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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A274496 Triangle read by rows: T(n,k) is the number of binary words of length n having degree of asymmetry equal to k (n >= 0; 0 <= k <= n/2).

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%I A274496 #16 Jan 13 2018 04:49:04
%S A274496 1,2,2,2,4,4,4,8,4,8,16,8,8,24,24,8,16,48,48,16,16,64,96,64,16,32,128,
%T A274496 192,128,32,32,160,320,320,160,32,64,320,640,640,320,64,64,384,960,
%U A274496 1280,960,384,64,128,768,1920,2560,1920,768,128
%N A274496 Triangle read by rows: T(n,k) is the number of binary words of length n having degree of asymmetry equal to k (n >= 0; 0 <= k <= n/2).
%C A274496 The degree of asymmetry of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the degree of asymmetry of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).
%C A274496 A sequence is palindromic if and only if its degree of asymmetry is 0.
%C A274496 Sum_{k>=0} k*T(n,k) = A274497(n).
%F A274496 T(n,k) = 2^ceiling(n/2)*binomial(floor(n/2),k).
%F A274496 G.f.:  G(t,z) = (1 + 2z)/(1 - 2(1 + t)z^2).
%F A274496 The row generating polynomials P[n] satisfy P[n] = 2(1 + t)P[n-2] (n >= 2). Easy to see if we note that the binary words of length n (n >= 2) are 0w0, 0w1, 1w0, and 1w1, where w is a binary word of length n-2.
%e A274496 From _Andrew Howroyd_, Jan 10 2018: (Start)
%e A274496 Triangle begins:
%e A274496    1;
%e A274496    2;
%e A274496    2,   2;
%e A274496    4,   4;
%e A274496    4,   8,   4;
%e A274496    8,  16,   8;
%e A274496    8,  24,  24,   8;
%e A274496   16,  48,  48,  16;
%e A274496   16,  64,  96,  64,  16;
%e A274496   32, 128, 192, 128,  32;
%e A274496   32, 160, 320, 320, 160, 32;
%e A274496   ...
%e A274496 (End)
%e A274496 T(4,0) = 4 because we have 0000, 0110, 1001, and 1111.
%e A274496 T(4,1) = 8 because we have 0001, 0010, 0100, 1000, 0111, 1011, 1101, and 1110.
%e A274496 T(4,2) = 4 because we have 0011, 0101, 1010, and 1100.
%p A274496 T := proc(n, k) options operator, arrow: 2^ceil((1/2)*n)*binomial(floor((1/2)*n), k) end proc: for n from 0 to 15 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
%t A274496 Table[2^Ceiling[n/2] Binomial[Floor[n/2], k], {n, 0, 13}, {k, 0, n/2}] // Flatten (* _Michael De Vlieger_, Jan 11 2018 *)
%o A274496 (PARI)
%o A274496 T(n,k) = 2^ceil(n/2)*binomial(floor(n/2), k);
%o A274496 for(n=0, 10, for(k=0, n\2, print1(T(n, k), ", ")); print); \\ _Andrew Howroyd_, Jan 10 2018
%Y A274496 Cf. A274497, A274498, A274499.
%K A274496 nonn,tabf
%O A274496 0,2
%A A274496 _Emeric Deutsch_, Jul 27 2016