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A267654
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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.
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1
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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, 12, 10, 12, 10, 12, 10, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14, 16
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OFFSET
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0,1
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COMMENTS
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Row sums = 2, 10, 26, 50, ... = A069894(n).
0, for b(n)
0, 1, 2, for c(n)
0, 1, 2, 3, 4, for d(n)
0, 1, 2, 3, 4, 5, 6,
etc,
a(n) is used for
1) b(n+1) = b(n) + (a(0)=2) i.e. 0, 2, 4, 6, ... = A005843(n).
2) c(n+3) = c(n) + (period 3:repeat 4, 2, 4) i.e. 0, 1, 2, 4, 3, 6, 8, ... = A265667(n).
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).
Etc.
a(n) has a companion with the same terms,differently distributed,yielding permutations of the nonnegative numbers. See A265672.
a(n) other writing (by pairs):
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, 12, 10, 12, 10, 12, 10, 12,
14, 12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12,
etc.
Row sums: 12, 20, 56, 72, ... = 4*A074378(n+1).
The last term of the successive rows is the number of their terms.
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LINKS
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FORMULA
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a(2n) = 4*n + 2 times 4*n + 2 = 2, 2, 6, 6, 6, 6, 6, 6, 10,....
a(2n+1) = 4*(n+1) times 4*(n+1) = 4, 4, 4, 4, 8, 8, 8, 8, 8, 8, 8, 8, 12, ....
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EXAMPLE
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The triangle is
2,
4, 2, 4,
6, 4, 6, 4, 6,
8, 6, 8, 6, 8, 6, 8,
etc.
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MATHEMATICA
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Table[2 (n - 1) + 2 (Boole@ OddQ@ k + 1), {n, 0, 7}, {k, 2 n + 1}] // Flatten (* Michael De Vlieger, Jan 19 2016 *)
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CROSSREFS
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Cf. A007395, A005843, A008586, A016825, A053186, A069894, A074378, A086520, A168273, A168276, A265667, A265672, A265734.
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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