A102364 -id:A102364 - cp's OEIS frontend

A080791 Number of nonleading 0's in binary expansion of n.

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

Offset: 0

Cino Hilliard, Mar 25 2003

In this version we consider the number zero to have no nonleading 0's, thus a(0) = 0. The variant A023416 has a(0) = 1.

Number of steps required to reach 1, starting at n + 1, under the operation: if x is even divide by 2 else add 1. This is the x + 1 problem (as opposed to the 3x + 1 problem).

			a(4) = 2 since 4 in binary is 100, which has two zeros.
a(5) = 1 since 5 in binary is 101, which has only one zero.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A000120, A007814, A023416, A029837, A036987, A080791-A080801, A048881, A054429, A092339, A102364, A120511, A233271, A233272, A233273.

seq(numboccur(0, Bits[Split](n)), n=0..100); # Robert Israel, Oct 26 2017

{0}~Join~Table[Last@ DigitCount[n, 2], {n, 120}] (* Michael De Vlieger, Mar 07 2016 *)
f[n_] := If[OddQ@ n, f[n -1] -1, f[n/2] +1]; f[0] = f[1] = 0; Array[f, 105, 0] (* Robert G. Wilson v, May 21 2017 *)
Join[{0}, Table[Count[IntegerDigits[n, 2], 0], {n, 1, 100}]] (* Vincenzo Librandi, Oct 27 2017 *)

PARI
```
a(n)=if(n,a(n\2)+1-n%2)
```

A080791(n)=if(n,logint(n,2)+1-hammingweight(n)) \\ M. F. Hasler, Oct 26 2017

def a(n): return bin(n)[2:].count("0") if n>0 else 0 # Indranil Ghosh, Apr 10 2017

;; with memoizing definec-macro from Antti Karttunen's IntSeq-library)
(define (A080791 n) (- (A029837 (+ 1 n)) (A000120 n)))
;; Alternative version based on a simple recurrence:
(definec (A080791 n) (if (zero? n) 0 (+ (A080791 (- n 1)) (A007814 n) (A036987 (- n 1)) -1)))
;; from Antti Karttunen, Dec 12 2013

From Antti Karttunen, Dec 12 2013: (Start)
a(n) = A029837(n+1) - A000120(n).
a(0) = 0, and for n > 0, a(n) = (a(n-1) + A007814(n) + A036987(n-1)) - 1.
For all n >= 1, a(A054429(n)) = A048881(n-1) = A000120(n) - 1.
Equally, for all n >= 1, a(n) = A000120(A054429(n)) - 1.
(End)
Recurrence: a(2n) = a(n) + 1 (for n > 0), a(2n + 1) = a(n). - Ralf Stephan from Cino Hilliard's PARI program, Dec 16 2013. Corrected by Alonso del Arte, May 21 2017 after consultation with Chai Wah Wu and Ray Chandler, "n > 0" added by M. F. Hasler, Oct 26 2017
a(n) = A023416(n) for all n > 0. - M. F. Hasler, Oct 26 2017
G.f. g(x) satisfies g(x) = (1+x)*g(x^2) + x^2/(1-x^2). - Robert Israel, Oct 26 2017

A117479 Number of zeros in the maximal Fibonacci bit-representation of n (A104326).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 1, 0, 2, 2, 1, 2, 1, 1, 1, 0, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 3, 3, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 4, 3, 3, 3, 2, 3, 3, 2, 3, 2, 2, 2, 1, 3, 3, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 1, 2, 2, 1, 2, 1, 1, 1, 0, 4

Offset: 0

Casey Mongoven, Mar 20 2006

nonn

			a(7) = 2 because A104326(7) = 1010 contains 2 zeros.

J. L. Brown, Jr., A new characterization of the Fibonacci numbers, Fibonacci Quarterly 3, no. 1 (1965) 1-8.
Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Cf. A112310, A102364, A104325, A104324, A104326.

A227352 Permutation of nonnegative integers: map each number by lengths of runs in its binary representation to the number in whose once left-shifted Zeckendorf representation occurs the same run lengths (in the same order) as the lengths of consecutive blocks of zeros.

Original entry on oeis.org

0, 1, 4, 2, 7, 12, 6, 3, 11, 19, 33, 20, 10, 17, 9, 5, 18, 30, 51, 31, 54, 88, 53, 32, 16, 27, 46, 28, 15, 25, 14, 8, 29, 48, 80, 49, 83, 135, 82, 50, 87, 142, 232, 143, 86, 140, 85, 52, 26, 43, 72, 44, 75, 122, 74, 45, 24, 40, 67, 41, 23, 38, 22, 13, 47, 77

Offset: 0

Antti Karttunen, Jul 08 2013

See the comments at the inverse permutation A227351 where the idea behind this mapping is explained.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A227351. Cf. A048680, A003188.

(define (A227352 n) (A048680 (A003188 n)))

a(n) = A048680(A003188(n)). [The defining formula]
Moreover, this permutation effects the following correspondences:
For n>=1 A000523(n) = A102364(a(n)).
For all n, A167489(n) = A227355(a(2n+1)).

A227355 Product of run lengths in Zeckendorf representation of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 2, 1, 5, 4, 3, 4, 2, 3, 2, 1, 6, 5, 4, 6, 3, 6, 4, 2, 4, 3, 2, 2, 1, 7, 6, 5, 8, 4, 9, 6, 3, 8, 6, 4, 4, 2, 5, 4, 3, 4, 2, 3, 2, 1, 8, 7, 6, 10, 5, 12, 8, 4, 12, 9, 6, 6, 3, 10, 8, 6, 8, 4, 6, 4, 2, 6, 5, 4, 6, 3, 6, 4, 2, 4, 3

Offset: 0

Antti Karttunen, Jul 08 2013

The same sequence also gives the product for the lengths of zero-runs only, as by definition, no two consecutive 1's can occur in Fibonacci number system (aka Zeckendorf representation), thus any 1's present contribute just *1 to the total product.

Antti Karttunen, Table of n, a(n) for n = 0..10946

Cf. A167489, A003714, A102364, A014417, A227350.

(define (A227355 n) (A167489 (A003714 n)))(define (A227355v2 n) (A227350 (A003714 n))) ;; Alternative definition.

a(n) = A167489(A003714(n)) = A227350(A003714(n)).
a(A227352(A005408(n))) = A167489(n).
For n>= 3, a(A000045(n)) = n-2.

A371590 Irregular table T(n, k), n >= 0, k = 1..max(2, 2^n), read by rows; the n-th row lists the nonnegative numbers whose Zeckendorf-binary representation has n nonleading zeros.

Original entry on oeis.org

0, 1, 2, 4, 3, 6, 7, 12, 5, 9, 10, 11, 17, 19, 20, 33, 8, 14, 15, 16, 18, 25, 27, 28, 30, 31, 32, 46, 51, 53, 54, 88, 13, 22, 23, 24, 26, 29, 38, 40, 41, 43, 44, 45, 48, 49, 50, 52, 67, 72, 74, 75, 80, 82, 83, 85, 86, 87, 122, 135, 140, 142, 143, 232

Offset: 0

Rémy Sigrist, Mar 28 2024

As a flat sequence, this is a permutation of the nonnegative integers with inverse A371591.

			Array T(n, k) begins:
    0, 1
    2, 4
    3, 6, 7, 12
    5, 9, 10, 11, 17, 19, 20, 33
    8, 14, 15, 16, 18, 25, 27, 28, 30, 31, 32, 46, 51, 53, 54, 88
    ...

Rémy Sigrist, PARI program
Index entries for sequences that are permutations of the natural numbers

See A371592 for a similar sequence.

Cf. A000045, A027941, A102364, A371591 (inverse).

PARI
```
\\ See Links section.
```

A102364(T(n, k)) = n.
T(n, 1) = A000045(n + 2) for any n > 0.
T(n, max(2, 2^n)) = A027941(n + 1) for any n >= 0.

A131340 Number of 0's in minimal Lucas representation (A130310) of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Jun 29 2007

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000
Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Cf. A131341, A000032, A102364.

a[n_] := Module[{s = {}, m = n, k = 1}, While[m > 0, If[m == 1, k = 1; AppendTo[s, k]; m = 0, If[m == 2, k = 0; AppendTo[s, k]; m = 0, While[LucasL[k] <= m, k++]; k--; AppendTo[s, k]; m -= LucasL[k]; k = 1]]]; Total[1 - IntegerDigits[Total[2^s], 2]]]; Array[a, 100] (* Amiram Eldar, Jul 05 2025 *)

More terms from Amiram Eldar, Jul 05 2025

A131341 Number of 0's in maximal Lucas representation (A130311) of n.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 2, 2, 1, 1, 1, 1, 0, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 0, 3, 3, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 0, 3, 3, 3, 2, 3, 3, 2, 2, 2, 2, 1, 3, 3, 2, 2, 2, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 1, 0, 4, 4, 3, 3, 3, 3, 2, 3, 3, 3, 2, 3

Offset: 1

Casey Mongoven, Jun 29 2007, corrected Mar 23 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000
Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Cf. A131340, A000032, A102364.

lazy = Select[IntegerDigits[Range[400], 2], SequenceCount[#, {0, 0}] == 0 &]; t = Total[#*Reverse@LucasL[Range[0, Length[#] - 1]]] & /@ lazy; Plus @@@ (1 - lazy[[TakeWhile[Flatten[FirstPosition[t, #] & /@ Range[Max[t]]], NumberQ]]]) (* Amiram Eldar, Jul 05 2025 *)

More terms from Amiram Eldar, Jul 05 2025

A356894 a(n) is the number of 0's in the maximal tribonacci representation of n (A352103).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 1, 0, 2, 2, 1, 2, 1, 1, 0, 3, 2, 3, 2, 2, 1, 2, 2, 1, 2, 1, 1, 0, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 3, 2, 3, 2, 2, 1, 2, 2, 1, 2, 1, 1, 0, 4, 4, 3, 4, 3, 3, 2, 4, 3, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 4, 3, 3, 2, 3, 3, 2, 3, 2, 2, 1, 3, 2, 3, 2

Offset: 0

Amiram Eldar, Sep 03 2022

			  n  a(n)  A352103(n)
  -  ----  ----------
  0     1           0
  1     0           1
  2     1          10
  3     0          11
  4     2         100
  5     1         101
  6     1         110
  7     0         111
  8     2        1001
  9     2        1010

Amiram Eldar, Table of n, a(n) for n = 0..10000

Cf. A000073, A352103, A352104, A356895.

Similar sequences: A023416, A102364, A117479, A278042.

t[1] = 1; t[2] = 2; t[3] = 4; t[n_] := t[n] = t[n - 1] + t[n - 2] + t[n - 3]; trib[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[t[k] <= m, k++]; k--; AppendTo[s, k]; m -= t[k]; k = 1]; IntegerDigits[Total[2^(s - 1)], 2]]; a[n_] := Module[{v = trib[n]}, nv = Length[v]; i = 1; While[i <= nv - 3, If[v[[i ;; i + 3]] == {1, 0, 0, 0}, v[[i ;; i + 3]] = {0, 1, 1, 1}; If[i > 3, i -= 4]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, 1, Count[v[[i[[1, 1]] ;; -1]], 0]]]; Array[a, 100, 0]

a(n) = A356895(n) - A352104(n).

A212278 Number of adjacent pairs of zeros (possibly overlapping) in the representation of n in base of Fibonacci numbers (A014417).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 1, 0, 4, 3, 2, 2, 1, 2, 1, 0, 5, 4, 3, 3, 2, 3, 2, 1, 3, 2, 1, 1, 0, 6, 5, 4, 4, 3, 4, 3, 2, 4, 3, 2, 2, 1, 4, 3, 2, 2, 1, 2, 1, 0, 7, 6, 5, 5, 4, 5, 4, 3, 5, 4, 3, 3, 2, 5, 4, 3, 3, 2, 3, 2, 1, 5, 4, 3, 3, 2, 3, 2, 1, 3, 2, 1, 1, 0, 8

Offset: 0

Alex Ratushnyak, May 13 2012

a(n) = 0 only if n = Fibonacci(k)-1.

			A014417(5) = 1000, two pairs of adjacent zeros, so a(5) = 2.

Alois P. Heinz, Table of n, a(n) for n = 0..10946

Cf. A000045, A003714, A014417, A007895, A102364.

F:= combinat[fibonacci]:
b:= proc(n) option remember; local j;
      if n=0 then 0
    else for j from 2 while F(j+1)<=n do od;
         b(n-F(j))+2^(j-2)
      fi
    end:
a:= proc(n) local c, h, m, t;
      c, t, m:= 0, 1, b(n);
      while m>0 do
        h:= irem(m, 2, 'm');
        if h=t and h=0 then c:=c+1 fi;
        t:=h
      od; c
    end:
seq(a(n), n=0..150);  # Alois P. Heinz, May 18 2012

cp's OEIS Frontend

A080791 Number of nonleading 0's in binary expansion of n.

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A117479 Number of zeros in the maximal Fibonacci bit-representation of n (A104326).

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A227352 Permutation of nonnegative integers: map each number by lengths of runs in its binary representation to the number in whose once left-shifted Zeckendorf representation occurs the same run lengths (in the same order) as the lengths of consecutive blocks of zeros.

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A227355 Product of run lengths in Zeckendorf representation of n.

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A371590 Irregular table T(n, k), n >= 0, k = 1..max(2, 2^n), read by rows; the n-th row lists the nonnegative numbers whose Zeckendorf-binary representation has n nonleading zeros.

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A131340 Number of 0's in minimal Lucas representation (A130310) of n.

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A131341 Number of 0's in maximal Lucas representation (A130311) of n.

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A356894 a(n) is the number of 0's in the maximal tribonacci representation of n (A352103).

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