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 1-5 of 5 results.

A256655 R(k), the minimal alternating Fibonacci representation of k, concatenated for k = 0, 1, 2,....

Original entry on oeis.org

0, 1, 2, 3, 5, -1, 5, 8, -2, 8, -1, 8, 13, -5, 1, 13, -3, 13, -2, 13, -1, 13, 21, -8, 1, 21, -8, 2, 21, -5, 21, -5, 1, 21, -3, 21, -2, 21, -1, 21, 34, -13, 1, 34, -13, 2, 34, -13, 3, 34, -13, 5, -1, 34, -8, 34, -8, 1, 34, -8, 2, 34, -5, 34, -5, 1, 34, -3, 34
Offset: 0

Views

Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

Suppose that b = (b(0), b(1), ... ) is an increasing sequence of positive integers satisfying b(0) = 1 and b(n+1) <= 2*b(n) for n >= 0. Let B(n) be the least b(m) >= n. Let R(0) = 1, and for n > 0, let R(n) = B(n) - R(B(n) - n). The resulting sum of the form R(n) = B(n) - B(m(1)) + B(m(2)) - ... + ((-1)^k)*B(k) is introduced here as the minimal alternating b-representation of n. The sum B(n) + B(m(2)) + ... we call the positive part of R(n), and the sum B(m(1)) + B(m(3)) + ... , the nonpositive part of R(n). The number ((-1)^k)*B(k) is the trace of n.
If b(n) = F(n+2), where F = A000045, then the sum is the minimal alternating Fibonacci-representation of n.

Examples

			R(0) = 0
R(1) = 1
R(2) = 2
R(3) = 3
R(4) = 5 - 1
R(9) = 13 - 5 + 1
R(25) = 34 - 13 + 5 - 1
R(64) = 89 - 34 + 13 - 5 + 1

Links

Crossrefs

Cf. A000045, A255973 (trace), A256656 (numbers with positive trace), A256657 (numbers with nonpositive trace), A256663 (positive part of R(n)), A256664 (nonpositive part of R(n)), A256654, A256696 (minimal alternating binary representations), A255974 (minimal alternating triangular-number representations), A256789 (minimal alternating squares representations).

Programs

  • Mathematica
    f[n_] = Fibonacci[n]; ff = Table[f[n], {n, 1, 70}];
s[n_] := Table[f[n + 2], {k, 1, f[n]}];
h[0] = {1}; h[n_] := Join[h[n - 1], s[n]];
g = h[12]; r[0] = {0};
r[n_] := If[MemberQ[ff, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]];
Flatten[Table[r[n], {n, 0, 60}]]

Formula

R(F(k)^2) = F(2k-1) - F(2k-3) + F(2k-5) - ... + d*F(5) + (-1)^k, where d = (-1)^(k+1).

A256657 Numbers for which the minimal alternating Fibonacci representation has negative trace.

Original entry on oeis.org

4, 6, 7, 10, 11, 12, 16, 18, 19, 20, 25, 26, 29, 31, 32, 33, 38, 40, 41, 42, 46, 47, 50, 52, 53, 54, 59, 61, 62, 65, 66, 67, 68, 72, 74, 75, 76, 80, 81, 84, 86, 87, 88, 93, 95, 96, 99, 100, 101, 105, 107, 108, 109, 110, 114, 116, 117, 120, 121, 122, 123, 127
Offset: 1

Views

Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

See A256655 for definitions. This sequence and A256656 partition the positive integers.

Examples

			Let R(k) be the minimal alternating Fibonacci representation of k.  The trace of R(k) is the last term.
R(1) = 1, trace = 1
R(2) = 2, trace = 2
R(3) = 3, trace = 3
R(4) = 5 - 1, trace = -1
R(5) = 5, trace = 5
R(6) = 6 - 2, trace = -2

Links

Crossrefs

Cf. A000045, A256655, A256656.

Programs

  • Mathematica
    b[n_] = Fibonacci[n]; bb = Table[b[n], {n, 1, 70}];
h[0] = {1}; h[n_] := Join[h[n - 1], Table[b[n + 2], {k, 1, b[n]}]];
g = h[18];  r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]]
t = Table[Last[r[n]], {n, 0, 1000}];  (* A256656 *)
Select[Range[200], Last[r[#]] > 0 &]  (* A256656 *)
Select[Range[200], Last[r[#]] < 0 &]  (* A256657 *)

A255973 Trace of the minimal alternating Fibonacci representation of n.

Original entry on oeis.org

0, 1, 2, 3, -1, 5, -2, -1, 8, 1, -3, -2, -1, 13, 1, 2, -5, 1, -3, -2, -1, 21, 1, 2, 3, -1, -8, 1, 2, -5, 1, -3, -2, -1, 34, 1, 2, 3, -1, 5, -2, -1, -13, 1, 2, 3, -1, -8, 1, 2, -5, 1, -3, -2, -1, 55, 1, 2, 3, -1, 5, -2, -1, 8, 1, -3, -2, -1, -21, 1, 2, 3, -1
Offset: 0

Views

Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

See A256655 for definitions.

Examples

			Let R(k) be the minimal alternating Fibonacci representation of k.  The trace of R(k) is the last term.
R(1) = 1, trace = 1
R(2) = 2, trace = 2
R(3) = 3, trace = 3
R(4) = 5 - 1, trace = -1
R(5) = 5, trace = 5
R(6) = 6 - 2, trace = -2

Links

Crossrefs

Cf. A000045, A256655 (representations R(n)), A256656 (numbers with positive trace), A256657 (numbers with nonpositive trace), A256663 (positive part of R(n)), A256664 (nonpositive part of R(n)), A256654.

Programs

  • Mathematica
    b[n_] = Fibonacci[n]; bb = Table[b[n], {n, 1, 70}];
h[0] = {1}; h[n_] := Join[h[n - 1], Table[b[n + 2], {k, 1, b[n]}]];
g = h[12];  r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]];
Table[Last[r[n]], {n, 0, 200}]  (* A255973 *)

A256658 Rectangular array by antidiagonals: row n consists of numbers k such that F(n+1) is the trace of the minimal alternating Fibonacci representation of k, where F = A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 9, 2, 14, 15, 3, 17, 23, 24, 5, 22, 28, 37, 39, 8, 27, 36, 45, 60, 63, 13, 30, 44, 58, 73, 97, 102, 21, 35, 49, 71, 94, 118, 157, 165, 34, 43, 57, 79, 115, 152, 191, 254, 267, 55, 48, 70, 92, 128, 186, 246, 309, 411, 432, 89, 51, 78, 113, 149, 207, 301
Offset: 1

Views

Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

See A256655 for definitions. This array and the array at A256659 partition the positive integers. The row differences are Fibonacci numbers. The columns satisfy the Fibonacci recurrence x(n) = x(n-1) + x(n-2).

Examples

			Northwest corner:
1    9     14    17    22    27    30    35    43
2    15    23    28    36    44    49    57    70
3    24    37    45    58    71    79    92    113
5    39    69    73    94    115   128   149   183
8    63    97    118   152   186   207   241   296
13   102   157   191   246   301   335   390   479

Crossrefs

Cf. A000045, A256655, A256659.

Programs

  • Mathematica
    b[n_] = Fibonacci[n]; bb = Table[b[n], {n, 1, 70}];
h[0] = {1}; h[n_] := Join[h[n - 1], Table[b[n + 2], {k, 1, b[n]}]];
g = h[18];  r[0] = {0};
 r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]];
t = Table[Last[r[n]], {n, 0, 1000}];  (* A256656 *)
TableForm[Table[Flatten[-1 + Position[t, b[n]]], {n, 2, 8}]]   (* A256658 *)
TableForm[Table[Flatten[-1 + Position[t, -b[n]]], {n, 2, 8}]]  (* A256659 *)

A256659 Rectangular array by antidiagonals: row n consists of numbers k such that -F(n+1) is the trace of the minimal alternating Fibonacci representation of k, where F = A000045 (Fibonacci numbers).

Original entry on oeis.org

4, 7, 6, 12, 11, 10, 20, 19, 18, 16, 25, 32, 31, 29, 26, 33, 40, 52, 50, 47, 42, 38, 53, 65, 84, 81, 76, 68, 41, 61, 86, 105, 136, 131, 123, 110, 46, 66, 99, 139, 170, 220, 212, 199, 178, 54, 74, 107, 160, 225, 275, 356, 343, 322, 288, 59, 87, 120, 173, 259
Offset: 1

Views

Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

See A256655 for definitions. This array and the array at A256658 partition the positive integers. The row differences are Fibonacci numbers. The columns satisfy the Fibonacci recurrence x(n) = x(n-1) + x(n-2).

Examples

			Northwest corner:
4    7    12    20    25    33    38    41    46
6    11   19    32    40    53    61    66    74
10   18   31    52    65    86    99    102   120
16   29   50    84    105   139   160   173   194
26   47   81    136   170   225   259   280   314

Crossrefs

Cf. A000045, A256655, A246658.

Programs

  • Mathematica
    b[n_] = Fibonacci[n]; bb = Table[b[n], {n, 1, 70}];
h[0] = {1}; h[n_] := Join[h[n - 1], Table[b[n + 2], {k, 1, b[n]}]];
g = h[18];  r[0] = {0};
r[n_] := If[MemberQ[bb, n], {n}, Join[{g[[n]]}, -r[g[[n]] - n]]];
t = Table[Last[r[n]], {n, 0, 1000}];  (* A256656 *)
TableForm[Table[Flatten[-1 + Position[t, b[n]]], {n, 2, 8}]]   (* A256658 *)
TableForm[Table[Flatten[-1 + Position[t, -b[n]]], {n, 2, 8}]]  (* A256659 *)
Showing 1-5 of 5 results.

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