A057890 -id:A057890 - cp's OEIS frontend

A233571 In balanced ternary notation, reverse digits of a(n) equals to -a(n).

0, 2, 8, 20, 26, 32, 56, 80, 104, 146, 164, 182, 224, 242, 260, 302, 320, 338, 416, 488, 560, 656, 728, 800, 896, 968, 1040, 1172, 1226, 1280, 1406, 1460, 1514, 1640, 1694, 1748, 1898, 1952, 2006, 2132, 2186, 2240, 2366, 2420, 2474, 2624, 2678, 2732, 2858

Offset: 1

Lei Zhou, Dec 13 2013

			In balanced ternary notation, 8=(10T)_bt, where we use T to represent -1.  Reverse digits of (10T)_bt is (T01)_bt = -8. So 8 is in this sequence.
Similarly, 2240 = (1001T00T)_bt, whose reverse digits is (T00T1001)_bt = -2240.  So 2240 is in this sequence.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A002113, A061917, A006995, A057890, A134027, A233010

BTDigits[m_Integer, g_] :=
Module[{n = m, d, sign, t = g},
  If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n];
   d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];
   While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign;
   t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
ct = 1; n = 0; m = 0; dg = 0; switch = 1; res = {0}; While[ct < 50, n++;
  bits = BTDigits[n, {0}]; If[lb = Length[bits]; lb > dg,
  If[switch == 0, n = m; switch = 1; OK = 0, dg = lb; m = n - 1;
   switch = 0; OK = 1], OK = 1];  If[OK == 1, rbt = -Reverse[bits];
  If[switch == 1, nb = Join[bits, {0}], nb = bits];
  nb = Join[nb, rbt]; nb = Reverse[nb]; data = 0;
  Do[data = data + 3^(i - 1)*nb[[i]], {i, 1, Length[nb]}]; ct++;
  AppendTo[res, data]]]; res

A233572 In balanced ternary notation, if prepending same numbers of zeros, reverse digits of a(n) equals to -a(n).

Original entry on oeis.org

0, 2, 6, 8, 18, 20, 24, 26, 32, 54, 56, 60, 72, 78, 80, 96, 104, 146, 162, 164, 168, 180, 182, 216, 224, 234, 240, 242, 260, 288, 302, 312, 320, 338, 416, 438, 486, 488, 492, 504, 540, 546, 560, 648, 656, 672, 702, 720, 726, 728, 780, 800, 864, 896, 906, 936

Offset: 1

Lei Zhou, Dec 13 2013

A233571 is a subset of this sequence.

			In balanced ternary notation, 18 = (1T00)_bt, where we use T to represent -1.  Patching two zeros before it, (1T00)_bt=(001T00)_bt.  The reverse digits of (001T00)_bt is (00T100)_bt = -18.  So 18 is in this sequence.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A002113, A061917, A006995, A057890, A134027, A233010, A233571

BTDigits[m_Integer, g_] :=
Module[{n = m, d, sign, t = g},
  If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n];
   d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];
   While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign;
   t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
BTrteQ[n_Integer] :=
Module[{t, trim = n/3^IntegerExponent[n, 3]},
  t = BTDigits[trim, {0}]; DeleteDuplicates[t + Reverse[t]] == {0}];
sb = Select[Range[0, 950], BTrteQ[#] &]

A273245 Non-palindromic binary numbers whose reversal is a palindrome.

Original entry on oeis.org

10, 100, 110, 1000, 1010, 1100, 1110, 10000, 10010, 10100, 11000, 11100, 11110, 100000, 100010, 100100, 101000, 101010, 110000, 110110, 111000, 111100, 111110, 1000000, 1000010, 1000100, 1001000, 1010000, 1010100, 1011010, 1100000, 1100110, 1101100, 1110000, 1111000, 1111100

Offset: 1

Giovanni Teofilatto, May 18 2016

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A006995, A273329, A272670 (this sequence written in base 10), A057890 (and divided by 2).

# see A272670 for the other subroutines
for n from 1 to 400 do
    if isA272670(n) then
        printf("%d,",A007088(n)) ;
    end if;
end do: # R. J. Mathar, May 20 2016

A273245_list = [int(m) for m in (bin(n)[2:] for n in range(1,10**4)) if m != m[::-1] and m.rstrip('0') == m[::-1].lstrip('0')] # Chai Wah Wu, May 21 2016

Consists of binary palindromes (A006995) times nontrivial powers of 2, that is, 2^i, i>0. - N. J. A. Sloane, May 19 2016

a(n) = A007088(A272670(n)). - R. J. Mathar, May 20 2016

A272670 Numbers whose binary expansion is not palindromic but which when reversed and leading zeros omitted, does form a palindrome.

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 30, 32, 34, 36, 40, 42, 48, 54, 56, 60, 62, 64, 66, 68, 72, 80, 84, 90, 96, 102, 108, 112, 120, 124, 126, 128, 130, 132, 136, 144, 146, 160, 168, 170, 180, 186, 192, 198, 204, 214, 216, 224, 238, 240, 248, 252, 254

Offset: 1

N. J. A. Sloane, May 19 2016

Decimal interpretation of A273245. Twice A057890.

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Cf. A006995, A057890, A273245, A273329, subsequence of A154809.

isPal := proc(L::list)
    local i;
    for i from 1 to nops(L)/2 do
        if op(i,L) <> op(-i,L) then
            return false;
        end if;
    end do:
    true ;
end proc:
isA272670 := proc(n)
    local bdgs ;
    bdgs := convert(n,base,2) ;
    if isPal(bdgs) then
        return false;
    else
        A000265(n) ;
        bdgs := convert(%,base,2) ;
        isPal(bdgs) ;
    end if;
end proc:
for n from 1 to 500 do
    if isA272670(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, May 20 2016

A272670_list = [n for n in range(1,10**4) if bin(n)[2:] != bin(n)[:1:-1] and bin(n)[2:].rstrip('0') == bin(n)[:1:-1].lstrip('0')] # Chai Wah Wu, May 21 2016

A274036 a(n) is the thickness of n (see Comments section for definition).

Original entry on oeis.org

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

Offset: 0

Gheorghe Coserea, Jun 07 2016

Let b_k..b_0 be the binary representation of n and B_n(x) = b_k*x^k + .. + b_0 the associated polynomial with n = B_n(2); we define the thickness of n to be the thickness of B_n, i.e., the magnitude of the largest coefficient in the expansion of B_n(x)^2 (see A169950).
The thickness histogram for numbers in the interval I_n = [2^n, 2^(n+1)-1] is given by row n of triangle A169950, i.e., A169950(n,k) = card {p, p in I_n and a(p) = k}.
In general a(n) <= A000120(n), with equality only if in base-2 n becomes a palindrome after trailing 0's (if any) are omitted, i.e., n = A057890(k) for some k; the number of such numbers in I_n having binary weight (and thickness) w is given by A207974(n,w-1), i.e., A207974(n,w-1) = card {k, A057890(k) in I_n and A000120(A057890(k)) = w}; the total number of these numbers in the interval I_n is given by A027383(n), i.e., card {p, p in I_n and a(p) = A000120(p)} = A027383(n) = 2^floor((n+2)/2) + 2^floor((n+1)/2) - 2.

			For n = 3 we have the base-2 representation 11, the associated polynomial B_3(x) = x + 1, B_3(x)^2 = x^2 + 2*x + 1 and the magnitude of the largest coefficient in the expansion of B_3(x)^2 is 2, therefore a(3) = 2.
For n = 4 we have the base-2 representation 100, the associated polynomial B_4(x) = x^2, B_4(x)^2 = x^4 and the magnitude of the largest coefficient in the expansion of B_4(x)^2 is 1, therefore a(4) = 1.

Gheorghe Coserea, Table of n, a(n) for n = 0..32768

Cf. A000120, A027383, A057890, A169950, A207974.

Table[Max@ CoefficientList[#, x] &[SeriesData[x, 0, #, 0, Length@ #, 1]^2] &@ Reverse@ IntegerDigits[n, 2], {n, 120}] (* Michael De Vlieger, Jun 08 2016 *)

a(n) = my(pol = Pol(binary(n))); return(vecmax(Vec(sqr(pol))));
concat(0, vector(100, n, a(n)))

bitrev(n) = subst(Pol(Vecrev(binary(n>>valuation(n,2))), 'x), 'x, 2);
a(n) = {
  my(e = logint(n, 2), r = bitrev(n) << e, v = vector(2*e+1));
  for (i = 1, #v, v[i] = hammingweight(bitand(r, n)); r >>= 1);
  return(vecmax(v));
};
concat(0, vector(100, n, a(n)))

a(n) <= A000120(n), with equality iff n = A057890(k).

A342121 a(n) is the remainder when the larger of n and its binary reverse is divided by the smaller.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 6, 0, 0, 9, 6, 0, 6, 4, 0, 0, 6, 0, 0, 0, 0, 0, 14, 0, 4, 13, 18, 0, 4, 0, 10, 5, 0, 17, 14, 0, 14, 12, 0, 8, 10, 0, 4, 0, 18, 12, 4, 0, 14, 0, 0, 0, 0, 0, 30, 0, 12, 21, 42, 0, 0, 33, 30, 1, 12, 21, 42, 0

Offset: 1

Rémy Sigrist, Feb 28 2021

The binary reverse of a number is given by A030101.

This sequence is the analog of A061467 for the binary base.

			For n = 43,
- the binary reverse of 43 ("101011" in binary) is 53 ("110101" in binary),
- so a(43) = 53 mod 43 = 10.

Rémy Sigrist, Table of n, a(n) for n = 1..8192
Index entries for sequences related to binary expansion of n

Cf. A030101, A057890, A061467, A342122, A342123..

rbr[n_]:=Module[{r=IntegerReverse[n,2]},If[r>n,Mod[r,n],Mod[n,r]]]; Array[rbr,100] (* Harvey P. Dale, Mar 18 2023 *)

a(n, base=2) = { my (r=fromdigits(Vecrev(digits(n, base)), base)); max(n, r) % min(n, r) }

def A342121(n):
    a, b = sorted([n,int(bin(n)[:1:-1],2)])
    return b % a if n > 0 else 0 # Chai Wah Wu, Mar 01 2021

a(n) = max(n, A030101(n)) mod min(n, A030101(n)).
a(n) = min(A342122(n), A342123(n)).
a(n) < n.
a(n) = 0 iff n belongs to A057890.

A342123 a(n) is the remainder when n is divided by its binary reverse.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 2, 0, 0, 0, 0, 0, 19, 0, 0, 9, 23, 0, 6, 4, 0, 0, 6, 0, 0, 0, 0, 0, 35, 0, 37, 13, 39, 0, 4, 0, 43, 5, 0, 17, 47, 0, 14, 12, 0, 8, 10, 0, 55, 0, 18, 12, 4, 0, 14, 0, 0, 0, 0, 0, 67, 0, 69, 21, 71, 0, 0, 33, 75, 1, 77, 21

Offset: 1

Rémy Sigrist, Feb 28 2021

The binary reverse of a number is given by A030101.

This sequence is the analog of A071955 for the binary base.

			For n = 43,
- the binary reverse of 43 ("101011" in binary) is 53 ("110101" in binary),
- so a(43) = 43 mod 53 = 43.

Rémy Sigrist, Table of n, a(n) for n = 1..8192
Index entries for sequences related to binary expansion of n

Cf. A030101, A057890, A071955, A161601, A342121, A342122.

a(n, base=2) = { my (r=fromdigits(Vecrev(digits(n, base)), base)); n%r }

def A342123(n): return n % int(bin(n)[:1:-1],2) if n > 0 else 0 # Chai Wah Wu, Mar 01 2021

a(n) = n mod A030101(n).
a(n) <= n with equality iff n belongs to A161601.
a(n) = 0 iff n belongs to A057890.

A366378 Numbers k such that A057889(k) == k (mod 3), where A057889 is the bijective bit-reverse.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 36, 38, 39, 40, 42, 45, 46, 48, 50, 51, 54, 56, 57, 58, 60, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 75, 76, 77, 78, 79, 80, 81, 83, 84, 85, 87, 89, 90, 91, 92, 93, 95, 96, 97, 99, 100, 101, 102, 103

Offset: 1

Antti Karttunen, Oct 22 2023

nonn

Cf. A030101, A057889, A366379 (complement), A366389.

Subsequences: A000079, A008585, A057890.

A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));
A057889(n) = if(!n,n,A030101(n/(2^valuation(n,2))) * (2^valuation(n, 2)));
isA366378(n) = !((A057889(n)-n)%3);

A161604 A positive integer k is included if the value of (the reversal of k's representation in binary) divides k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 27, 28, 30, 31, 32, 33, 34, 36, 40, 42, 45, 48, 51, 54, 56, 60, 62, 63, 64, 65, 66, 68, 72, 73, 80, 84, 85, 90, 93, 96, 99, 102, 107, 108, 112, 119, 120, 124, 126, 127, 128, 129, 130, 132, 136, 144, 146, 153

Offset: 1

Leroy Quet, Jun 14 2009

By "reversal" of k's representation in binary, it is meant: write k in binary, reverse the order of its digits, and read the result as a binary value.
It seems (verified for the first 120000 entries) that a(n) = A057890(n+1). - R. J. Mathar, Jun 18 2009
Indeed, this is A057890 (palindromes with optional trailing zeros) without the initial term. In other bases this does not have to be so, as illustrated by A071687. - Ivan Neretin, Sep 04 2015

			24 in binary is 11000. The reversal of this is 11 (ignoring leading 0's), which is 3 in decimal. Since 3 divides 24, then 24 is included in this sequence.

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Cf. A030101, A057890, A071687.

A030101 := proc(n) local bdgs ; bdgs := convert(n,base,2) ; add( op(-i,bdgs)*2^(i-1),i=1..nops(bdgs)) ; end: isA161604 := proc(n) if ( n mod A030101(n) ) = 0 then true ; else false; fi; end: for n from 1 to 600 do if isA161604(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Jun 18 2009

Select[Range@ 153, Divisible[#, FromDigits[Reverse@ IntegerDigits[#, 2], 2]] &] (* Michael De Vlieger, Sep 04 2015 *)

is(n, base=2) = my (r=fromdigits(Vecrev(digits(n, base)), base)); n%r==0 \\ Rémy Sigrist, Apr 04 2020

Extended by R. J. Mathar, Jun 18 2009

A207974 Triangle related to A152198.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 5, 2, 4, 1, 1, 1, 6, 3, 6, 3, 2, 1, 1, 7, 3, 9, 3, 5, 1, 1, 1, 8, 4, 12, 6, 8, 4, 2, 1, 1, 9, 4, 16, 6, 14, 4, 6, 1, 1, 1, 10, 5, 20, 10, 20, 10, 10, 5, 2, 1

Offset: 0

Philippe Deléham, Feb 22 2012

Row sums are A027383(n).
Diagonal sums are alternately A014739(n) and A001911(n+1).
The matrix inverse starts
1;
-1,1;
1,-2,1;
1,-1,-1,1;
-1,2,0,-2,1;
-1,1,2,-2,-1,1;
1,-2,-1,4,-1,-2,1;
1,-1,-3,3,3,-3,-1,1;
-1,2,2,-6,0,6,-2,-2,1;
-1,1,4,-4,-6,6,4,-4,-1,1;
1,-2,-3,8,2,-12,2,8,-3,-2,1;
apparently related to A158854. - R. J. Mathar, Apr 08 2013
From Gheorghe Coserea, Jun 11 2016: (Start)
T(n,k) is the number of terms of the sequence A057890 in the interval [2^n,2^(n+1)-1] having binary weight k+1.
T(n,k) = A007318(n,k) (mod 2) and the number of odd terms in row n of the triangle is 2^A000120(n).
(End)

			Triangle begins :
n\k  [0] [1] [2] [3] [4] [5] [6] [7] [8] [9]
[0]  1;
[1]  1,  1;
[2]  1,  2,  1;
[3]  1,  3,  1,  1;
[4]  1,  4,  2,  2,  1;
[5]  1,  5,  2,  4,  1,  1;
[6]  1,  6,  3,  6,  3,  2,  1;
[7]  1,  7,  3,  9,  3,  5,  1,  1;
[8]  1,  8,  4,  12, 6,  8,  4,  2,  1;
[9]  1,  9,  4,  16, 6,  14, 4,  6,  1,  1;
[10] ...

Gheorghe Coserea, Rows n = 0..200, flattened

Cf. Columns : A000012, A000027, A004526, A002620, A008805, A006918, A058187
Cf. Diagonals : A000012, A000034, A052938, A097362
Cf. A007318, A110813, A135278, A152201
Related to thickness: A000120, A027383, A057890, A274036.

A207974 := proc(n,k)
    if k = 0 then
        1;
    elif k < 0 or k > n then
        0 ;
    else
        procname(n-1,k-1)-(-1)^k*procname(n-1,k) ;
    end if;
end proc: # R. J. Mathar, Apr 08 2013

seq(N) = {
  my(t = vector(N+1, n, vector(n, k, k==1 || k == n)));
  for(n = 2, N+1, for (k = 2, n-1,
      t[n][k] = t[n-1][k-1] + (-1)^(k%2)*t[n-1][k]));
  return(t);
};
concat(seq(10))  \\ Gheorghe Coserea, Jun 09 2016

P(n) = ((2+x+(n%2)*x^2) * (1+x^2)^(n\2) - 2)/x;
concat(vector(11, n, Vecrev(P(n-1)))) \\ Gheorghe Coserea, Mar 14 2017

T(n,k) = T(n-1,k-1) - (-1)^k*T(n-1,k), k>0 ; T(n,0) = 1.
T(2n,2k) = T(2n+1,2k) = binomial(n,k) = A007318(n,k).
T(2n+1,2k+1) = A110813(n,k).
T(2n+2,2k+1) = 2*A135278(n,k).
T(n,2k) + T(n,2k+1) = A152201(n,k).
T(n,2k) = A152198(n,k).
T(n+1,2k+1) = A152201(n,k).
T(n,k) = T(n-2,k-2) + T(n-2,k).
T(2n,n) = A128014(n+1).
T(n,k) = card {p, 2^n <= A057890(p) <= 2^(n+1)-1 and A000120(A057890(p)) = k+1}. - Gheorghe Coserea, Jun 09 2016
P_n(x) = Sum_{k=0..n} T(n,k)*x^k = ((2+x+(n mod 2)*x^2)*(1+x^2)^(n\2) - 2)/x. - Gheorghe Coserea, Mar 14 2017

cp's OEIS Frontend

A233571 In balanced ternary notation, reverse digits of a(n) equals to -a(n).

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A233572 In balanced ternary notation, if prepending same numbers of zeros, reverse digits of a(n) equals to -a(n).

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A273245 Non-palindromic binary numbers whose reversal is a palindrome.

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A272670 Numbers whose binary expansion is not palindromic but which when reversed and leading zeros omitted, does form a palindrome.

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A274036 a(n) is the thickness of n (see Comments section for definition).

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A342121 a(n) is the remainder when the larger of n and its binary reverse is divided by the smaller.

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A342123 a(n) is the remainder when n is divided by its binary reverse.

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A366378 Numbers k such that A057889(k) == k (mod 3), where A057889 is the bijective bit-reverse.

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