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.

A296481 Decimal expansion of ratio-sum for A295952; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 05 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 ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + ..., assuming that this series converges. For A = A295952, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			ratio-sum = 4.845853683514362071350020567372501789034...
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 5; b[0] = 2; b[1 ] = 3; b[2] = 4;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
    j = 1; While[j < 13, 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}]; (* A295952 *)
    g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
    Take[RealDigits[s, 10][[1]], 100]  (* A296481 *)

A296482 Decimal expansion of limiting power-ratio for A295952; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Jan 06 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 = A295952, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See the guide at A296469 for related sequences.

Examples

			limiting power-ratio = 7.090700687355142881167747526503371215921...
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[1] = 5; b[0] = 2; b[1 ] = 3; b[2] = 4;
    a[n_] := a[n] = a[n - 1] + a[n - 2] + b[n];
    j = 1; While[j < 13, 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}]; (* A295952 *)
    z = 2000; 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]   (* A296482 *)

A296469 Decimal expansion of ratio-sum for A295862; see Comments.

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Dec 18 2017

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 ratio-sum for A is |a(1)/a(0) - g| + |a(2)/a(1) - g| + ..., assuming that this series converges. For A = A295862, we have g = (1 + sqrt(5))/2, the golden ratio (A001622). See A296425-A296434 for related ratio-sums and A296452-A296461 for related limiting power-ratios. Guide to more ratio-sums and limiting power-ratios:
****
Sequence A ratio-sum for A limiting power-ratio for A

Examples

			ratio-sum = 6.21032710946618494227967...
		

Crossrefs

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 < 13, 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 *)
    g = GoldenRatio; s = N[Sum[- g + a[n]/a[n - 1], {n, 1, 1000}], 200]
    Take[RealDigits[s, 10][[1]], 100]  (* A296469 *)

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, ...)
		

Crossrefs

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.
Showing 1-4 of 4 results.