A048679 -id:A048679 - cp's OEIS frontend

A048680 Nonnegative integers A001477 expanded with rewrite 0->0, 01->1, then interpreted as Zeckendorffian expansions (as numbers of Fibonacci number system).

0, 1, 2, 4, 3, 6, 7, 12, 5, 9, 10, 17, 11, 19, 20, 33, 8, 14, 15, 25, 16, 27, 28, 46, 18, 30, 31, 51, 32, 53, 54, 88, 13, 22, 23, 38, 24, 40, 41, 67, 26, 43, 44, 72, 45, 74, 75, 122, 29, 48, 49, 80, 50, 82, 83, 135, 52, 85, 86, 140, 87, 142, 143, 232, 21, 35, 36, 59, 37, 61

Offset: 0

Antti Karttunen, Jul 14 1999

A permutation of the nonnegative integers (A001477). Inverse permutation to A048679, i.e. A048679[ A048680[ n ] ] = n for all n and vice versa.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Ron Knott, About Fibonacci Number System
Index entries for sequences that are permutations of the natural numbers

Cf. A003714, A005203, A048678, A048679.

Equals A074049(n+1) - 1.

rewrite_0to0_1to01 := proc(n) option remember; if(n < 2) then RETURN(n); else RETURN(((2^(1+(n mod 2))) * rewrite_0to0_1to01(floor(n/2))) + (n mod 2)); fi; end; interpret_as_zeckendorf_expansion := n -> sum('(bit_i(n,i)*fib(i+2))','i'=0..floor_log_2(n));

a(n)=my(k=1,s);while(n,if(n%2,s+=fibonacci(k++));k++;n>>=1);s \\ Charles R Greathouse IV, Nov 17 2013

a(n) = interpret_as_zeckendorf_expansion(rewrite_0to0_1to01(n)) (where rewrite_0to0_1to01(n)=A048678[ n ])

A048678 Binary expansion of nonnegative integers expanded to "Zeckendorffian format" with rewrite rules 0->0, 1->01.

Original entry on oeis.org

0, 1, 2, 5, 4, 9, 10, 21, 8, 17, 18, 37, 20, 41, 42, 85, 16, 33, 34, 69, 36, 73, 74, 149, 40, 81, 82, 165, 84, 169, 170, 341, 32, 65, 66, 133, 68, 137, 138, 277, 72, 145, 146, 293, 148, 297, 298, 597, 80, 161, 162, 325, 164, 329, 330, 661, 168, 337, 338, 677, 340

Offset: 0

Antti Karttunen

No two adjacent 1-bits. Permutation of A003714.

Replace 1 with 01 in binary. - Ralf Stephan, Oct 07 2003

			11=1011 in binary, thus is rewritten as 100101 = 37 in decimal.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Transforms

Cf. A003714, A005203, A048679, A048680.
MASKTRANS transform of A053644.
Cf. A084471, A088697, A088698.
Cf. A124108.

a048678 0 = 0
a048678 x = 2 * (b + 1) * a048678 x' + b
            where (x', b) = divMod x 2
-- Reinhard Zumkeller, Mar 31 2015

rewrite_0to0_1to01 := proc(n) option remember; if(n < 2) then RETURN(n); else RETURN(((2^(1+(n mod 2))) * rewrite_0to0_1to01(floor(n/2))) + (n mod 2)); fi; end;

f[n_] := FromDigits[ Flatten[IntegerDigits[n, 2] /. {1 -> {0, 1}}], 2]; Table[f@n, {n, 0, 60}] (* Robert G. Wilson v, Dec 11 2009 *)

a(n)=if(n<1,0,(3-(-1)^n)*a(floor(n/2))+(1-(-1)^n)/2)

a(n) = if(n == 0, 0, my(A = -2); sum(i = 0, logint(n, 2), A++; if(bittest(n, i), 1 << (A++)))) \\ Mikhail Kurkov, Mar 14 2024

def a(n):
    return 0 if n==0 else (3 - (-1)**n)*a(n//2) + (1 - (-1)**n)//2
print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 30 2017

def A048678(n): return int(bin(n)[2:].replace('1','01'),2) # Chai Wah Wu, Mar 18 2024

a(n) = rewrite_0to0_1to01(n) [ Each 0->1, 1->10 in binary expansion of n ].
a(0)=0; a(n) = (3-(-1)^n)*a(floor(n/2))+(1-(-1)^n)/2. - Benoit Cloitre, Aug 31 2003
a(0)=0, a(2n) = 2a(n), a(2n+1) = 4a(n) + 1. - Ralf Stephan, Oct 07 2003

A060112 Sums of nonconsecutive factorial numbers.

Original entry on oeis.org

0, 1, 2, 6, 7, 24, 25, 26, 120, 121, 122, 126, 127, 720, 721, 722, 726, 727, 744, 745, 746, 5040, 5041, 5042, 5046, 5047, 5064, 5065, 5066, 5160, 5161, 5162, 5166, 5167, 40320, 40321, 40322, 40326, 40327, 40344, 40345, 40346, 40440, 40441, 40442

Offset: 1

Antti Karttunen, Mar 01 2001

Zeckendorf (Fibonacci) expansion of n (A003714) reinterpreted as a factorial expansion.
Also positions in A055089, A060117 and A060118 of the permutations that are composed of disjoint adjacent transpositions only. (That these positions are same can be seen by comparing algorithms PermRevLexUnrankAMSD, PermUnrank3R, PermUnrank3L in the respective sequences). Thus also positions of the fixed terms in A065181-A065184. See comment at A065163.
Written as disjoint cycles the permutations are: (), (1 2), (2 3), (3 4), (1 2)(3 4), (4 5), (1 2)(4 5), (2 3)(4 5), etc. Apart from the first one (the identity), these are the only kind of permutations used in campanology when moving from one "change" to next.

			Zeckendorf Expansions of first few natural numbers and the corresponding values when interpreted as factorial expansions: 0 = 0 = 0, 1 = 1 = 1, 2 = 10 = 2, 3 = 100 = 6, 4 = 101 = 7, 5 = 1000 = 24, 6 = 1001 = 25, 7 = 1010 = 26, 8 = 10000 = 120, etc.,

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Arthur T. White, Ringing the Changes, Math. Proc. Camb. Phil. Soc., September 1983, Vol. 94, part 2, pp. 203-215.
Index entries for sequences related to bell ringing

Subset of A059590. Cf. also A001611, A064640.

For PermRevLexRank, see A056019, for fibbinary see A048679 and A003714.

CampanoPerm := proc(n) local z,p,i; p := []; z := fibbinary(n); i := 1; while(z > 0) do if(1 = (z mod 2)) then p := permul(p,[[i,i+1]]); fi; i := i+1; z := floor(z/2); od; RETURN(convert(p,'permlist',i)); end;

With[{b = MixedRadix[Range[12, 2, -1]]}, FromDigits[#, b] & /@ Select[Tuples[{0, 1}, 8], SequenceCount[#, {1, 1}] == 0 &]] (* Michael De Vlieger, Jun 26 2017 *)

fill(lim,k,val)=if(k>#f, return); my(t=val+f[k]); if(t<=lim, listput(v,t); fill(lim,k+2,t)); fill(lim,k+1,val)
list(lim)=my(k,t=1); local(f=List(),v=List([0])); while((t*=k++)<=lim, listput(f,t)); f=Vecrev(f); fill(lim,1,0); Set(v) \\ Charles R Greathouse IV, Jun 25 2017

first(n) = my(res = [0, 1], k = 1, t = 1, p = 1); while(#res < n, k++; t++; p *= t; res = concat(res, vector(fibonacci(k), i, res[i]+p))); vector(n, i, res[i]) \\ David A. Corneth, Jun 26 2017

a(n) = PermRevLexRank(CampanoPerm(n))

a(A001611(n)) = (n-1)! for n > 2. - David A. Corneth, Jun 25 2017

A371176 Numbers k such that A000120(k) <= A001511(k).

Original entry on oeis.org

1, 2, 4, 6, 8, 10, 12, 16, 18, 20, 24, 28, 32, 34, 36, 40, 44, 48, 52, 56, 64, 66, 68, 72, 76, 80, 84, 88, 96, 100, 104, 112, 120, 128, 130, 132, 136, 140, 144, 148, 152, 160, 164, 168, 176, 184, 192, 196, 200, 208, 216, 224, 232, 240, 256, 258, 260, 264, 268

Offset: 1

Mikhail Kurkov, Mar 14 2024

It appears that this sequence is obtained when ordering Schreier sets as explained in the Bird link. See decM(n) PARI code. - Michel Marcus, May 31 2024
That is correct since the binary representation of these numbers can be put into 1-to-1 correspondence with Schreier sets, which satisfy |X| <= min X, using the indicator function of X as the bits (starting from the right, LSB). The reason is that A000120 then computes |X| and A001511 computes min X. For example, the Schreier set X = {2, 5} can be mapped to 10010_2 = 18. - Michael S. Branicky, May 31 2024
From David A. Corneth, May 31 2024: (Start)
If k is in the sequence then so is 2*k.
a(A000045(k)) = 2^(k-2) for k >= 2. (End)
Apart from a(1), all terms are even. - Paolo Xausa, May 31 2024
Zeckendorf representation of n with rewrite 0 -> 0, {0, 1} -> 1 and k-1 zeros appended to the right side (where k is the number of ones in the given representation) and then interpreted as binary expansion is the same as a(n) (see the first formula). - Mikhail Kurkov, Oct 21 2024

Michel Marcus, Table of n, a(n) for n = 1..10945
Alistair Bird, Jozef Schreier, Schreier sets and the Fibonacci sequence, Out Of The Norm blog, May 13 2012.
Vaclav Kotesovec, Plot of log(a(n))/log(n) for n = 2..2129243
Vaclav Kotesovec, Plot of a(n) / 2^(log(sqrt(5)*n)/log((1 + sqrt(5))/2) - 2) for n = 2..2129243

Cf. A000045, A000120, A001316, A001511, A003849, A005206, A007895, A048679, A066628, A072649, A213911.
Cf. A355489, A373345, A373347 (complement), A373557.
Cf. A104287.

filter:= proc(n) convert(convert(n,base,2),`+`) <= 1+padic:-ordp(n,2) end proc:
select(filter, [1,seq(i,i=2..1000,2)]); # Robert Israel, Oct 20 2024

Join[{1}, Select[Range[2, 1000, 2], DigitSum[#, 2] <= IntegerExponent[#, 2] + 1 &]] (* Paolo Xausa, Aug 12 2025 *)

isok(n) = hammingweight(n) <= (valuation(n, 2) + 1)

M(n) = my(list=List()); for (i=1, n, forsubset(i, s, my(bOk = if (#s && (vecmax(s) == n), #s <= vecmin(s), 0)); if (bOk, listput(list, vecsort(Vec(s),,4))););); Vec(list);
decM(nn) = my(v = vector(nn, k, M(k)), list=List()); for (i=1, #v, my(vi = v[i]); for (j=1, #vi, my(s = vecsort(vi[j]), slist=List(), m = vecmax(s)); forstep(k=m, 1, -1, listput(slist, sign(vecsearch(s, k)))); listput(list, fromdigits(Vec(slist), 2)););); vecsort(Vec(list)); \\ Michel Marcus, May 31 2024

def ok(n): return n.bit_count() <= (-n&n).bit_length()
print([k for k in range(1, 300) if ok(k)]) # Michael S. Branicky, May 31 2024

# Assuming the list starts with 0.
def a():
    n = na = nb = 1
    while True:
        yield not(nb < (na - 1) << 1)
        nb, na = na, n.bit_count()
        n += 1
aList = a(); print([n for n in range(77) if next(aList)]) # Peter Luschny, Jun 07 2024

a(n) = b(n)*A001316(b(n))/2 where b(n) = A048679(n).
a(n) = Sum_{i=0..n-1} 2^A213911(i).
a(n) = 2^(A072649(n) - 1) + [c(n) > 0]*2*a(c(n)) where c(n) = A066628(n).
a(n) = 2*a(A005206(n)) - A003849(n)*2^A007895(n-1) for n > 1 with a(1) = 1.
Conjecture: lim sup_{n -> oo} log(a(n))/log(n) = log(2) / log((1 + sqrt(5))/2) = 1.440420090412556479... = A104287. - Vaclav Kotesovec, Aug 12 2025

A200714 Stolarsky representation interpreted as binary to decimal integers.

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 10, 9, 23, 12, 27, 29, 30, 63, 8, 21, 19, 22, 47, 26, 25, 55, 28, 59, 61, 62, 127, 20, 17, 43, 18, 39, 45, 46, 95, 24, 53, 51, 54, 111, 58, 57, 119, 60, 123, 125, 126, 255, 16, 41, 35, 42, 87, 37, 38, 79, 44, 91, 93

Offset: 1

Casey Mongoven, Nov 20 2011

See explanation of Stolarsky representations in the C. Mongoven link.

			The Stolarsky representation of 19 is 11101. In binary this is equal to 29. So a(19) = 29.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A048679, A200648, A200649, A200650, A200651.

Cf. A000120, A005811, A023416, A070939.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := FromDigits[stol[n], 2]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

a(n) = {if (n == 1, return (0)); tau = (1 + sqrt(5))/2; mn = 0; while ((m = round(mn*tau)) < n, mn++;); if (m == n, return (2*a(mn)+1)); mn = 0; while ((m = floor(mn*(1+tau)-tau/2)) < n, mn++;); if (m == n, return (2*a(mn))); error("neither A nor B !!");} \\ (cf C. Mongoven link) Michel Marcus, May 21 2013, Sep 02 2013

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = fromdigits(stol(n), 2); \\ Amiram Eldar, Jul 07 2023

From Amiram Eldar, Jul 07 2023: (Start)
A000120(a(n)) = A200649(n).
A023416(a(n)) = A200650(n).
A070939(a(n)) = A200648(n).
A005811(a(n)) = A200651(n). (End)
Conjecture: a(n) = A367306(A358654(n-1)). - Mikhail Kurkov, Oct 17 2024

More terms from Amiram Eldar, Jul 07 2023

A214973 Number of terms in greedy representation of n using Fibonacci and Lucas numbers.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Oct 20 2012

nonn

Consider the sequence b = A116470 consisting of all the Fibonacci numbers and Lucas numbers.  For n>=0, let k(1) be the greatest k in the basis b = {b(k)} such that b(k) <= n, let k(2) be the greatest k in b such that k <= n-b(k(1)), and so on, resulting in the greedy b-representation (to be abbreviated as "rep") of n.  For comparison with the rep using A000045 as basis (called Zeckendorf, or Fibonacci, rep) and also with the rep using A000032 as basis (called the Lucas rep), it is natural to ask this:  what terms in b can possibly follow a given term b(k)?  The answer follows.
If b(k) < 21 = b(11), the terms that can follow b(k) are easily found and not recorded here.  Otherwise, if k is odd, then b(k) can be followed by b(k-i) for some i>=5; if k is even, then b(k) can be followed by b(k-i) for some i>=8.  In Zeckendorf and Lucas reps, the "lag" is k-i for i>=2.
Conjecture: a(A049651(n)) = n and this is the first instance of n in the sequence for all n > 0. In other words, apart from its initial term, A049651 is the RECORDS transform of this sequence. - Charles R Greathouse IV, Oct 14 2021

			Let F, L, U denote the Fibonacci (aka Zeckendorf), Lucas, and greedy F-union-L representations.  Then 45 = 34+8+3 (F) = 29+11+4+1 (L) = 34+11 (U), which shows that a(45) = 2 and that the U representation of 45 requires fewer terms than the others; 45 is the least number having this property.

Clark Kimberling, Table of n, a(n) for n = 1..10000
Jon Maiga, Computer-generated formulas for A214973, Sequence Machine.

Cf. A000032, A000045, A007895, A116470.

s = Reverse[Union[Flatten[Table[{Fibonacci[n + 1], LucasL[n - 1]}, {n, 1, 22}]]]]; Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]]
[[2, 1]], # > 0 &]] &, Range[120]]
(* Peter J. C. Moses, Oct 18 2012 *)

w(n)=if(n%2, fibonacci(n\2+3), fibonacci(n\2) + fibonacci(n\2+2));
k(n)=if(n<9, return(if(n==6,5,n))); for(i=8,n, if(w(i)>n, return(w(i-1))));
a(n)=my(s); while(n, n-=k(n); s++); s; \\ Charles R Greathouse IV, Oct 14 2021

Conjecture: a(n) = A329320(A048679(n)) for n > 0 (noticed by Sequence Machine). - Mikhail Kurkov, Oct 13 2021

A106151 In binary representation of n: delete one zero in each contiguous block of zeros.

Original entry on oeis.org

1, 1, 3, 2, 3, 3, 7, 4, 5, 3, 7, 6, 7, 7, 15, 8, 9, 5, 11, 6, 7, 7, 15, 12, 13, 7, 15, 14, 15, 15, 31, 16, 17, 9, 19, 10, 11, 11, 23, 12, 13, 7, 15, 14, 15, 15, 31, 24, 25, 13, 27, 14, 15, 15, 31, 28, 29, 15, 31, 30, 31, 31, 63, 32, 33, 17, 35, 18, 19, 19, 39, 20, 21, 11, 23, 22, 23

Offset: 1

Reinhard Zumkeller, May 07 2005

Equivalently, change bits 10 -> 0. - Michael S. Branicky, Nov 12 2021

			n=144 = '10010000' -> '101000' = 40 = a(144);
n=145 = '10010001' -> '101001' = 41 = a(145);
n=146 = '10010010' -> '10101'  = 21 = a(146).

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A007088, A030308, A038573, A090078, A175047, A048679, A348710.

import Data.List (group)
a106151 = foldr (\b v -> 2 * v + b) 0 . concatMap
   (\bs'@(b:bs) -> if b == 0 then bs else bs') . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

A106151(n) = if(n<=1, n, if(n%2, 1+(2*A106151((n-1)/2)), A106151(n>>valuation(n, 2))<<(valuation(n, 2)-1))); \\ Antti Karttunen, May 13 2018

A106151(n) = { my(s=0, i=0); while(n, if(2!=(n%4), s += (n%2)<>= 1); (s); }; \\ Antti Karttunen, Jul 01 2024

def a(n): return int(bin(n).replace("b", "").replace("10", "1"), 2)
print([a(n) for n in range(1, 78)]) # Michael S. Branicky, Nov 12 2021

a(n) <= n; a(n) = n iff n = 2^k-1: a(A000225(n))=A000225(n);
A000120(a(n)) = A000120(n);
A023416(a(n)) = A023416(n) - A087116(n).
a(n) = b(n, 0), where b(n, r) = if n = 1 then 1 else b(floor(n/2), 1 - n mod 2)*(1 + floor((1 + r + n mod 2)/2)) + n mod 2.
For n <= 1, a(n) = n, and for n > 1, if n is odd, then a(n) = 1+2*a((n-1)/2), otherwise, when n is even, a(n) = (2^(A007814(n)-1)) * a(A000265(n)). - Antti Karttunen, May 13 2018

A277006 a(n) = A005940(1+A003714(n)); permutation of squarefree numbers.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 15, 11, 14, 21, 35, 30, 13, 22, 33, 55, 42, 77, 70, 105, 17, 26, 39, 65, 66, 91, 110, 165, 143, 154, 231, 385, 210, 19, 34, 51, 85, 78, 119, 130, 195, 187, 182, 273, 455, 330, 221, 286, 429, 715, 462, 1001, 770, 1155, 23, 38, 57, 95, 102, 133, 170, 255, 209, 238, 357, 595, 390, 247, 374, 561, 935, 546, 1309, 910, 1365, 323

Offset: 0

Antti Karttunen, Oct 07 2016

nonn

Permutation of A005117 (squarefree numbers).

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

Cf. A000040, A000045, A003714, A005117, A005940, A048675, A048679, A087808.

Cf. also A019565, A277010, A277196.

(define (A277006 n) (A005940 (+ 1 (A003714 n))))

a(n) = A005940(1+A003714(n)).
Other identities.
For n >= 0, A048675(a(n)) = A087808(A003714(n)) = A048679(n).
For n >= 1, a(A000045(n+1)) = A000040(n).

A358654 a(n) = A025480(A353654(n+1) - 1).

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 9, 10, 23, 12, 27, 29, 30, 63, 8, 19, 21, 22, 47, 25, 26, 55, 28, 59, 61, 62, 127, 17, 18, 39, 20, 43, 45, 46, 95, 24, 51, 53, 54, 111, 57, 58, 119, 60, 123, 125, 126, 255, 16, 35, 37, 38, 79, 41, 42, 87, 44, 91, 93

Offset: 0

Mikhail Kurkov, Nov 25 2022

Permutation of the nonnegative integers.

Conjecture: A247648(n) with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n).

Cf. A025480, A048679, A247648, A343152, A348366, A353654, A355489.

Conjecture: a(n) = A348366(A343152(n)) for n > 0 with a(0) = 1.

A355489 Numbers k such that A000120(k) = A007814(k) + 2.

Original entry on oeis.org

3, 5, 9, 14, 17, 22, 26, 33, 38, 42, 50, 60, 65, 70, 74, 82, 92, 98, 108, 116, 129, 134, 138, 146, 156, 162, 172, 180, 194, 204, 212, 228, 248, 257, 262, 266, 274, 284, 290, 300, 308, 322, 332, 340, 356, 376, 386, 396, 404, 420, 440, 452, 472, 488, 513, 518

Offset: 1

Mikhail Kurkov, Jul 04 2022 [verification needed]

Each term k, doubled, can be put into a one-to-one correspondence with a maximal Schreier set (a subset of the positive integers with cardinality equal to the minimum element in the set) by interpreting the 1-based position of the ones in the binary expansion of 2*k (where position 1 corresponds to the least significant bit) as the elements of the corresponding maximal Schreier set. See A373556 for more information. Cf. also A371176. - Paolo Xausa, Jun 13 2024

Cf. A000045, A000120, A007814, A010056, A025480, A048679, A072649, A371176, A373556.

Select[Range[500], DigitCount[#, 2, 1] == IntegerExponent[#, 2] + 2 &] (* Amiram Eldar, Jul 04 2022 *)

r=quadgen(5);
A355489_upto(nMax)={my(v1,v2,v3,v4); v1=vector(nMax,i,0); v1[1]=1; for(i=1,nMax-1,v1[i+1]=v1[i\r+1]+1); v2=vector(nMax,i,0); v2[1]=1; for(i=2,nMax,v2[i]=v1[i]-v1[i-1]); v3=vector(nMax,i,0); for(i=1,3,v3[i]=2^(i-1)); for(i=4,nMax,v3[i]=if(v2[i-1]==1,5,2*v3[i-fibonacci(v1[i-1]+1)]-if(v2[i]==1,1,0))); v4=vector(nMax,i,0); v4[1]=3; for(i=2,nMax,v4[i]=v4[i-1]+v3[i]); v4}

isok(k) = hammingweight(k) == valuation(k, 2) + 2; \\ Michel Marcus, Jul 06 2022
(Python 3.10+)
from itertools import count, islice
def A355489_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:n.bit_count()==(n&-n).bit_length()+1,count(max(startvalue,1)))
A355480_list = list(islice(A355489_gen(),30)) # Chai Wah Wu, Jul 15 2022

a(n) = a(n-1) + b(n) for n > 1 with a(1) = 3 where b(n) = {2^(n-1) if n < 4; 5 if c(n-1) = 1; otherwise 2*b(n - A000045(A072649(n-1) + 1)) - [c(n) = 1]} and where c(n) = A010056(n).
A025480(a(n)-1) = A048679(n) for n > 0.
a(A000045(n)) = 2^(n-1) + 1 for n > 1.

cp's OEIS Frontend

A048680 Nonnegative integers A001477 expanded with rewrite 0->0, 01->1, then interpreted as Zeckendorffian expansions (as numbers of Fibonacci number system).

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A048678 Binary expansion of nonnegative integers expanded to "Zeckendorffian format" with rewrite rules 0->0, 1->01.

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A060112 Sums of nonconsecutive factorial numbers.

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A371176 Numbers k such that A000120(k) <= A001511(k).

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A200714 Stolarsky representation interpreted as binary to decimal integers.

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A214973 Number of terms in greedy representation of n using Fibonacci and Lucas numbers.

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