n, 0, 1000}]]  (* A297827 *)
w = Flatten[Position[du, 0]]; Map[a, w] (* A297829 *)

A297830 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n, 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, 9, 12, 15, 18, 21, 26, 28, 33, 35, 40, 42, 47, 49, 54, 56, 59, 62, 67, 71, 73, 76, 79, 84, 88, 90, 93, 96, 101, 105, 107, 110, 113, 118, 122, 124, 127, 130, 135, 139, 141, 146, 148, 153, 155, 158, 161, 166, 168, 171, 176, 180, 182, 187, 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. Conjecture: a(n) - (2 +sqrt(2))*n < 3 for n >= 1.
Guide to related sequences having initial values a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, where (b(n)) is the increasing sequence of positive integers not in (a(n)):
***
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n (a(n)) = A297826; (b(n)) = A297997
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n (a(n)) = A297830; (b(n)) = A298003
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 3*n (a(n)) = A297836; (b(n)) = A298004
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 4*n (a(n)) = A297837; (b(n)) = A298005
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 1 (a(n)) = A297831; (b(n)) = A298006
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 2 (a(n)) = A297832; (b(n)) = A298007
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 3 (a(n)) = A297833; (b(n)) = A298108
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 4 (a(n)) = A297834; (b(n)) = A298109
a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n + 1 (a(n)) = A297835;
a(n) = a(1)*b(n-1) - a(0)*b(n-2)+floor(5*n/2) (a(n)) = A297998;
***
For sequences (a(n)) and (b(n)) associated with equations of the form a(n) = a(1)*b(n) - a(0)*b(n-1), see the guide at A297800.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 9.
Complement: (b(n)) = (3,4,5,6,8,10,11,13,14,16,17,19,...)

Links

Crossrefs

Cf. A297826, A297836, A297837.

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;
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}]  (* A297830 *)

A297836 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 3*n, 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, 11, 15, 19, 23, 27, 31, 35, 41, 44, 48, 54, 57, 61, 67, 70, 74, 80, 83, 87, 93, 96, 100, 106, 109, 113, 119, 122, 126, 130, 134, 140, 143, 149, 152, 156, 162, 165, 169, 173, 177, 183, 186, 192, 195, 199, 205, 208, 212, 216, 220, 226, 229, 235, 238, 242
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. For a guide to related sequences, see A297830.
Conjectures: a(n) - (5 + sqrt(13))*n/2 < 2 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 11.
Complement: (b(n)) = (3,4,5,6,7,8,9,10,12,13,14,16,17,18,20,...)

Links

Crossrefs

Cf. A297826, A297830, A297837.

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] + 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}]  (* A297836 *)

A298000 Solution of the complementary equation a(n) = a(1)*b(n) - a(0)*b(n-1) + 2*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, 10, 13, 16, 19, 22, 27, 29, 34, 36, 41, 43, 48, 50, 55, 57, 60, 63, 68, 72, 74, 77, 80, 85, 89, 91, 94, 97, 102, 106, 108, 111, 114, 119, 123, 125, 128, 131, 136, 140, 142, 147, 149, 154, 156, 159, 162, 167, 169, 172, 177, 181, 183, 188, 190, 195, 197
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.
Conjectures: a(n) - (2 +sqrt(2))*n < 4 for n >= 1. Guide to related sequences having initial values a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, where (b(n)) is the increasing sequence of positive integers not in (a(n)):
***
a(n) = a(1)*b(n) - a(0)*b(n-1) + n (a(n)) = A297999; (b(n)) = A298110
a(n) = a(1)*b(n) - a(0)*b(n-1) + 2*n (a(n)) = A298000; (b(n)) = A298111
a(n) = a(1)*b(n) - a(0)*b(n-1) + 3*n (a(n)) = A298001; (b(n)) = A298112
a(n) = a(1)*b(n) - a(0)*b(n-1) + 4*n (a(n)) = A298002; (b(n)) = A298113

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 10.
Complement: (b(n)) = (3,4,5,6,8,9,11,12,14,15,17,18,20,...)

Links

Crossrefs

Cf. 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] + 2 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}]  (* A298000 *)

A297837 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 4*n, 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, 13, 18, 23, 28, 33, 38, 43, 48, 53, 60, 64, 69, 74, 81, 85, 90, 95, 102, 106, 111, 116, 123, 127, 132, 137, 144, 148, 153, 158, 165, 169, 174, 179, 186, 190, 195, 200, 207, 211, 216, 221, 228, 232, 237, 242, 247, 252, 259, 263, 268, 275, 279, 284, 289
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. For a guide to related sequences, see A297830.
Conjecture: a(n) - (3 + sqrt(5))*n < 3 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 13.
Complement: (b(n)) = (3,4,5,6,7,8,9,10,11,12,14,15,16,17,19,20,...)

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] + 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}]  (* A297836 *)

A297997 Solution (b(n)) of the near-complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n, 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

3, 4, 5, 6, 8, 10, 12, 13, 14, 16, 17, 19, 20, 23, 25, 26, 27, 30, 31, 32, 35, 36, 37, 38, 39, 41, 44, 46, 47, 48, 49, 50, 52, 54, 55, 56, 57, 58, 60, 62, 64, 66, 67, 68, 70, 71, 72, 73, 75, 78, 80, 82, 84, 85, 86, 88, 89, 92, 94, 96, 98, 99, 100, 102, 103
Offset: 0

Views

Author

Clark Kimberling, Feb 04 2018

Keywords

Comments

The sequence (a(n)) generated by the equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + n, with initial values as shown, includes duplicates; e.g. a(18) = a(19) = 51. If the duplicates are removed from (a(n)), the resulting sequence and (b(n)) are complementary. Conjectures:
(1) 1 <= b(k) - b(k-1) <= 3 for k>=1;
(2) if d is in {1,2,3}, then b(k) = b(k-1) + d for infinitely many k.
***
See A297830 for a guide to related sequences.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 7.
Complement: (b(n)) = (3, 4, 5, 6, 8,10,12,13,14,16, ...)

Links

Crossrefs

Cf. A297997, A297830.

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]];
Table[a[n], {n, 0, 300}]  (* A297826 *)
Table[b[n], {n, 0, 300}]  (* A297997 *)
(* Peter J. C. Moses, Jan 03 2017 *)

A297832 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*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, 7, 10, 13, 18, 20, 25, 27, 32, 34, 37, 40, 45, 49, 51, 54, 57, 62, 66, 68, 71, 74, 79, 83, 85, 90, 92, 97, 99, 102, 105, 110, 112, 115, 120, 124, 126, 131, 133, 138, 140, 143, 146, 151, 153, 156, 161, 165, 167, 172, 174, 179, 181, 184, 187, 192, 194
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.
a(n) - (2+sqrt(2))*n < 2 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 7.
Complement: (b(n)) = (3,4,5,7,8,10,12,13,15,17,18,19,...)

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 - 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}]  (* A297832 *)

A297831 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, 8, 11, 14, 17, 22, 24, 29, 31, 36, 38, 43, 45, 48, 51, 56, 60, 62, 65, 68, 73, 77, 79, 82, 85, 90, 94, 96, 99, 102, 107, 111, 113, 118, 120, 125, 127, 130, 133, 138, 140, 143, 148, 152, 154, 159, 161, 166, 168, 171, 174, 179, 181, 184, 189, 193, 195
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 < 5/2 for n >= 1.

Examples

			a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 8.
Complement: (b(n)) = (3,4,5,6,7,9,10,12,13,15,16,18,19,...)

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}]  (* A297831 *)

A297833 Solution of the complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 2*n - 3, 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, 6, 9, 14, 16, 21, 23, 26, 29, 34, 38, 40, 43, 46, 51, 55, 57, 62, 64, 69, 71, 74, 77, 82, 84, 87, 92, 96, 98, 103, 105, 110, 112, 115, 118, 123, 125, 128, 133, 137, 139, 142, 145, 150, 154, 156, 159, 162, 167, 171, 173, 178, 180, 185, 187, 190, 193
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: -2 < 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) = 6.
Complement: (b(n)) = (3,4,5,7,8,10,12,13,15,17,18,19,...)

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 - 3;
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}]  (* A297833 *)
Showing 1-10 of 16 results. Next

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