A007895 -id:A007895 - cp's OEIS frontend

A214974 Numbers k for which A116543(k) = A007895(k); i.e., the Lucas and Zeckendorf representations of k have the same length.

1, 2, 3, 6, 9, 10, 14, 15, 17, 22, 27, 28, 36, 38, 41, 43, 44, 46, 52, 58, 59, 61, 62, 66, 69, 74, 75, 81, 84, 94, 95, 96, 98, 107, 112, 114, 117, 119, 120, 122, 128, 131, 136, 139, 148, 152, 153, 154, 155, 159, 161, 164, 173, 175, 176, 181, 182, 184, 185

Offset: 1

Clark Kimberling, Oct 20 2012

nonn

			k...Lucas.....Zeckendorf....counter
1...1.........1.............a(1)= 1
2...2.........2.............a(2)= 2
3...3.........3.............a(3)= 3
4...4.........3+1
5...4+1.......5
6...4+2.......5+1...........a(4)= 6
7...7.........5+2
8...7+1.......8
9...7+2.......8+1...........a(5)= 9

Clark Kimberling, Table of n, a(n) for n = 1..8000

Cf. A116543, A007895, A214975, A214976, A214973.

u = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 50}]]];
u1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, u]][[2,1]], # > 0 &]] &, Range[1000]];
v = Reverse[Table[Fibonacci[n + 1], {n, 1, 50}]];
v1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, v]][[2,1]], # > 0 &]] &, Range[1000]];
s[n_] := If[u1[[n]] == v1[[n]], 1, 0];
s1 = Table[s[n], {n, 1, 200}];
f1 = Flatten[Position[s1, 1]] (* A214974 *)
s[n_] := If[u1[[n]] < v1[[n]], 1, 0];
s2 = Table[s[n], {n, 1, 200}];
f2 = Flatten[Position[s2, 1]] (* A214975 *)
s[n_] := If[u1[[n]] > v1[[n]], 1, 0];
s3 = Table[s[n], {n, 1, 200}];
f3 = Flatten[Position[s3, 1]] (* A214976 *)
(* Peter J. C. Moses *)

A214975 Numbers k for which A116543(k) < A007895(k); i.e., the Lucas representation of k is shorter than the Zeckendorf representation.

Original entry on oeis.org

4, 7, 11, 12, 18, 19, 20, 25, 29, 30, 31, 32, 33, 40, 47, 48, 49, 50, 51, 53, 54, 65, 67, 72, 76, 77, 78, 79, 80, 82, 83, 85, 86, 87, 88, 101, 105, 106, 108, 109, 116, 123, 124, 125, 126, 127, 129, 130, 132, 133, 134, 135, 137, 138, 140, 141, 142, 143, 156

Offset: 1

Clark Kimberling, Oct 20 2012

nonn

			k...Lucas.....Zeckendorf....counter
1...1.........1
2...2.........2
3...3.........3
4...4.........3+1...........a(1) = 4
5...4+1.......5
6...4+2.......5+1
7...7.........5+2...........a(2) = 7
8...7+1.......8
9...7+2.......8+1

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A116543, A007895, A214974, A214976.

Mathematica
```
(See A214974.)
```

A214976 Numbers k for which A116543(k) > A007895(k); i.e., the Lucas representation of k is longer than the Zeckendorf representation.

Original entry on oeis.org

5, 8, 13, 16, 21, 23, 24, 26, 34, 35, 37, 39, 42, 45, 55, 56, 57, 60, 63, 64, 68, 70, 71, 73, 89, 90, 91, 92, 93, 97, 99, 100, 102, 103, 104, 110, 111, 113, 115, 118, 121, 144, 145, 146, 147, 149, 150, 151, 157, 158, 160, 162, 165, 166, 167, 168, 169, 178

Offset: 1

Clark Kimberling, Oct 20 2012

nonn

			k...Lucas.....Zeckendorf....counter
1...1.........1
2...2.........2
3...3.........3
4...4.........3+1
5...4+1.......5.............a(1) = 5
6...4+2.......5+1
7...7.........5+2
8...7+1.......8.............a(2) = 8
9...7+2.......8+1

Clark Kimberling, Table of n, a(n) for n = 1..10000

Cf. A116543, A007895, A214974, A214975.

Mathematica
```
(See A214974.)
```

A361755 Irregular triangle T(n, k), n >= 0, k = 1..2^A007895(n), read by rows; the n-th row lists the numbers k such that the Fibonacci numbers that appear in the Zeckendorf representation of k also appear in that of n.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 3, 0, 1, 3, 4, 0, 5, 0, 1, 5, 6, 0, 2, 5, 7, 0, 8, 0, 1, 8, 9, 0, 2, 8, 10, 0, 3, 8, 11, 0, 1, 3, 4, 8, 9, 11, 12, 0, 13, 0, 1, 13, 14, 0, 2, 13, 15, 0, 3, 13, 16, 0, 1, 3, 4, 13, 14, 16, 17, 0, 5, 13, 18, 0, 1, 5, 6, 13, 14, 18, 19, 0, 2, 5, 7, 13, 15, 18, 20

Offset: 0

Rémy Sigrist, Mar 23 2023

In other words, the n-th row lists the numbers k such that A003714(n) AND A003714(k) = A003714(k) (where AND denotes the bitwise AND operator).

The Zeckendorf representation is also known as the greedy Fibonacci representation (see A356771 for further details).

			Triangle T(n, k) begins:
  n   n-th row
  --  ------------------------
   0  0
   1  0, 1
   2  0, 2
   3  0, 3
   4  0, 1, 3, 4
   5  0, 5
   6  0, 1, 5, 6
   7  0, 2, 5, 7
   8  0, 8
   9  0, 1, 8, 9
  10  0, 2, 8, 10
  11  0, 3, 8, 11
  12  0, 1, 3, 4, 8, 9, 11, 12

Rémy Sigrist, Table of n, a(n) for n = 0..10924 (rows for n = 0..610 flattened)
Rémy Sigrist, PARI program
Index entries for sequences related to Zeckendorf expansion of n

See A361756 for a similar sequence.

Cf. A003714, A007895, A139764, A295989, A356771.

PARI
```
See Links section.
```

T(n, 1) = 0.
T(n, 2) = A139764(n) for any n > 0.
T(n, 2^A007895(n)) = n.

A377208 a(n) is the number of iterations that n requires to reach a noninteger or a Fibonacci number under the map x -> x / z(x), where z(k) = A007895(k) is the number of terms in the Zeckendorf representation of k; a(n) = 0 if n is a Fibonacci number.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 1, 1, 0, 2, 1, 3, 1, 1, 2, 1, 1, 2, 1, 1, 1, 0, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 0, 2, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 2, 1, 1, 1, 1, 0, 3, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1

Offset: 1

Amiram Eldar, Oct 20 2024

The Fibonacci numbers are the fixed points of the map, since z(Fibonacci(k)) = 1 for all k >= 1. Therefore they are arbitrarily assigned the value a(Fibonacci(k)) = 0.

Each number n starts a chain of a(n) integers: n, n/z(n), (n/z(n))/z(n/z(n)), ..., of them the first a(n)-1 integers are Zeckendorf-Niven numbers (A328208).

			a(12) = 2 since 12/z(12) = 4 and 4/z(4) = 2 is a Fibonacci number that is reached after 2 iterations.
a(36) = 3 since 36/z(36) = 18, 18/z(18) = 9 and 9/z(9) = 9/2 is a noninteger that is reached after 3 iterations.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000005, A000045, A007895, A328208, A376615 (binary analog), A377209, A377210.

zeck[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
a[n_] := a[n] = Module[{z = zeck[n]}, If[z == 1, 0, If[!Divisible[n, z], 1, 1 + a[n/z]]]]; Array[a, 100]

zeck(n) = if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s) \\ Charles R Greathouse IV at A007895
a(n) = {my(z = zeck(n)); if(z == 1, 0, if(n % z, 1, 1 + a(n/z)));}

a(n) = 0 if and only if n is in A000045 (by definition).
a(n) >= 2 if and only if n is in A328208 \ A000079 (i.e., n is a Zeckendorf-Niven number that is not a Fibonacci number).
a(n) >= 3 if and only if n is in A377209 \ A000079.
a(n) >= 4 if and only if n is in A377210 \ A000079.
a(n) < A000005(n).

A379175 Irregular triangle T(n, k), n >= 0, k = 1..ceiling(2^(A007895(n)-1)); the n-th row lists the nonnegative integers m such that A184617(m) = A003714(n).

Original entry on oeis.org

0, 1, 2, 4, 3, 5, 8, 7, 9, 6, 10, 16, 15, 17, 14, 18, 12, 20, 11, 13, 19, 21, 32, 31, 33, 30, 34, 28, 36, 27, 29, 35, 37, 24, 40, 23, 25, 39, 41, 22, 26, 38, 42, 64, 63, 65, 62, 66, 60, 68, 59, 61, 67, 69, 56, 72, 55, 57, 71, 73, 54, 58, 70, 74, 48, 80, 47, 49, 79, 81

Offset: 0

Rémy Sigrist, Dec 17 2024

Also the nonnegative terms of A379147, in order of appearance.
This sequence is a permutation of the nonnegative integers with inverse A379176.
This sequence shares graphical features with A368225.

			Triangle T(n, k) begins:
  n   n-th row
  --  --------------
   0  0
   1  1
   2  2
   3  4
   4  3, 5
   5  8
   6  7, 9
   7  6, 10
   8  16
   9  15, 17
  10  14, 18
  11  12, 20
  12  11, 13, 19, 21
  13  32
  14  31, 33
  15  30, 34

Rémy Sigrist, Table of n, a(n) for n = 0..10922 (rows for n = 0..986 flattened)
Index entries for sequences related to Zeckendorf expansion of n
Index entries for sequences that are permutations of the natural numbers

Cf. A003714, A007895, A184617, A368225, A379147, A379176 (inverse).

tozeck(n) = { for (i=0, oo, if (n<=fibonacci(2+i), my (v=0, f); forstep (j=i, 0, -1, if (n>=f=fibonacci(2+j), n-=f; v+=2^j;); if (n==0, return (v););););); }
row(n) = { my (z = tozeck(n), r = [0], b); while (z, z -= b = 2^valuation(z, 2); r = concat([v - b | v <- r], [v + b | v <- r]);); return (select(v -> v >= 0, r)); }

T(n, ceiling(2^(A007895(n)-1))) = A003714(n).

A324901 a(n) = A007895(A324900(n)).

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 2, 3, 3, 4, 2, 4, 2, 4, 3, 6, 2, 5, 2, 5, 4, 4, 2, 6, 3, 4, 6, 5, 2, 3, 2, 7, 4, 4, 3, 7, 2, 4, 4, 5, 2, 7, 2, 6, 6, 4, 2, 7, 3, 6, 4, 6, 2, 5, 4, 4, 4, 4, 2, 5, 2, 4, 5, 7, 4, 7, 2, 6, 4, 5, 2, 8, 2, 4, 5, 6, 3, 8, 2, 9, 8, 4, 2, 5, 4, 4, 4, 11, 2, 6, 4, 6, 4, 4, 4, 10, 2, 6, 5, 5, 2, 8, 2, 11, 8

Offset: 1

Antti Karttunen, Apr 15 2019

nonn

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences computed from indices in prime factorization

Cf. A000032, A007895, A324900, A324905, A324907.

A000032(n) = (fibonacci(n+1)+fibonacci(n-1));
A324900(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = A000032(2*(1+primepi(f[i, 1])))); factorback(f); };
A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A324901(n) = A007895(A324900(n));

a(n) = A007895(A324900(n)).

A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).

Original entry on oeis.org

0, 5, 6, 7, 11, 18, 20, 25, 39, 52, 61, 96, 104, 157, 176, 199, 206, 210, 279, 326, 333, 339, 369, 380, 397, 411, 426, 473, 542, 576, 743, 860, 898, 921, 961, 970, 993, 1024, 1043, 1049, 1100, 1121, 1176, 1184, 1193, 1199, 1206, 1230, 1253, 1376, 1380, 1387, 1435

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A020908(k) = A020908(k+1).

The corresponding values of A020908(k) are 1, 3, 3, 3, 6, 7, 8, 9, 18, 20, 28, 44, 37, ... .

			0 is a term since the Zeckendorf representation of 2^0 = 1 is A014417(1) = 1, and the Zeckendorf representation of 2^1 = 2 is A014417(2) = 10, so A020908(0) = A020908(1) = 1.
5 is a term since the Zeckendorf representation of 2^5 = 32 is A014417(32) = 1010100, and the Zeckendorf representation of 2^6 = 64 is A014417(64) = 100010001, so A020908(5) = A020908(6) = 3.

Amiram Eldar, Table of n, a(n) for n = 1..550 (terms below 10^5)

Cf. A007895, A014417, A020908, A353986.

A374961 is a subsequence.

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
s[n_] := s[n] = z[2^n]; Select[Range[0, 1500], s[#] == s[# + 1] &]

A007895(n)=if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
lista(kmax) = {my(z1 = A007895(1), z2); for(k = 1, kmax, z2 = A007895(2^k); if(z1 == z2, print1(k-1 , ", ")); z1 = z2);}

A379147 Irregular triangle T(n, k), n >= 0, k = 1..2^A007895(n), read by rows; the n-th row lists the integers m such that A184617(abs(m)) = A003714(n).

Original entry on oeis.org

0, -1, 1, -2, 2, -4, 4, -5, -3, 3, 5, -8, 8, -9, -7, 7, 9, -10, -6, 6, 10, -16, 16, -17, -15, 15, 17, -18, -14, 14, 18, -20, -12, 12, 20, -21, -19, -13, -11, 11, 13, 19, 21, -32, 32, -33, -31, 31, 33, -34, -30, 30, 34, -36, -28, 28, 36

Offset: 0

Rémy Sigrist, Dec 16 2024

A permutation of the integers (Z).
For any n >= 0:
- in the Zeckendorf expansion of n,
- replace each Fibonacci number, say A000045(2+i) with i >= 0, by 2^i or -2^i,
- the various values obtained make up the n-th row.

			Triangle T(n, k) begins:
  n   n-th row
  --  ----------------------------------
   0  0
   1  -1, 1
   2  -2, 2
   3  -4, 4
   4  -5, -3, 3, 5
   5  -8, 8
   6  -9, -7, 7, 9
   7  -10, -6, 6, 10
   8  -16, 16
   9  -17, -15, 15, 17
  10  -18, -14, 14, 18
  11  -20, -12, 12, 20
  12  -21, -19, -13, -11, 11, 13, 19, 21
  13  -32, 32
  14  -33, -31, 31, 33
  15  -34, -30, 30, 34

Rémy Sigrist, Table of n, a(n) for n = 0..10922 (rows for n = 0..609 flattened)
Index entries for sequences related to Zeckendorf expansion of n

Cf. A000045, A003714, A007895, A184617, A379175.

tozeck(n) = { for (i=0, oo, if (n<=fibonacci(2+i), my (v=0, f); forstep (j=i, 0, -1, if (n>=f=fibonacci(2+j), n-=f; v+=2^j;); if (n==0, return (v););););); }
row(n) = { my (z = tozeck(n), r = [0], b); while (z, z -= b = 2^valuation(z, 2); r = concat([v - b | v <- r], [v + b | v <- r]);); return (r); }

T(n, 1) = -A003714(n).
T(n, 2^A007895(n)) = A003714(n).
T(n, k) = -T(n, 2^A007895(n)+1-k) for k = 1..2^A007895(n).

A374961 Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of terms in their Zeckendorf representation (A007895).

Original entry on oeis.org

5, 6, 1931, 4127, 26584

Offset: 1

Amiram Eldar, Jul 25 2024

Numbers k such that A020908(k) = A020908(k+1) = A020908(k+2).
The corresponding values of A020908(k) are 3, 3, 763, 1660, 10596, ... .
a(6) > 10^5, if it exists.

			5 is a term since A020908(5) = A020908(6) = A020908(7) = 3.
763 is a term since A020908(1931) = A020908(1932) = A020908(1933) = 763.

Cf. A007895, A020908, A353987.

Subsequence of A374960.

z[n_] := Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5]*# + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]]; (* Alonso del Arte at A007895 *)
s[n_] := s[n] = z[2^n]; Select[Range[0, 4200], s[#] == s[# + 1] == s[# + 2] &]

A007895(n)=if(n<4, n>0, my(k=2, s, t); while(fibonacci(k++)<=n, ); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s); \\ Charles R Greathouse IV at A007895
lista(kmax) = {my(z1 = A007895(1), z2 = A007895(2), z3); for(k = 2, kmax, z3 = A007895(2^k); if(z1 == z2 && z2 == z3, print1(k-2 , ", ")); z1 = z2; z2 = z3);}

cp's OEIS Frontend

A214974 Numbers k for which A116543(k) = A007895(k); i.e., the Lucas and Zeckendorf representations of k have the same length.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A214975 Numbers k for which A116543(k) < A007895(k); i.e., the Lucas representation of k is shorter than the Zeckendorf representation.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A214976 Numbers k for which A116543(k) > A007895(k); i.e., the Lucas representation of k is longer than the Zeckendorf representation.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

A361755 Irregular triangle T(n, k), n >= 0, k = 1..2^A007895(n), read by rows; the n-th row lists the numbers k such that the Fibonacci numbers that appear in the Zeckendorf representation of k also appear in that of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A377208 a(n) is the number of iterations that n requires to reach a noninteger or a Fibonacci number under the map x -> x / z(x), where z(k) = A007895(k) is the number of terms in the Zeckendorf representation of k; a(n) = 0 if n is a Fibonacci number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A379175 Irregular triangle T(n, k), n >= 0, k = 1..ceiling(2^(A007895(n)-1)); the n-th row lists the nonnegative integers m such that A184617(m) = A003714(n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A324901 a(n) = A007895(A324900(n)).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

A374960 Numbers k such that 2^k and 2^(k+1) have the same number of terms in their Zeckendorf representation (A007895).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs