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 1-2 of 2 results.

A296843 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(n+1), where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, b(2) = 5, b(3) = 6, and (a(n)) and (b(n)) are increasing complementary sequences.

Original entry on oeis.org

1, 2, 9, 18, 35, 63, 109, 184, 306, 504, 825, 1345, 2187, 3551, 5758, 9330, 15110, 24463, 39597, 64085, 103708, 167820, 271556, 439405, 710991, 1150427, 1861450, 3011910, 4873394, 7885340, 12758771, 20644149, 33402959, 54047148, 87450148, 141497338
Offset: 0

Views

Author

Clark Kimberling, Jan 12 2018

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). See A296245 for a guide to related sequences.

Examples

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

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5; b[3] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n + 1];
    j = 1; While[j < 16, k = a[j] - j - 1;
    While[k < a[j + 1] - j + 1, b[k] = j + k + 2; k++]; j++];
    u = Table[a[n], {n, 0, k}];  (* A296843 *)
    Table[b[n], {n, 0, 20}] (* complement *)

A296845 Decimal expansion of limiting power-ratio for A296843; see Comments.

Original entry on oeis.org

6, 8, 5, 7, 7, 3, 9, 7, 3, 7, 0, 4, 2, 3, 0, 5, 2, 4, 8, 6, 4, 0, 8, 0, 9, 7, 8, 2, 0, 2, 2, 1, 3, 6, 6, 2, 8, 4, 1, 1, 8, 6, 5, 7, 9, 6, 7, 3, 7, 9, 4, 5, 7, 5, 9, 2, 7, 6, 6, 9, 4, 2, 2, 7, 4, 3, 0, 3, 4, 8, 8, 2, 6, 9, 2, 0, 1, 5, 5, 0, 4, 7, 6, 4, 5, 3
Offset: 1

Views

Author

Clark Kimberling, Jan 12 2018

Keywords

Comments

Suppose that A = (a(n)), for n >= 0, is a sequence, and g is a real number such that a(n)/a(n-1) -> g. The limiting power-ratio for A is the limit as n->oo of a(n)/g^n, assuming that this limit exists. For A = A296843, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			limiting power-ratio = 6.136385518666220790955343949152636124460...
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4; b[2] = 5; b[3] = 6;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n + 1];
    j = 1; While[j < 16, 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}];  (* A296843 *)
    z = 1700; g = GoldenRatio; h = Table[N[a[n]/g^n, z], {n, 0, z}];
    StringJoin[StringTake[ToString[h[[z]]], 41], "..."]
    Take[RealDigits[Last[h], 10][[1]], 120] (* A296845 *)
Showing 1-2 of 2 results.