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 1-8 of 8 results.

A105424 The part of n in base phi left of the decimal point, using a greedy algorithm representation (more precisely, using the Bergman-canonical representation).

Original entry on oeis.org

0, 1, 10, 100, 101, 1000, 1010, 10000, 10001, 10010, 10100, 10101, 100000, 100010, 100100, 100101, 101000, 101010, 1000000, 1000001, 1000010, 1000100, 1000101, 1001000, 1001010, 1010000, 1010001, 1010010, 1010100, 1010101, 10000000
Offset: 0

Views

Author

Bryan Jacobs (bryanjj(AT)gmail.com), Apr 08 2005

Keywords

Examples

			2 = 10.01 in base phi, so left of the decimal point is 10.
The first few numbers written in base phi:
0 = 0.
1 = 1.
2 = 10.01
3 = 100.01
4 = 101.01
5 = 1000.1001
6 = 1010.0001
7 = 10000.0001
8 = 10001.0001
9 = 10010.0101
10 = 10100.0101
11 = 10101.0101
12 = 100000.101001
13 = 100010.001001
14 = 100100.001001
15 = 100101.001001
16 = 101000.100001
17 = 101010.000001
18 = 1000000.000001
19 = 1000001.000001
20 = 1000010.010001
21 = 1000100.010001
22 = 1000101.010001
23 = 1001000.100101
24 = 1001010.000101
...
		

Crossrefs

See A341722 for the part to the right of the decimal point.
Cf. A105116 (base e), A344939 (base Pi).

Programs

  • Mathematica
    nn = 1000; len = 2*Ceiling[Log[GoldenRatio, nn]]; Table[d = RealDigits[n, GoldenRatio, len]; FromDigits[Take[d[[1]], d[[2]]]], {n, 0, nn}] (* T. D. Noe, May 20 2011 *)

Extensions

Definition clarified by N. J. A. Sloane, May 27 2023

A362692 Length of the "integer part" of the phi-expansion of n.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10
Offset: 0

Views

Author

Jeffrey Shallit, May 01 2023

Keywords

Comments

The phi-representation of n is the (essentially) unique way to write n = Sum_{j=L..R} b(j)*phi^j, where b(j) is in {0,1} and -oo < L <= 0 <= R, where phi = (1+sqrt(5))/2, subject to the condition that b(j)b(j+1) != 1. The "integer" part is the string of bits b(R)b(R-1)...b(1)b(0), and its length is thus R+1.
The gaps between consecutive terms are all either 0 or 1, and a gap of 1 occurs if and only if n = 1 or n = L(2i) or n = L(2i-1) + 1 for i >= 1. This is equivalent to Theorem 2.1 of Sanchis and Sanchis (2001).

Examples

			For n = 20 we have n = phi^6 + phi^1 + phi^(-2) + phi^(-6), and the "integer part" has largest term phi^6, so a(20) = 7.
		

Crossrefs

Programs

Formula

There is a linear representation of rank 9 for a(n).
a(n) = ceiling(log_phi(n)) for n >= 2.

Extensions

a(0) changed to 1 by N. J. A. Sloane, May 26 2023

A362917 The part of n to the left of the decimal point in the Dekking-van-Loon-canonical base phi representation of n.

Original entry on oeis.org

0, 1, 10, 11, 101, 1000, 1010, 1011, 10001, 10010, 10011, 10101, 100000, 100010, 100011, 100101, 101000, 101010, 101011, 1000001, 1000010, 1000011, 1000101, 1001000, 1001010, 1001011, 1010001, 1010010, 1010011, 1010101, 10000000
Offset: 0

Views

Author

N. J. A. Sloane, May 26 2023

Keywords

Comments

The part to the right of the decimal point, reversed, is given by A341722, that is, it is the same as in the Bergman-canonical representation. I asked Jeffrey Shallit to confirm this, and he provided the following verification using the Walnut Theorem-Prover:
[Walnut]$ eval sloane "?msd_fib An,x1,x2,y1,y2 ($saka(n,x1,y1) & $dvl(n,x2,y2)) => $equal(y1,y2)":
(saka(n,x1,y1))&dvl(n,x2,y2))):54 states - 66ms
((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2))):2 states - 25ms
(A n , x1 , x2 , y1 , y2 ((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2)))):1 states - 81ms
Total computation time: 264ms.
TRUE

Examples

			The canonical base phi representations of the numbers 0 through 12 are:
0 = 0.
1 = 1.
2 = 10.01
3 = 11.01
4 = 101.01
5 = 1000.1001
6 = 1010.0001
7 = 1011.0001
8 = 10001.0001
9 = 10010.0101
10 = 10011.0101
11 = 10101.0101
12 = 100000.101001
		

References

  • Dekking, Michel, and Ad van Loon. "On the representation of the natural numbers by powers of the golden mean." arXiv preprint arXiv:2111.07544 (2021); Fib. Quart. 61:2 (May 2023), 105-118.

Crossrefs

Differs from A105424 at positions given by A003231.

Extensions

a(13)-a(32) from Hugo Pfoertner, May 26 2023

A342089 Numbers that have two representations as the sum of distinct non-consecutive Lucas numbers (A000032).

Original entry on oeis.org

5, 12, 16, 23, 30, 34, 41, 45, 52, 59, 63, 70, 77, 81, 88, 92, 99, 106, 110, 117, 121, 128, 135, 139, 146, 153, 157, 164, 168, 175, 182, 186, 193, 200, 204, 211, 215, 222, 229, 233, 240, 244, 251, 258, 262, 269, 276, 280, 287, 291, 298, 305, 309, 316, 320, 327
Offset: 1

Views

Author

Amiram Eldar, Feb 27 2021

Keywords

Comments

Brown (1969) proved that every positive number has a unique representation as a sum of non-consecutive Lucas numbers, if L(0) = 2 and L(2) = 3 do not appear simultaneously in the representation.
Chu et al. (2020) proved that if L(0) and L(2) are allowed to appear simultaneously, then each positive number can have at most two representations. The terms with two representations are listed in this sequence. They found that the number of terms that do not exceed 10^k, for k = 1, 2, ..., are 1, 17, 171, 1708, 17082, 170820, ..., and proved that the asymptotic density of this sequence is 1/(3*phi+1) = 0.1708203932... (A176015 - 1), where phi is the golden ratio (A001622).
A number n appears in the sequence if and only if the coefficient of phi^{-1} in the base-phi expansion of n is 1. Alternatively, the last bit of the n-th term of A341722 is 1. - Jeffrey Shallit, May 03 2023

Examples

			5 is a term since it has two representations: L(0) + L(2) = 2 + 3 and L(1) + L(3) = 1 + 4.
12 is a term since it has two representations: L(1) + L(5) = 1 + 11 and L(0) + L(2) + L(4) = 2 + 3 + 7.
		

Crossrefs

Programs

  • Java
    See David C. Luo's GitHub link.
  • Maple
    L:= [seq(combinat:-fibonacci(n+1)+combinat:-fibonacci(n-1), n=0..40)]:
    f1:= proc(n, m) option remember;
          if n = 0 then return 1 fi;
          if m <= 0 then 0
          elif L[m] <= n then procname(n - L[m],m-2) + procname(n, m-1)
          else procname(n,m-1)
          fi
    end proc:
    filter:= n -> f1(n,ListTools:-BinaryPlace(L,n+1))=2:
    select(filter, [$1..1000]); # Robert Israel, Mar 10 2021
  • Mathematica
    L = Table[Fibonacci[n+1] + Fibonacci[n-1], {n, 0, 40}];
    f1[n_, m_] := f1[n, m] = If[n == 0, Return[1], Which[m <= 0, 0, L[[m]] <= n, f1[n-L[[m]], m-2] + f1[n, m-1], True, f1[n, m-1]]];
    filterQ[n_] := f1[n, FirstPosition[L, b_ /; b > n+1][[1]]-1] == 2;
    Select[Range[1000], filterQ] (* Jean-François Alcover, Aug 27 2022, after Robert Israel *)

A362781 Natural numbers n for which some base-phi representation of n is anti-palindromic.

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 8, 11, 13, 14, 15, 16, 21, 23, 29, 31, 33, 35, 37, 39, 41, 43, 45, 53, 55, 61, 63, 76, 78, 80, 86, 88, 89, 91, 97, 99, 100, 102, 108, 110, 111, 113, 119, 121, 136, 138, 144, 146, 158, 160, 166, 168, 199, 201, 203, 209, 211, 223, 225, 230, 231
Offset: 1

Views

Author

Jeffrey Shallit, May 03 2023

Keywords

Comments

Here "anti-palindromic" means the expansion is of the form x.y, where the complement of y is the reverse of x (allowing leading or trailing zeros). Here we do not insist that the base-phi representation be "canonical" (that is, we do not insist that xy contains no 11).

Examples

			For example, one base-phi representation of 13 is 00100001.01111011.
		

Crossrefs

Formula

There is a 193-state automaton accepting the Zeckendorf representation of the members of this sequence.

A362872 Length of the "fractional part" of the phi-representation of n.

Original entry on oeis.org

0, 0, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 10, 10, 10
Offset: 0

Views

Author

Jeffrey Shallit, May 07 2023

Keywords

Comments

The phi-representation of n is the (essentially) unique way to write n = Sum_{j=L..R} b(j)*phi^j, where b(j) is in {0,1} and -oo < L <= 0 <= R, where phi = (1+sqrt(5))/2, subject to the condition that b(j)b(j+1) != 1. The "fractional" part is the string of bits b(L)b(L+1)...b(-1), and its length is thus L.
The gaps between consecutive terms are all either 0 or 2, and a gap of 2 occurs if and only if n = L(2i+1) for i >= 0. This is equivalent to Theorem 2.1 of Sanchis and Sanchis (2001).

Examples

			The phi-representation of 20 is 1000010.010001, so a(20) = 6.
		

Crossrefs

Formula

There is a linear representation of rank 11 for a(n).

A362918 Length of the part of n to the left of the decimal point in the Dekking-van-Loon-canonical base phi representation of n.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, May 26 2023

Keywords

Comments

a(n) = length of the binary string A362917(n).

References

  • Dekking, Michel, and Ad van Loon. "On the representation of the natural numbers by powers of the golden mean." arXiv preprint arXiv:2111.07544 (2021); Fib. Quart. 61:2 (May 2023), 105-118.

Crossrefs

A362921 The Dekking-van-Loon-canonical base-phi representation of n described in A362917 written as a binary string, omitting the dot.

Original entry on oeis.org

0, 1, 1001, 1101, 10101, 10001001, 10100001, 10110001, 100010001, 100100101, 100110101, 101010101, 100000101001, 100010001001, 100011001001, 100101001001, 101000100001, 101010000001, 101011000001, 1000001000001, 1000010010001, 1000011010001, 1000101010001, 1001000100101, 1001010000101
Offset: 0

Views

Author

N. J. A. Sloane, May 27 2023

Keywords

Crossrefs

Showing 1-8 of 8 results.