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

%I A233573 #13 Jan 10 2025 06:08:45
%S A233573 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,
%T A233573 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,
%U A233573 3,5,3,2,6,3,1,4,3,1,2,4,2,9,4,4,7,3,2
%N A233573 Number of ways n can be partitioned as A233010(i)+A233572(j), where i,j >= 1.
%C A233573 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.
%C A233573 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.
%C A233573 This sequence gives the number of possibilities of such partitions.
%C A233573 In the first 10000 terms, 152 zeros found.
%H A233573 Lei Zhou, <a href="/A233573/b233573.txt">Table of n, a(n) for n = 0..10000</a>
%e A233573 0=0+0=A233010(1)+A233572(1). This is the only valid partition by definition. So a(0)=1.
%e A233573 3=3+0=A233010(3)+A233572(1), as well 3=1+2=A233010(2)+A233572(2).
%e A233573   Two valid partitions found. So a(3)=2.
%e A233573 9=9+0=7+2=3+6=1+8, four valid partitions found. So a(9)=4.
%t A233573 BTDigits[m_Integer, g_] :=
%t A233573 (*This is to determine digits of a number in balanced ternary notation.*)
%t A233573 Module[{n = m, d, sign, t = g},
%t A233573   If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n];
%t A233573    d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];
%t A233573    While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign;
%t A233573      t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
%t A233573 BTpaleQ[n_Integer] :=
%t A233573 (*This is to query if a number is an element of sequence A233010.*)
%t A233573 Module[{t, trim = n/3^IntegerExponent[n, 3]},
%t A233573   t = BTDigits[trim, {0}]; t == Reverse[t]];
%t A233573 BTrteQ[n_Integer] :=
%t A233573 (*This is to query if a number is an element of sequence A233572.*)
%t A233573 Module[{t, trim = n/3^IntegerExponent[n, 3]},
%t A233573   t = BTDigits[trim, {0}]; DeleteDuplicates[t + Reverse[t]] == {0}];
%t A233573 sa = Select[Range[0, 11000], BTpaleQ[#] &];
%t A233573 (*This is to generate a limited list of A233010.*)
%t A233573 sb = Select[Range[0, 11000], BTrteQ[#] &];
%t A233573 (*This is to generate a limited list of A233572.*)
%t A233573 range = 86; Table[ct = 0; i1 = 0;
%t A233573 While[i1++; sa[[i1]] <= n, i2 = 0;
%t A233573   While[i2++; (sa[[i1]] + sb[[i2]]) <= n,
%t A233573    If[(sa[[i1]] + sb[[i2]]) == n, ct++]]]; ct, {n, 0, range}]
%Y A233573 Cf. A002113, A061917, A006995, A057890, A134027, A233010, A233571, A233572
%K A233573 nonn,easy,base
%O A233573 0,4
%A A233573 _Lei Zhou_, Dec 13 2013