A069010 -id:A069010 - cp's OEIS frontend

A384890 Number of maximal anti-runs (increasing by more than 1) in the binary indices of n.

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

Offset: 0

Gus Wiseman, Jun 17 2025

nonn

First differs from A272604 at a(51) = 3, A272604(51) = 2.
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.
Do all constant runs in this sequence have lengths 1, 2, or 3?

			The binary indices of 51 are {1,2,5,6}, with maximal anti-runs ((1),(2,5),(6)), so a(51) = 3.

For runs instead of anti-runs we have A069010 = run-lengths of A245563 (reverse A245562).
Row-lengths of A384877, firsts A384878.
For prime indices instead of binary indices we have A384906.
A000120 counts binary indices.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A052499, A164707, A243815, A300820, A328592, A384177, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length[Split[bpe[n],#2!=#1+1&]],{n,0,100}]

A372433 Binary weight (number of ones in binary expansion) of the n-th squarefree number.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, May 04 2024

Harvey P. Dale, Table of n, a(n) for n = 1..1000
MathOverflow, Are there primes of every Hamming weight?
Wikipedia, Hamming weight.

Restriction of A000120 to A005117.
For prime instead of squarefree we have A014499, zeros A035103.
Counting zeros instead of ones gives A372472, cf. A023416, A372473.
For binary length instead of weight we have A372475.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A145037 counts ones minus zeros in binary expansion, cf. A031443, A031444, A031448, A097110.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
A372516 counts ones minus zeros in binary expansion of primes, cf. A177718, A177796, A372538, A372539.
Cf. A039004, A049093, A049094, A059015, A069010, A070939, A073642, A211997, A368494, A372474.

DigitCount[Select[Range[100],SquareFreeQ],2,1]
Total[IntegerDigits[#,2]]&/@Select[Range[200],SquareFreeQ] (* Harvey P. Dale, Feb 14 2025 *)

from math import isqrt
from sympy import mobius
def A372433(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return int(m).bit_count() # Chai Wah Wu, Aug 02 2024

a(n) = A000120(A005117(n)).

a(n) + A372472(n) = A372475(n) = A070939(A005117(n)).

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

Original entry on oeis.org

1, 0, 2, 0, 5, 2, 3, 0, 9, 5, 6, 2, 7, 3, 4, 0, 14, 9, 10, 5, 11, 6, 7, 2, 12, 7, 8, 3, 9, 4, 5, 0, 20, 14, 15, 9, 16, 10, 11, 5, 17, 11, 12, 6, 13, 7, 8, 2, 18, 12, 13, 7, 14, 8, 9, 3, 15, 9, 10, 4, 11, 5, 6, 0, 27, 20, 21, 14, 22, 15, 16, 9, 23, 16, 17, 10

Offset: 0

Gus Wiseman, Jan 03 2023

			The binary expansion of 100 is (1,1,0,0,1,0,0), with zeros at positions {3,4,6,7}, so a(100) = 20.

The number of zeros is A023416, partial sums A059015.
For positions of 1's we have A230877, reversed A029931.
The reversed version is A359400.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A065359, A069010, A070939, A073642, A083652, A328594, A328595, A359402, A359495.

Table[Total[Join@@Position[IntegerDigits[n,2],0]],{n,0,100}]

a(n>0) = binomial(A029837(n)+1,2) - A230877(n).

A359400 Sum of positions of zeros in the reversed binary expansion of n, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 1, 0, 6, 5, 4, 3, 3, 2, 1, 0, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 15, 14, 13, 12, 12, 11, 10, 9, 11, 10, 9, 8, 8, 7, 6, 5, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 21, 20, 19, 18, 18, 17, 16, 15, 17, 16, 15, 14, 14, 13

Offset: 0

Gus Wiseman, Jan 05 2023

			The reversed binary expansion of 100 is (0,0,1,0,0,1,1), with zeros at positions {1,2,4,5}, so a(100) = 12.

The number of zeros is A023416, partial sums A059015.
Row sums of A368494.
For positions of 1's we have A029931, non-reversed A230877.
The non-reversed version is A359359.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A069010, A070939, A073642, A328594, A328595, A344618, A359402, A359495.

long A359400(long n) {
  long result = 0, counter = 1;
  do {
    if (n % 2 == 0)
      result += counter;
    counter++;
    n /= 2;
  } while (n > 0);
  return result; } // Frank Hollstein, Jan 06 2023

Table[Total[Join@@Position[Reverse[IntegerDigits[n,2]],0]],{n,0,100}]

def a(n): return sum(i for i, bi in enumerate(bin(n)[:1:-1], 1) if bi=='0')
print([a(n) for n in range(78)]) # Michael S. Branicky, Jan 09 2023

a(n) = binomial(A029837(n)+1, 2) - A029931(n), for n>0.

A033264 Number of blocks of {1,0} in the binary expansion of n.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

Number of i such that d(i) < d(i-1), where Sum_{d(i)*2^i: i=0,1,....,m} is base 2 representation of n.

This is the base-2 down-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Eric Weisstein's World of Mathematics, Digit Block.
Index entries for sequences related to binary expansion of n

Cf. A014081, A014082, A037800, A056974, A056975, A056976, A056977, A056978, A056979, A056980, A196521.
a(n) = A005811(n) - ceiling(A005811(n)/2) = A005811(n) - A069010(n).
Equals (A072219(n+1)-1)/2.
Cf. also A175047, A030308.
Essentially the same as A087116.

a033264 = f 0 . a030308_row where
   f c [] = c
   f c (0 : 1 : bs) = f (c + 1) bs
   f c (_ : bs) = f c bs
-- Reinhard Zumkeller, Feb 20 2014, Jun 17 2012

f:= proc(n) option remember; local k;
k:= n mod 4;
if k = 2 then procname((n-2)/4) + 1
elif k = 3 then procname((n-3)/4)
else procname((n-k)/2)
fi
end proc:
f(1):= 0: f(0):= q:
seq(f(i),i=1..100); # Robert Israel, Aug 31 2015

Table[Count[Partition[IntegerDigits[n, 2], 2, 1], {1, 0}], {n, 102}] (* Michael De Vlieger, Aug 31 2015, after Robert G. Wilson v at A014081 *)
Table[SequenceCount[IntegerDigits[n,2],{1,0}],{n,110}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 26 2017 *)

a(n) = { hammingweight(bitand(n>>1, bitneg(n))) }; \\ Gheorghe Coserea, Aug 30 2015

def A033264(n): return ((n>>1)&~n).bit_count() # Chai Wah Wu, Jun 25 2025

G.f.: 1/(1-x) * Sum_(k>=0, t^2/(1+t)/(1+t^2), t=x^2^k). - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - (n mod 2). - Ralf Stephan, Sep 10 2003
a(4n) = a(4n+1) = a(2n), a(4n+2) = a(n)+1, a(4n+3) = a(n). - Ralf Stephan, Aug 20 2003
a(n) = A087116(n) for n > 0, since strings of 0's alternate with strings of 1's, which end in (1,0). - Jonathan Sondow, Jan 17 2016
Sum_{n>=1} a(n)/(n*(n+1)) = Pi/4 - log(2)/2 (A196521) (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A372683 Least squarefree number >= 2^n.

Original entry on oeis.org

1, 2, 5, 10, 17, 33, 65, 129, 257, 514, 1027, 2049, 4097, 8193, 16385, 32770, 65537, 131073, 262145, 524289, 1048577, 2097154, 4194305, 8388609, 16777217, 33554433, 67108865, 134217730, 268435457, 536870913, 1073741826, 2147483649, 4294967297, 8589934594

Offset: 0

Gus Wiseman, May 26 2024

nonn

			The terms together with their binary expansions and binary indices begin:
       1:                    1 ~ {1}
       2:                   10 ~ {2}
       5:                  101 ~ {1,3}
      10:                 1010 ~ {2,4}
      17:                10001 ~ {1,5}
      33:               100001 ~ {1,6}
      65:              1000001 ~ {1,7}
     129:             10000001 ~ {1,8}
     257:            100000001 ~ {1,9}
     514:           1000000010 ~ {2,10}
    1027:          10000000011 ~ {1,2,11}
    2049:         100000000001 ~ {1,12}
    4097:        1000000000001 ~ {1,13}
    8193:       10000000000001 ~ {1,14}
   16385:      100000000000001 ~ {1,15}
   32770:     1000000000000010 ~ {2,16}
   65537:    10000000000000001 ~ {1,17}
  131073:   100000000000000001 ~ {1,18}
  262145:  1000000000000000001 ~ {1,19}
  524289: 10000000000000000001 ~ {1,20}

For primes instead of powers of two we have A112926, opposite A112925, sum A373197, length A373198.
Counting zeros instead of all bits gives A372473, firsts of A372472.
These are squarefree numbers at indices A372540, firsts of A372475.
Counting ones instead of all bits gives A372541, firsts of A372433.
The opposite (greatest squarefree number <= 2^n) is A372889.
The difference from 2^n is A373125.
For prime instead of squarefree we have:
- bits A372684, firsts of A035100
- zeros A372474, firsts of A035103
- ones A372517, firsts of A014499
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308, length A070939 or A029837.
A061398 counts squarefree numbers between primes (exclusive).
A077643 counts squarefree terms between powers of 2, run-lengths of A372475.
A143658 counts squarefree numbers up to 2^n.
Cf. A029931, A048793, A049095, A049096, A059015, A069010, A076259, A077641, A211997, A230877.

Table[NestWhile[#+1&,2^n,!SquareFreeQ[#]&],{n,0,10}]

a(n) = my(k=2^n); while (!issquarefree(k), k++); k; \\ Michel Marcus, May 29 2024

from itertools import count
from sympy import factorint
def A372683(n): return next(i for i in count(1<Chai Wah Wu, Aug 26 2024

a(n) = A005117(A372540(n)).

a(n) = A067535(2^n). - R. J. Mathar, May 31 2024

A372684 Least k such that prime(k) >= 2^n.

Original entry on oeis.org

1, 3, 5, 7, 12, 19, 32, 55, 98, 173, 310, 565, 1029, 1901, 3513, 6543, 12252, 23001, 43391, 82026, 155612, 295948, 564164, 1077872, 2063690, 3957810, 7603554, 14630844, 28192751, 54400029, 105097566, 203280222, 393615807, 762939112, 1480206280, 2874398516, 5586502349

Offset: 1

Gus Wiseman, May 30 2024

nonn

			The numbers prime(a(n)) together with their binary expansions and binary indices begin:
        2:                       10 ~ {2}
        5:                      101 ~ {1,3}
       11:                     1011 ~ {1,2,4}
       17:                    10001 ~ {1,5}
       37:                   100101 ~ {1,3,6}
       67:                  1000011 ~ {1,2,7}
      131:                 10000011 ~ {1,2,8}
      257:                100000001 ~ {1,9}
      521:               1000001001 ~ {1,4,10}
     1031:              10000000111 ~ {1,2,3,11}
     2053:             100000000101 ~ {1,3,12}
     4099:            1000000000011 ~ {1,2,13}
     8209:           10000000010001 ~ {1,5,14}
    16411:          100000000011011 ~ {1,2,4,5,15}
    32771:         1000000000000011 ~ {1,2,16}
    65537:        10000000000000001 ~ {1,17}
   131101:       100000000000011101 ~ {1,3,4,5,18}
   262147:      1000000000000000011 ~ {1,2,19}
   524309:     10000000000000010101 ~ {1,3,5,20}
  1048583:    100000000000000000111 ~ {1,2,3,21}
  2097169:   1000000000000000010001 ~ {1,5,22}
  4194319:  10000000000000000001111 ~ {1,2,3,4,23}
  8388617: 100000000000000000001001 ~ {1,4,24}

The opposite (greatest k such that prime(k) <= 2^n) is A007053.
Positions of first appearances in A035100.
The distance from prime(a(n)) to 2^n is A092131.
Counting zeros instead of all bits gives A372474, firsts of A035103.
Counting ones instead of all bits gives A372517, firsts of A014499.
For primes between powers of 2:
- sum A293697
- length A036378
- min A104080 or A014210
- max A014234, delta A013603
For squarefree numbers between powers of 2:
- sum A373123
- length A077643, run-lengths of A372475
- min A372683, delta A373125, indices A372540
- max A372889, delta A373126, indices A143658
For squarefree numbers between primes:
- sum A373197
- length A373198 = A061398 - 1
- min A000040
- max A112925, opposite A112926
Cf. A000120, A029837, A029931, A030190, A049095, A069010, A070939, A077641, A080791, A211997.

Table[PrimePi[If[n==1,2,NextPrime[2^n]]],{n,30}]

a(n) = primepi(nextprime(2^n)); \\ Michel Marcus, May 31 2024

a(n>1) = A007053(n) + 1.
a(n) = A000720(A104080(n)).
prime(a(n)) = A104080(n).
prime(a(n)) - 2^n = A092131(n).

More terms from Michel Marcus, May 31 2024

A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

Original entry on oeis.org

1, 6, 34, 178, 882, 4210, 19570, 89202, 400498, 1776754

Offset: 1

Gus Wiseman, Jun 23 2025

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 set of binary indices of each nonnegative integer and its partition into anti-runs begins:
  0: {}      {{}}
  1: {1}     {{1}}
  2: {2}     {{2}}
  3: {1,2}   {{1},{2}}
  4: {3}     {{3}}
  5: {1,3}   {{1,3}}
  6: {2,3}   {{2},{3}}
  7: {1,2,3} {{1},{2},{3}}
The flattened version begins: {}, {1}, {2}, {1}, {2}, {3}, {1,3}, {2}, {3}, {1}, {2}, {3}. Of these sets, the first of length 2 is the sixth (starting with 0), so we have a(2) = 6.

For runs instead of anti-runs we have A001792.
The unflattened version is A052499.
Positions of first appearances in A384877, see A000120, A245562, A245563, A384890.
A023758 lists differences of powers of 2.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
q=Join@@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}];
Table[Position[q,i][[1,1]],{i,mnrm[q]}]

A252754 Inverse of "Tree of Eratosthenes" permutation: a(1) = 0, after which, a(2n) = 1 + 2a(n), a(2n+1) = 2 a(A268674(2n+1)).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 14, 33, 256, 23, 12, 65, 18, 35, 512, 21, 1024, 31, 22, 129, 20, 27, 2048, 257, 34, 39, 4096, 29, 8192, 67, 30, 513, 16384, 47, 24, 25, 26, 131, 32768, 37, 28, 71, 38, 1025, 65536, 43, 131072, 2049, 66, 63, 36, 45, 262144, 259, 46, 41, 524288, 55, 1048576, 4097, 130, 515, 40

Offset: 1

Antti Karttunen, Jan 02 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..1024
Index entries for sequences that are permutations of the natural numbers

Inverse: A252753.
Fixed points of a(n)+1: A253789.
Cf. A250470, A253556, A253557, A253558, A253559, A268674, A302041.
Similar permutations: A156552, A252756, A054429, A250246, A269388.
Differs from A156552 for the first time at n=21, where a(21) = 14, while A156552(21) = 18.

A252754(n) = if(1==n,0,if(!(n%2),1 + 2*A252754(n/2),2*A252754(A268674(n)))); \\ Antti Karttunen, Mar 31 2018

;; With memoization-macro definec.
(definec (A252754 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A252754 (/ n 2))))) (else (* 2 (A252754 (A250470 n))))))

a(1) = 0, after which, a(2n) = 1 + 2*a(n), a(2n+1) = 2 * a(A268674(2n+1)).
As a composition of related permutations:
a(n) = A054429(A252756(n)).
a(n) = A156552(A250246(n)).
From Antti Karttunen, Mar 31 2018: (Start)
A000120(a(n)) = A253557(n).
A069010(a(n)) = A302041(n).
A132971(a(n)) = A302050(n).
A106737(a(n)) = A302051(n).
(End)

Name edited and formula corrected by Antti Karttunen, Mar 31 2018

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