A198275 -id:A198275 - cp's OEIS frontend

A036044 BCR(n): write in binary, complement, reverse.

1, 0, 2, 0, 6, 2, 4, 0, 14, 6, 10, 2, 12, 4, 8, 0, 30, 14, 22, 6, 26, 10, 18, 2, 28, 12, 20, 4, 24, 8, 16, 0, 62, 30, 46, 14, 54, 22, 38, 6, 58, 26, 42, 10, 50, 18, 34, 2, 60, 28, 44, 12, 52, 20, 36, 4, 56, 24, 40, 8, 48, 16, 32, 0, 126, 62, 94, 30, 110, 46, 78, 14, 118, 54, 86

Offset: 0

N. J. A. Sloane

a(0) could be considered to be 0 if the binary representation of zero were chosen to be the empty string. - Jason Kimberley, Sep 19 2011
From Bernard Schott, Jun 15 2021: (Start)
Except for a(0) = 1, every term is even.
For each q >= 0, there is one and only one odd number h such that a(n) = 2*q iff n = h*2^m-1 for m >= 1 when q = 0, and for m >= 0 when q >= 1 (see A345401 and some examples below).
a(n) =  0 iff n =    2^m-1 for m >= 1 (Mersenne numbers) (A000225).
a(n) =  2 iff n =  3*2^m-1 for m >= 0 (A153893).
a(n) =  4 iff n =  7*2^m-1 for m >= 0 (A086224).
a(n) =  6 iff n =  5*2^m-1 for m >= 0 (A153894).
a(n) =  8 iff n = 15*2^m-1 for m >= 0 (A196305).
a(n) = 10 iff n = 11*2^m-1 for m >= 0 (A086225).
a(n) = 12 iff n = 13*2^m-1 for m >= 0 (A198274).
For k >= 1, a(n) = 2^k iff n = (2^(k+1)-1)*2^m - 1 for m >= 0.
Explanation for a(n) = 2:
   For m >= 0, A153893(m) = 3*2^m-1 -> 1011...11 -> 0100...00 -> 10 -> 2 where 1011...11_2 is 10 followed by m 1's. (End)

			4 -> 100 -> 011 -> 110 -> 6.

Indranil Ghosh, Table of n, a(n) for n = 0..10000 (first 1024 terms from T. D. Noe)

Cf. A030101, A056539, A329369, A345401.
Cf. A035928 (fixed points), A195063, A195064, A195065, A195066.
Indices of terms 0, 2, 4, 6, 8, 10, 12, 14, 18, 22, 26, 30: A000225 \ {0}, A153893, A086224, A153894, A196305, A086225, A198274, A052996\{1,3}, A291557, A198276, A171389, A198275.

import Data.List (unfoldr)
a036044 0 = 1
a036044 n = foldl (\v d -> 2 * v + d) 0 (unfoldr bc n) where
   bc 0 = Nothing
   bc x = Just (1 - m, x') where (x',m) = divMod x 2
-- Reinhard Zumkeller, Sep 16 2011

A036044:=func; // Jason Kimberley, Sep 19 2011

A036044 := proc(n)
    local bcr ;
    if n = 0 then
        return 1;
    end if;
    convert(n,base,2) ;
    bcr := [seq(1-i,i=%)] ;
    add(op(-k,bcr)*2^(k-1),k=1..nops(bcr)) ;
end proc:
seq(A036044(n),n=0..200) ; # R. J. Mathar, Nov 06 2017

dtn[ L_ ] := Fold[ 2#1+#2&, 0, L ]; f[ n_ ] := dtn[ Reverse[ 1-IntegerDigits[ n, 2 ] ] ]; Table[ f[ n ], {n, 0, 100} ]
Table[FromDigits[Reverse[IntegerDigits[n,2]/.{1->0,0->1}],2],{n,0,80}] (* Harvey P. Dale, Mar 08 2015 *)

a(n)=fromdigits(Vecrev(apply(n->1-n,binary(n))),2) \\ Charles R Greathouse IV, Apr 22 2015

def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
print([BCR(n) for n in range(75)]) # Michael S. Branicky, Jun 14 2021

def A036044(n): return -int((s:=bin(n)[-1:1:-1]),2)-1+2**len(s) # Chai Wah Wu, Feb 04 2022

a(2n) = 2*A059894(n), a(2n+1) = a(2n) - 2^floor(log_2(n)+1). - Ralf Stephan, Aug 21 2003

Conjecture: a(n) = (-1)^A023416(n)*b(n) for n > 0 with a(0) = 1 where b(2^m) = (-1)^m*(2^(m+1) - 2) for m >= 0, b(2n+1) = b(n) for n > 0, b(2n) = b(n) + b(n - 2^f(n)) + b(2n - 2^f(n)) for n > 0 and where f(n) = A007814(n) (see A329369). - Mikhail Kurkov, Dec 13 2024

A075300 Array A read by antidiagonals upwards: A(n, k) = array A054582(n,k) - 1 = 2^n(2k+1) - 1 with n,k >= 0.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 7, 11, 9, 6, 15, 23, 19, 13, 8, 31, 47, 39, 27, 17, 10, 63, 95, 79, 55, 35, 21, 12, 127, 191, 159, 111, 71, 43, 25, 14, 255, 383, 319, 223, 143, 87, 51, 29, 16, 511, 767, 639, 447, 287, 175, 103, 59, 33, 18, 1023, 1535, 1279, 895, 575, 351, 207, 119

Offset: 0

Antti Karttunen, Sep 12 2002

From Philippe Deléham, Feb 19 2014: (Start)
A(0,k)  = 2*k = A005843(k),
A(1,k)  = 4*k + 1 = A016813(k),
A(2,k)  = 8*k + 3 = A017101(k),
A(n,0)  = A000225(n),
A(n,1)  = A153893(n),
A(n,2)  = A153894(n),
A(n,3)  = A086224(n),
A(n,4)  = A052996(n+2),
A(n,5)  = A086225(n),
A(n,6)  = A198274(n),
A(n,7)  = A238087(n),
A(n,8)  = A198275(n),
A(n,9)  = A198276(n),
A(n,10) = A171389(n). (End)
A permutation of the nonnegative integers. - Alzhekeyev Ascar M, Jun 05 2016
The values in array row n, when expressed in binary, have n trailing 1-bits. - Ruud H.G. van Tol, Mar 18 2025

			The array A begins:
   0    2    4    6    8   10   12   14   16   18 ...
   1    5    9   13   17   21   25   29   33   37 ...
   3   11   19   27   35   43   51   59   67   75 ...
   7   23   39   55   71   87  103  119  135  151 ...
  15   47   79  111  143  175  207  239  271  303 ...
  31   95  159  223  287  351  415  479  543  607 ...
  ... - _Philippe Deléham_, Feb 19 2014
From _Wolfdieter Lang_, Jan 31 2019: (Start)
The triangle T begins:
   n\k   0    1    2   3   4   5   6   7  8  9 10 ...
   0:    0
   1:    1    2
   2:    3    5    4
   3:    7   11    9   6
   4:   15   23   19  13   8
   5    31   47   39  27  17  10
   6:   63   95   79  55  35  21  12
   7:  127  191  159 111  71  43  25  14
   8:  255  383  319 223 143  87  51  29 16
   9:  511  767  639 447 287 175 103  59 33 18
  10: 1023 1535 1279 895 575 351 207 119 67 37 20
  ...
T(3, 1) = 2^2*(2*1+1) - 1 = 12 - 1 = 11.  (End)

Index entries for sequences that are permutations of the natural numbers

Inverse permutation: A075301. Transpose: A075302. The X-projection is given by A007814(n+1) and the Y-projection A025480.

Cf. A002262, A025581, A054582, A241957.

A075300bi := (x,y) -> (2^x * (2*y + 1))-1;
A075300 := n -> A075300bi(A025581(n), A002262(n));
A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2);
A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1);

Table[(2^# (2 k + 1)) - 1 &[m - k], {m, 0, 10}, {k, 0, m}] (* Michael De Vlieger, Jun 05 2016 *)

From Wolfdieter Lang, Jan 31 2019: (Start)
Array A(n, k) = 2^n*(2*k+1) - 1, for n >= 0 and m >= 0.
The triangle is T(n, k) = A(n-k, k) = 2^(n-k)*(2*k+1) - 1, n >= 0, k=0..n.
See also A054582 after subtracting 1. (End)
From Ruud H.G. van Tol, Mar 17 2025: (Start)
A(0, k) is even. For n > 0, A(n, k) is odd and (3 * A(n, k) + 1) / 2 = A(n-1, 3*k+1).
A(n, k) = 2^n - 1 (mod 2^(n+1)) (equivalent to the comment about trailing 1-bits). (End)

A365802 Numbers k such that A163511(k) is a fifth power.

Original entry on oeis.org

0, 16, 33, 67, 135, 271, 512, 543, 1025, 1056, 1087, 2051, 2113, 2144, 2175, 4103, 4227, 4289, 4320, 4351, 8207, 8455, 8579, 8641, 8672, 8703, 16384, 16415, 16911, 17159, 17283, 17345, 17376, 17407, 32769, 32800, 32831, 33792, 33823, 34319, 34567, 34691, 34753, 34784, 34815, 65539, 65601, 65632, 65663, 67585, 67616

Offset: 1

Antti Karttunen, Oct 01 2023

nonn

Equivalently, numbers k for which A332214(k), and also A332817(k) are fifth powers.
The sequence is defined inductively as:
 (a) it contains 0 and 16, and
 (b) for any nonzero term a(n), (2*a(n)) + 1 and 32*a(n) are also included as terms.
When iterating n -> 2n+1 mod 31, starting from 16 we obtain five distinct remainders 16, 2, 5, 11, 23, before the cycle starts again from 16. (see A153893), while x^5 mod 31 may obtain only these values: 0, 1, 5, 6, 25, 26, 30. The only common element of these sets is 5. We have x^5 == 5 (mod 31) whenever x == 7, 14, 19, 25, 28 mod 31, with all other x leaving a remainder that is not in the set [16, 2, 5, 11, 23].
On the other hand, when iterating n -> 2n+1 mod 33, starting from 16 we obtain ten distinct remainders 16, 0, 1, 3, 7, 15, 31, 30, 28, 24, before the cycle starts again from 16, while x^5 mod 33 obtain only these values: 0, 1, 10, 11, 12, 21, 22, 23, 32. We have x^5 == 0 (mod 33) iff x == 0 (mod 33) and x^5 == 1 (mod 33) whenever x == 1, 4, 16, 25, 31 mod 33. In the n->2n+1 cycles of 5 and 10 elements starting from 16, the 5's (of every second cycle) in the former and the 1's in the latter are aligned with each other.
In any case, this sequence do not contain any fifth powers after the initial zero. See A365805. - Antti Karttunen, Nov 23 2023

Index entries for sequences related to binary expansion of n

Positions of multiples of 5 in A365805.
Sequence A243071(n^5), n >= 1, sorted into ascending order.
Subsequences: A013825, A198275.
Cf. A000584, A153893, A163511, A332214, A332817.
Cf. also A365801, A365808, A366287, A366391.

A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
isA365802(n) = ispower(A163511(n),5);

isA365802(n) = if(n<=16, !(n%16), if(n%2, isA365802((n-1)/2), if(n%32, 0, isA365802(n/32))));

A206371 a(n) = 31*2^n + 1.

Original entry on oeis.org

32, 63, 125, 249, 497, 993, 1985, 3969, 7937, 15873, 31745, 63489, 126977, 253953, 507905, 1015809, 2031617, 4063233, 8126465, 16252929, 32505857, 65011713, 130023425, 260046849, 520093697, 1040187393, 2080374785, 4160749569, 8321499137, 16642998273

Offset: 0

Brad Clardy, Feb 07 2012

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000079, A083686, A198275, A257548.

Magma
```
[31*2^n +1 : n in [0..35]];
```

1+31*2^Range[0,50] (* G. C. Greubel, Jan 05 2023 *)

a(n)=31<Charles R Greathouse IV, Jun 01 2015

[1+31*2^n for n in range(51)] # G. C. Greubel, Jan 05 2023

a(n) = 31*2^n + 1.
a(n) = A198275(n) + 2*A083686(n).
From G. C. Greubel, Jan 05 2023: (Start)
a(n) = 31*A000079(n) + 1.
a(n) = A257548(n) + 1, for n > 5.
G.f.: (32-33*x)/((1-x)*(1-2*x)).
E.g.f.: exp(x) + 31*exp(2*x). (End)

cp's OEIS Frontend

A036044 BCR(n): write in binary, complement, reverse.

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A075300 Array A read by antidiagonals upwards: A(n, k) = array A054582(n,k) - 1 = 2^n*(2*k+1) - 1 with n,k >= 0.

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A365802 Numbers k such that A163511(k) is a fifth power.

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A206371 a(n) = 31*2^n + 1.

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A075300 Array A read by antidiagonals upwards: A(n, k) = array A054582(n,k) - 1 = 2^n(2k+1) - 1 with n,k >= 0.