A246029 -id:A246029 - cp's OEIS frontend

A083368 a(n) is the position of the highest one in A003754(n).

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

Offset: 1

Gary W. Adamson, Jun 04 2003

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.

Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Cf. A003714, A003754.
Cf. A000045, A000071.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.
Cf. A001629, A037800, A044813, A048793, A069010, A200649, A384878, A385886.

a083368 n = a083368_list !! (n-1)
a083368_list = concat $ h $ drop 2 a000071_list where
   h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
-- Reinhard Zumkeller, Aug 10 2014

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).

a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Edited by Don Reble, Nov 12 2005

Shorter name from Joerg Arndt, Jul 27 2025

A329133 Numbers whose augmented differences of prime indices are an aperiodic sequence.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74

Offset: 1

Gus Wiseman, Nov 09 2019

nonn

The augmented differences aug(y) of an integer partition y of length k are given by aug(y)i = y_i - y{i + 1} + 1 if i < k and aug(y)_k = y_k. For example, aug(6,5,5,3,3,3) = (2,1,3,1,1,3).
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
A finite sequence is aperiodic if its cyclic rotations are all different.

			The sequence of terms together with their augmented differences of prime indices begins:
    1: ()
    2: (1)
    3: (2)
    5: (3)
    6: (2,1)
    7: (4)
    9: (1,2)
   10: (3,1)
   11: (5)
   12: (2,1,1)
   13: (6)
   14: (4,1)
   17: (7)
   18: (1,2,1)
   19: (8)
   20: (3,1,1)
   21: (3,2)
   22: (5,1)
   23: (9)
   24: (2,1,1,1)

Complement of A329132.
These are the Heinz numbers of the partitions counted by A329136.
Aperiodic binary words are A027375.
Aperiodic compositions are A000740.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose prime signature is aperiodic are A329139.
Numbers whose differences of prime indices are aperiodic are A329135.
Cf. A056239, A112798, A121016, A124010, A152061, A246029, A325351, A325389, A329134, A329140.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
aperQ[q_]:=Array[RotateRight[q,#1]&,Length[q],1,UnsameQ];
aug[y_]:=Table[If[i

A385818 The number k such that the k-th composition in standard order lists the maximal run lengths of each nonnegative integer's binary indices, with duplicates removed.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 13, 17, 14, 18, 20, 24, 32, 15, 19, 21, 25, 33, 22, 26, 34, 28, 36, 40, 48, 64, 23, 27, 35, 29, 37, 41, 49, 65, 30, 38, 42, 50, 66, 44, 52, 68, 56, 72, 80, 96, 128, 31, 39, 43, 51, 67, 45, 53, 69, 57, 73, 81, 97

Offset: 0

Gus Wiseman, Jul 18 2025

A permutation of the nonnegative integers.
A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)) with lengths (1,1,2), which is the 14th composition in standard order, so A385889(53) = 14, and after removing duplicate rows a(16) = 14.

John Tyler Rascoe, Table of n, a(n) for n = 0..8192

For anti-runs instead of runs we appear to have A348366.
See also A385816 (standard compositions of rows of A384877), reverse A209859.
The compositions themselves are listed by A385817.
Before removing duplicates we had A385889.
A245563 lists run lengths of binary indices (ranks A246029), rev A245562, strict A328592.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A029931, A048793, A052499, A069010, A200648, A200649, A200650, A384890, A385886, A385887.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@DeleteDuplicates[Table[Length/@Split[bpe[n],#2==#1+1&],{n,0,100}]]

A153013 Starting with input 0, find the binary value of the input. Then interpret resulting string of 1's and 0's as prime-based numbers, as follows: 0's are separators, uninterrupted strings of 1's are interpreted from right to left as exponents of the prime numbers. Output is returned as input for the next number in sequence.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 9, 10, 15, 16, 11, 12, 25, 50, 147, 220, 6125, 1968750, 89142864525, 84252896510182189218, 34892570216750728458698250328871491829901861750593684043

Offset: 0

Mark Zegarelli (mtzmtz(AT)gmail.com), Dec 16 2008

nonn

From Antti Karttunen, Oct 15 2016: (Start)
Iterates of map f : n -> A005940(1+n), (Doudna-sequence, but with starting offset zero) starting from the initial value 0. Conversely, the unique infinite sequence such that a(n) = A156552(a(n+1)) and a(0) = 0.
Note that map f can also form cycles, like 7 <-> 8 (A005940(1+7) = 8, A005940(1+8) = 7).
On the other hand, this sequence cannot ever fall into a loop because 0 is not in the range of map f, for n=0.., while f is injective on [1..]. Thus the values obtained by this sequence are not bounded, although there might be more nonmonotonic positions like for example there is from a(10) = 16 to a(11) = 11.
The formula A008966(a(n+1)) = A085357(a(n)) tells that the squarefreeness of the next term a(n+1) is determined by whether the previous term a(n) is a fibbinary number (A003714) or not. Numerous other such correspondences hold, and they hold also for any other trajectories outside of this sequence.
Even and odd terms alternate. No two squares can occur in succession because A106737 obtains even values for all squares > 1 and A000005 is odd for all squares. More directly this is seen from the fact that the rightmost 1-bit in the binary expansion of any square is always alone.
(End)

			101 is interpreted as 3^1 * 2^1 = 6. 1110011 is interpreted as 5^3 * 2^2 = 500.

Yang Haoran, Table of n, a(n) for n = 0..23

Cf. A000005, A003714, A005940, A008966, A085357, A106737, A156552, A181819, A246029.

Cf. also A109162, A328316 for similar iteration sequences.

NestList[Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, #] &@ Flatten@ MapIndexed[If[Total@ #1 == 0, ConstantArray[0, Boole[First@ #2 == 1] + Length@ #1 - 1], Length@ #1] &, Reverse@ Split@ IntegerDigits[#, 2]] &, 0, 21] (* Michael De Vlieger, Oct 17 2016 *)

step(n)=my(t=1,v); forprime(p=2,, v=valuation(n+1,2); t*=p^v; n>>=v+1; if(!n, return(t)))
t=0; concat(0,vector(20,n, t=step(t))) \\ Charles R Greathouse IV, Sep 01 2015

;; With memoization-macro definec.
(definec (A153013 n) (if (zero? n) n (A005940 (+ 1 (A153013 (- n 1))))))
;; Antti Karttunen, Oct 15 2016

From Antti Karttunen, Oct 15 2016: (Start)
a(0) = 0; for n >= 1, a(n) = A005940(1+a(n-1)).
A008966(a(n+1)) = A085357(a(n)). [See the comment.]
A181819(a(1+n)) = A246029(a(n)).
A000005(a(n+1)) = A106737(a(n)).
(End)

a(20)-a(22) from Yang Haoran, Aug 31 2015

A385887 The number k such that the k-th composition in standard order is the reversed sequence of lengths of maximal runs of binary indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 4, 1, 3, 3, 6, 2, 5, 4, 8, 1, 3, 3, 6, 3, 7, 6, 12, 2, 5, 5, 10, 4, 9, 8, 16, 1, 3, 3, 6, 3, 7, 6, 12, 3, 7, 7, 14, 6, 13, 12, 24, 2, 5, 5, 10, 5, 11, 10, 20, 4, 9, 9, 18, 8, 17, 16, 32, 1, 3, 3, 6, 3, 7, 6, 12, 3, 7, 7, 14, 6, 13, 12, 24

Offset: 0

Gus Wiseman, Jul 17 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with reversed lengths (2,1), which is the 5th composition in standard order, so a(100) = 5.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

Removing duplicates appears to give A232559, see also A348366, A358654, A385818.
Sorted positions of firsts appearances appear to be A247648+1.
The non-reverse version is A385889.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384877 lists anti-run lengths of binary indices (ranks A385816), reverse A209859.
Cf. A000120, A029931, A044813, A048793, A069010, A384175, A385817, A385886.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Reverse[Length/@Split[bpe[n],#2==#1+1&]]],{n,0,100}]

A385892 In the sequence of run lengths of binary indices of each positive integer (A245563), remove all duplicate rows after the first and take the last term of each remaining row.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Jul 18 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2), which is the 16th row of A385817, so a(16) = 2.

In the following references, "before" is short for "before removing duplicate rows".
Positions of firsts appearances appear to be A000071.
Without the removals we have A090996.
For sum instead of last we have A200648, before A000120.
For length instead of last we have A200650+1, before A069010 = A037800+1.
Last term of row n of A385817 (ranks A385818, before A385889), first A083368.
A245563 gives run lengths of binary indices, see A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, A385816.
Cf. A001629, A044813, A048793, A164707, A200649, A384878, A384890, A385886.

Last/@DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,100}]]

A385890 Positions of first appearances in A245563 = run lengths of binary indices.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 14, 16, 22, 24, 28, 30, 32, 44, 46, 48, 54, 56, 60, 62, 64, 86, 88, 92, 94, 96, 108, 110, 112, 118, 120, 124, 126, 128, 172, 174, 176, 182, 184, 188, 190, 192, 214, 216, 220, 222, 224, 236, 238, 240, 246, 248, 252, 254, 256, 342, 344, 348

Offset: 1

Gus Wiseman, Jul 18 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

These are positions of firsts appearances in A245563, ranks A385889, reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 lists anti-run lengths of binary indices, ranks A385816.
Cf. A000120, A023758, A048793, A069010, A200649, A243815, A246029, A328592, A384890.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
q=Table[Length/@Split[bpe[n],#2==#1+1&],{n,0,1000}];
Select[Range[Length[q]-1],!MemberQ[Take[q,#-1],q[[#]]]&]

cp's OEIS Frontend

A083368 a(n) is the position of the highest one in A003754(n).

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A329133 Numbers whose augmented differences of prime indices are an aperiodic sequence.

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A385818 The number k such that the k-th composition in standard order lists the maximal run lengths of each nonnegative integer's binary indices, with duplicates removed.

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A385887 The number k such that the k-th composition in standard order is the reversed sequence of lengths of maximal runs of binary indices of n.

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A385892 In the sequence of run lengths of binary indices of each positive integer (A245563), remove all duplicate rows after the first and take the last term of each remaining row.

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A385890 Positions of first appearances in A245563 = run lengths of binary indices.

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