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.

A295862 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n), where a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, b(2) = 5, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 3, 9, 18, 34, 60, 104, 175, 291, 479, 784, 1278, 2078, 3373, 5470, 8863, 14354, 23239, 37616, 60879, 98520, 159425, 257972, 417425, 675426, 1092881, 1768338, 2861251, 4629622, 7490908, 12120566, 19611511, 31732115, 51343665, 83075820, 134419526, 217495388
Offset: 0

Views

Author

Clark Kimberling, Dec 08 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622). Following is a guide to related sequences:
*****
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n); initial values (a(0), a(1); b(0), b(1), b(2)):
A295862: (1,3; 2,4,5)
A295947: (2,4; 1,3,5)
A295948: (3,4; 1,2,5)
A295949: (1,2; 3,4,5)
A295950: (1,4; 2,3,5)
A295951: (2,3; 1,4,5)
A295952: (1,5; 2,3,4)
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) + 1; initial values (a(0), a(1); b(0), b(1), b(2)):
A295953: (1,3; 2,4,5)
A295954: (2,4; 1,3,5)
A295955: (3,4; 1,2,5)
A295956: (1,2; 3,4,5)
A295957: (1,4; 2,3,5)
A295958: (2,3; 1,4,5)
A295959: (1,5; 2,3,4)
Complementary equation: a(n) = a(n-1) + a(n-2) + b(n) - 1; initial values (a(0), a(1); b(0), b(1), b(2)):
A295860: (1,3; 2,4,5)
A295961: (2,4; 1,3,5)
A295962: (3,4; 1,2,5)
A295963: (1,2; 3,4,5)
A295964: (1,4; 2,3,5)
A295965: (2,3; 1,4,5)
A295966: (1,5; 2,3,4)

Examples

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

Links

Crossrefs

Cf. A001622, A000045, A295947.

Programs

  • Mathematica
    a[0] = 1; a[1] = 3; b[0] = 2; b[1] = 4; b[2] = 5;
a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
j = 1; While[j < 6, k = a[j] - j - 1;
 While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
Table[a[n], {n, 0, k}]  (*A295862*)
Table[b[n], {n, 0, 20}] (*complement*)

Formula

a(n) = H + R, where H = f(n-1)*a(0) + f(n)*a(1) and R = f(n-1)*b(2) + f(n-2)*b(3) + ... + f(2)*b(n-1) + f(1)*b(n), where f(n) = A000045(n), the n-th Fibonacci number.

A295859 a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = 0, a(2) = 1, a(3) = 1.

Original entry on oeis.org

-2, 0, 1, 1, 8, 9, 29, 38, 91, 129, 268, 397, 761, 1158, 2111, 3269, 5764, 9033, 15565, 24598, 41699, 66297, 111068, 177365, 294577, 471942, 778807, 1250749, 2054132, 3304881, 5408165, 8713046, 14219515, 22932561, 37348684, 60281245, 98023145, 158304390
Offset: 0

Views

Author

Clark Kimberling, Jan 07 2018

Keywords

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

Links

Crossrefs

Cf. A001622, A000045, A295860.

Programs

  • Mathematica
    LinearRecurrence[{1, 3, -2, -2}, {-2, 0, 1, 1}, 100]

Formula

a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = -2, a(1) = 0, a(2) = 1, a(3) = 1.
G.f.: (-2 + 2 x + 7 x^2 - 4 x^3)/(1 - x - 3 x^2 + 2 x^3 + 2 x^4).

A295998 Solution of the complementary equation a(n) = 2*a(n-2) + b(n-2), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 5, 8, 16, 23, 41, 56, 93, 124, 199, 262, 413, 541, 844, 1101, 1708, 2223, 3438, 4470, 6901, 8966, 13829, 17960, 27687, 35950, 55405, 71932, 110843, 143898, 221721, 287832, 443479, 575702, 886997, 1151444, 1774036, 2302931, 3548116, 4605907, 7096278
Offset: 0

Views

Author

Clark Kimberling, Dec 02 2017

Keywords

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. a(n)/a(n-1) -> 1.298123759410105...
See A295860 for a guide to related sequences.

Links

Crossrefs

Cf. A001622, A000045, A294860.

Programs

  • Mathematica
    mex[t_] := NestWhile[# + 1 &, 1, MemberQ[t, #] &];
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = 2 a[n - 2] + b[n - 2];  (* A295998 *)
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 100}];
Table[b[n], {n, 0, 30}]

Formula

a(0) = 1, a(1) = 2, b(0) = 3, so that a(2) = 5, b(1) = 4.
Complement: (b(n)) = (3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, ...)

A295861 a(n) = a(n-1) + 3*a(n-2) -2*a(n-3) - 2*a(n-4), where a(0) = -2, a(1) = -1, a(2) = 0, a(3) = 1.

Original entry on oeis.org

-2, -1, 0, 1, 7, 12, 31, 51, 106, 173, 327, 532, 955, 1551, 2698, 4377, 7459, 12092, 20319, 32923, 54778, 88725, 146575, 237348, 390067, 631511, 1033866, 1673569, 2732011, 4421964, 7203127, 11657859, 18959290, 30682685, 49838583, 80652340, 130884139
Offset: 0

Views

Author

Clark Kimberling, Jan 07 2018

Keywords

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

Links

Crossrefs

Cf. A001622, A000045, A295859, A295860.

Programs

  • Mathematica
    LinearRecurrence[{1, 3, -2, -2}, {-2, -1, 0, 1}, 100]

Formula

a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = -2, a(1) = -1, a(2) = 0, a(3) = 1.
G.f.: (-2 + x + 7 x^2)/(1 - x - 3 x^2 + 2 x^3 + 2 x^4).
Showing 1-4 of 4 results.

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