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

Author

Lei Zhou, Dec 13 2013

Keywords

• nonn
• easy
• base

Examples

			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.
		

Links

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

Crossrefs

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

Programs

Mathematica

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