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A267654 Irregular triangle of palindromic subsequences. Every row has 2*n+1 terms. From the second row, there are only two alternated numbers: 2*n+4 and 2*n+2.

%I A267654 #19 Mar 04 2016 07:25:41
%S A267654 2,4,2,4,6,4,6,4,6,8,6,8,6,8,6,8,10,8,10,8,10,8,10,8,10,12,10,12,10,
%T A267654 12,10,12,10,12,10,12,14,12,14,12,14,12,14,12,14,12,14,12,14,16,14,16,
%U A267654 14,16,14,16,14,16,14,16,14,16,14,16
%N A267654 Irregular triangle of palindromic subsequences. Every row has 2*n+1 terms. From the second row, there are only two alternated numbers: 2*n+4 and 2*n+2.
%C A267654 Row sums = 2, 10, 26, 50, ... = A069894(n).
%C A267654 Starting from A053186(n) =
%C A267654 0,                            for b(n)
%C A267654 0, 1, 2,                              for c(n)
%C A267654 0, 1, 2, 3, 4,                                for d(n)
%C A267654 0, 1, 2, 3, 4, 5, 6,
%C A267654 etc,
%C A267654 a(n) is used for
%C A267654 1) b(n+1) = b(n) + (a(0)=2) i.e. 0, 2, 4, 6, ... = A005843(n).
%C A267654 2) c(n+3) = c(n) + (period 3:repeat 4, 2, 4) i.e. 0, 1, 2, 4, 3, 6, 8, ... =  A265667(n).
%C A267654 3) d(n+5) = d(n) + (period 5:repeat 6, 4, 6, 4, 6) i.e. 0, 1, 2, 3, 4, 6, 5, 8, 7, 10, ... = A265734(n).
%C A267654 Etc.
%C A267654 a(n) has a companion with the same terms,differently distributed,yielding permutations of the nonnegative numbers. See A265672.
%C A267654 a(n) other writing (by pairs):
%C A267654 2,   4,  2,  4,
%C A267654 6,   4,  6,  4,
%C A267654 6,   8,  6,  8,  6,  8,  6,  8,
%C A267654 10   8, 10,  8, 10,  8, 10,  8,
%C A267654 10, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10, 12,
%C A267654 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12,
%C A267654 etc.
%C A267654 First column: A168276(n+2). Second column: A168273(n+2).
%C A267654 Row sums: 12, 20, 56, 72, ... = 4*A074378(n+1).
%C A267654 The last term of the successive rows is the number of their terms.
%C A267654 Main diagonal: A005843(n+1).
%F A267654 a(n) = 2 * A086520(n+2).
%F A267654 a(2n) = 4*n + 2 times 4*n + 2 = 2, 2, 6, 6, 6, 6, 6, 6, 10,....
%F A267654 a(2n+1) = 4*(n+1) times 4*(n+1) = 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 12, ....
%e A267654 The triangle is
%e A267654 2,
%e A267654 4, 2, 4,
%e A267654 6, 4, 6, 4, 6,
%e A267654 8, 6, 8, 6, 8, 6, 8,
%e A267654 etc.
%t A267654 Table[2 (n - 1) + 2 (Boole@ OddQ@ k + 1), {n, 0, 7}, {k, 2 n + 1}] // Flatten (* _Michael De Vlieger_, Jan 19 2016 *)
%Y A267654 Cf. A007395, A005843, A008586, A016825, A053186, A069894, A074378, A086520, A168273, A168276, A265667, A265672, A265734.
%K A267654 nonn,tabf
%O A267654 0,1
%A A267654 _Paul Curtz_, Jan 19 2016