A253315 -id:A253315 - cp's OEIS frontend

A253317 Indices in A261283 where records occur.

0, 1, 2, 3, 8, 9, 10, 11, 128, 129, 130, 131, 136, 137, 138, 139, 32768, 32769, 32770, 32771, 32776, 32777, 32778, 32779, 32896, 32897, 32898, 32899, 32904, 32905, 32906, 32907, 2147483648, 2147483649, 2147483650, 2147483651, 2147483656, 2147483657

Offset: 1

Philippe Beaudoin, Dec 30 2014

From Gus Wiseman, Dec 29 2023: (Start)
These are numbers whose binary indices are all powers of 2, where a binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. For example, the terms together with their binary expansions and binary indices begin:
       0 ~ {}
       1 ~ {1}
      10 ~ {2}
      11 ~ {1,2}
    1000 ~ {4}
    1001 ~ {1,4}
    1010 ~ {2,4}
    1011 ~ {1,2,4}
10000000 ~ {8}
10000001 ~ {1,8}
10000010 ~ {2,8}
10000011 ~ {1,2,8}
10001000 ~ {4,8}
10001001 ~ {1,4,8}
10001010 ~ {2,4,8}
10001011 ~ {1,2,4,8}
For powers of 3 we have A368531.
(End)

Michael De Vlieger, Table of n, a(n) for n = 1..4096
Lorenzo Sauras-Altuzarra, Some arithmetical problems that are obtained by analyzing proofs and infinite graphs, arXiv:2002.03075 [math.NT], 2020.

Cf. A053644 (most significant bit).
A048793 lists binary indices, length A000120, sum A029931.
A070939 gives length of binary expansion.
A096111 gives product of binary indices.
Cf. A058891, A062050, A072639, A326031, A326675, A326702, A367771, A367912, A368183, A368109, A368531.

a := proc(n) local k, A:
A := [seq(0,i=1..n)]: A[1]:=0:
for k from 1 to n-1 do
   A[k+1] := A[k-2^ilog2(k)+1]+2^(2^ilog2(k)-1): od:
return A[n]: end proc: # Lorenzo Sauras Altuzarra, Dec 18 2019
# second Maple program:
a:= n-> (l-> add(l[i+1]*2^(2^i-1), i=0..nops(l)-1))(Bits[Split](n-1)):
seq(a(n), n=1..38);  # Alois P. Heinz, Dec 13 2023

Nest[Append[#1, #1[[-#2]] + 2^(#2 - 1)] & @@ {#, 2^(IntegerLength[Length[#], 2] - 1)} &, {0, 1}, 36] (* Michael De Vlieger, May 08 2020 *)

a(n)={if(n<=1, 0, my(t=1<Andrew Howroyd, Dec 20 2019

a(1) = 0 and a(n) = a(n-A053644(n-1)) + 2^(A053644(n-1)-1). - Lorenzo Sauras Altuzarra, Dec 18 2019
a(n) = A358126(n-1) / 2. - Tilman Piesk, Dec 18 2022
a(2^n+1) = 2^(2^n-1) = A058891(n+1). - Gus Wiseman, Dec 29 2023
a(2^n) = A072639(n). - Gus Wiseman, Dec 29 2023
G.f.: 1/(1-x) * Sum_{k>=0} (2^(-1+2^k))*x^2^k/(1+x^2^k). - John Tyler Rascoe, May 22 2024

Corrected reference in name from A253315 to A261283. - Tilman Piesk, Dec 18 2022

A261283 a(n) = bitwise XOR of all the bit numbers for the bits that are set in n, using number 1 for the LSB.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Aug 14 2015, following the original version A253315 by Philippe Beaudoin, Dec 30 2014

If the least significant bit is numbered 0, then a(2n) = a(2n+1) if one uses the "natural" definition reading "...set in n": see A253315 for that version. To avoid the duplication, we chose here to start numbering the bits with 1 for the LSB; equivalently, we can start numbering the bits with 0 but use the definition "...bits set in 2n". In any case, a(n) = A253315(2n) = A253315(2n+1).

Since the XOR operation is associative, one can define XOR of an arbitrary number of terms in a recursive way, there is no ambiguity about the order in which the operations are performed.

			a(7) = a(4+2+1) = a(2^2+2^1+2^0) = (2+1) XOR (1+1) XOR (0+1) = 3 XOR 3 = 0.
a(12) = a(8+4) = a(2^3+2^2) = (3+1) XOR (2+1) = 4+3 = 7.

Philippe Beaudoin, Table of n, a(n) for n = 0..8190
Oliver Nash, Yet another prisoner puzzle, coins on a chessboard problem.
Tilman Piesk, Illustration of the first 128 terms

Cf. A075926 (indices of 0's), A253315, A327041 (OR equivalent).

A261283[n_] := If[n == 0, 0, BitXor @@ PositionIndex[Reverse[IntegerDigits[n, 2]]][1]]; Array[A261283, 100, 0] (* Paolo Xausa, May 29 2024 *)

A261283(n,b=bittest(n,0))={for(i=1,#binary(n),bittest(n,i)&&b=bitxor(b,i+1));b}

A358126 Replace 2^k in binary expansion of n with 2^(2^k).

Original entry on oeis.org

0, 2, 4, 6, 16, 18, 20, 22, 256, 258, 260, 262, 272, 274, 276, 278, 65536, 65538, 65540, 65542, 65552, 65554, 65556, 65558, 65792, 65794, 65796, 65798, 65808, 65810, 65812, 65814, 4294967296, 4294967298, 4294967300

Offset: 0

Tilman Piesk, Oct 30 2022

Sums of distinct terms of A001146.
The name "ballooned integers" is proposed for this sequence.
a(n) is the index of the first occurrence of n in A253315.

			Let    n   =     25  =  1 +   8 +    16  =     2^0  +    2^3  +    2^4.
Then a(n)  =  65794  =  2 + 256 + 65536  =  2^(2^0) + 2^(2^3) + 2^(2^4).
The binary indices of n are {0, 3, 4}. Those of a(n) are {1, 8, 16}.

Tilman Piesk, Table of n, a(n) for n = 0..4095
Tilman Piesk, Boolean Walsh functions and corresponding balloons
Tilman Piesk, First 32 entries in little-endian binary

Cf. A253317, A001146, A060803, A133457, A197819, A228539, A253315.

a := proc(n) select(d -> d[2] <> 0, ListTools:-Enumerate(convert(n,base,2))):
add(2^(2^(%[j][1] - 1)), j = 1..nops(%)) end: seq(a(n), n = 0..34); # Peter Luschny, Oct 31 2022

a[n_] := Total[2^(2^Range[If[n == 0, 1, IntegerLength[n,2]] - 1, 0, -1]) * IntegerDigits[n, 2]]; Array[a, 35, 0] (* Amiram Eldar, Oct 31 2022 *)

a(n) = my(d=Vecrev(digits(n,2))); for (k=1, #d, d[k] *= 2^(2^(k-1))); vecsum(d); \\ Michel Marcus, Oct 31 2022

def a(n):
    binary_string = "{0:b}".format(n)[::-1]  # little-endian
    result = 0
    for i, binary_digit in enumerate(binary_string):
        if binary_digit == '1':
            result += 1 << (1 << i)  # 2 ** (2 ** i)
    return result

If n = Sum_{i=0..k} 2^s_i, then a(n) = Sum_{i=0..k} 2^(2^s_i).
a(n) = 2 * A253317(n+1).
a(2^n-1) = A060803(n-1) for n >= 1.
a(2^n) = A001146(n).
A197819[m, a(n)] = A228539[m, n]. (Compare link about Boolean Walsh functions.)

A341242 Numbers whose binary representation encodes a subset S of the natural numbers such that the XOR of the binary representations of all s in S gives 0.

Original entry on oeis.org

0, 1, 14, 15, 50, 51, 60, 61, 84, 85, 90, 91, 102, 103, 104, 105, 150, 151, 152, 153, 164, 165, 170, 171, 194, 195, 204, 205, 240, 241, 254, 255, 770, 771, 780, 781, 816, 817, 830, 831, 854, 855, 856, 857, 868, 869, 874, 875, 916, 917, 922, 923, 934, 935, 936

Offset: 1

Marc A. A. van Leeuwen, Feb 07 2021

The numbers for which the set S of positions of bits 1 in the binary representation, interpreted as a set of distinct-sized Nim heaps (including a possible uninteresting size 0 heap for the least significant bit) is losing for the player to move.
Viewing the list as a set of valid code words, every natural number N can be "corrected" to a valid code word by changing exactly one bit, in exactly one way. The position of that bit is found by computing for N the XOR of its raised-bit positions of the title (if the result is 0, then N is already valid but flipping the irrelevant bit 0 makes it valid again).
The "error correcting" interpretation, applied to 64-bit numbers interpreted as orientation of 64 coins, corresponds to a solution of the "coins on a chessboard" puzzle described in the Nick Berry's blog, and also mentioned at A253315.
Numbers 2*n and 2*n+1 for n = A075926(m).
Numbers m such that A253315(m) = 0. - Rémy Sigrist, Feb 09 2021

Nick Berry, Impossible Escape?, DataGenetics blog, December 2014.

Cf. A075926, A253315.

def ok(n):
  xor, b = 0, (bin(n)[2:])[::-1]
  for i, c in enumerate(b):
    if c == '1': xor ^= i
  return xor == 0
print([m for m in range(937) if ok(m)]) # Michael S. Branicky, Feb 07 2021

a(2*n+1) = 2*A075926(n), a(2*n+2) = 2*A075926(n) + 1 for any n >= 0. - Rémy Sigrist, Feb 09 2021

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A253317 Indices in A261283 where records occur.

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A261283 a(n) = bitwise XOR of all the bit numbers for the bits that are set in n, using number 1 for the LSB.

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A358126 Replace 2^k in binary expansion of n with 2^(2^k).

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A341242 Numbers whose binary representation encodes a subset S of the natural numbers such that the XOR of the binary representations of all s in S gives 0.

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