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A233270 a(n) = A233271(n) - A179016(n).

%I A233270 #14 Jan 03 2014 10:43:32
%S A233270 0,0,-1,0,0,0,1,0,0,2,1,2,0,0,3,3,4,3,4,3,3,0,0,4,4,5,4,6,5,5,3,5,5,6,
%T A233270 4,5,4,4,0,0,5,8,9,10,13,13,15,16,17,18,18,17,17,19,19,17,17,18,18,17,
%U A233270 16,15,13,13,10,9,8,5,0,0,6,9,14,17,18,20,22,21
%N A233270 a(n) = A233271(n) - A179016(n).
%C A233270 For all n>=2, a(1+A213710(n)) = n-2.
%C A233270 Except for a(2)=-1 (which seems to be the only negative term in the sequence), the sequences A218600 and A213710 give the positions of zeros.
%C A233270 Furthermore, each subrange [A213710(n)..A218600(n+1)] is palindromic. A233268 gives the middle points of those ranges, the sequence A234018 gives the values at those points, while A234019 gives the maximum term in that range in this sequence.
%H A233270 Antti Karttunen, <a href="/A233270/b233270.txt">Rows 0..16, flattened</a>
%F A233270 a(n) = A233271(n) - A179016(n).
%F A233270 a(A218602(n)) = a(n). [This is just a claim that each row is palindrome]
%e A233270 This irregular table begins as:
%e A233270 0;
%e A233270 0;
%e A233270 -1;
%e A233270 0, 0;
%e A233270 0, 1, 0;
%e A233270 0, 2, 1, 2, 0;
%e A233270 0, 3, 3, 4, 3, 4, 3, 3, 0;
%e A233270 0, 4, 4, 5, 4, 6, 5, 5, 3, 5, 5, 6, 4, 5, 4, 4, 0;
%e A233270 ...
%e A233270 After zero, each row n is A213709(n-1) elements long.
%o A233270 (Scheme)
%o A233270 (define (A233270 n) (- (A233271 n) (A179016 n)))
%Y A233270 Except for a(2)=-1 (which seems to be the only negative term in the sequence), the sequences A218600 and A213710 give the positions of zeros.
%Y A233270 Cf. A080468, A179016, A233271, A233268, A233274, A234018, A234019, A234020.
%K A233270 sign,tabf
%O A233270 0,10
%A A233270 _Antti Karttunen_, Dec 14 2013