A233573 Number of ways n can be partitioned as A233010(i)+A233572(j), where i,j >= 1.

1, 1, 1, 2, 1, 1, 2, 2, 1, 4, 2, 2, 3, 2, 1, 3, 2, 1, 4, 2, 2, 4, 2, 2, 3, 2, 1, 6, 3, 4, 5, 2, 3, 5, 3, 3, 7, 1, 3, 4, 2, 3, 4, 1, 2, 5, 2, 3, 5, 1, 2, 2, 2, 2, 6, 1, 4, 3, 3, 3, 6, 4, 2, 7, 3, 3, 4, 4, 3, 5, 3, 2, 6, 3, 1, 4, 3, 1, 2, 4, 2, 9, 4, 4, 7, 3, 2

Offset: 0

Author

Lei Zhou, Dec 13 2013

Keywords

• nonn
• easy
• base

Comments

Number in sequence A233010 takes palindrome form when written in balanced ternary notation and left patch same number of zeros as number of trailing zeros, which means that if we slice the balanced ternary notation number with left zero padding in the middle, the left part is a mirrored image of the right part. This is a feature like an even function.

Number in sequence A233571 takes "reversed palindrome form" when written in balanced ternary notation and left patch same number of zeros as number of trailing zeros, which means that if we slice the balanced ternary notation number with left zero padding in the middle, the left part is a mirrored image of the right part with sign changes. This is a feature like an odd function.

This sequence gives the number of possibilities of such partitions.

In the first 10000 terms, 152 zeros found.

Examples

			0=0+0=A233010(1)+A233572(1). This is the only valid partition by definition. So a(0)=1.
3=3+0=A233010(3)+A233572(1), as well 3=1+2=A233010(2)+A233572(2).
  Two valid partitions found. So a(3)=2.
9=9+0=7+2=3+6=1+8, four valid partitions found. So a(9)=4.
		

Links

• Lei Zhou, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A002113, A061917, A006995, A057890, A134027, A233010, A233571, A233572

Programs

Mathematica

BTDigits[m_Integer, g_] :=
(*This is to determine digits of a number in balanced ternary notation.*)
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];
BTpaleQ[n_Integer] :=
(*This is to query if a number is an element of sequence A233010.*)
Module[{t, trim = n/3^IntegerExponent[n, 3]},
  t = BTDigits[trim, {0}]; t == Reverse[t]];
BTrteQ[n_Integer] :=
(*This is to query if a number is an element of sequence A233572.*)
Module[{t, trim = n/3^IntegerExponent[n, 3]},
  t = BTDigits[trim, {0}]; DeleteDuplicates[t + Reverse[t]] == {0}];
sa = Select[Range[0, 11000], BTpaleQ[#] &];
(*This is to generate a limited list of A233010.*)
sb = Select[Range[0, 11000], BTrteQ[#] &];
(*This is to generate a limited list of A233572.*)
range = 86; Table[ct = 0; i1 = 0;
While[i1++; sa[[i1]] <= n, i2 = 0;
  While[i2++; (sa[[i1]] + sb[[i2]]) <= n,
   If[(sa[[i1]] + sb[[i2]]) == n, ct++]]]; ct, {n, 0, range}]