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%I #15 Jan 13 2023 10:37:27
%S 1,1,2,1,4,2,7,3,13,6,23,10,42,19,75,33,136,61,244,108,441,197,793,
%T 352,1431,638,2576,1145,4645,2069,8366,3721,15080,6714,27167,12087,
%U 48961,21794,88215,39254,158970,70755,286439,127469,516164,229725,930072
%N Number of palindromic compositions of n with parts in {1,2,4,6,8,10,...}.
%D S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
%H V. E. Hoggatt, Jr., and Marjorie Bicknell, <a href="http://www.fq.math.ca/Scanned/13-4/hoggatt1.pdf">Palindromic compositions</a>, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,1,0,2,0,-1).
%F G.f.: g(z) =(1+z^2 )*(1+z-z^3)/(1-z^2-2z^4+z^6). In the more general situation of compositions into a[1]<a[2]<a[3]<..., denoting F(z) = Sum(z^{a[j]},j>=1}, we have g(z)=(1+F(z))/(1-F(z^2)) (see Theorem 1.2 in the Hoggatt et al. reference).
%F a(2n) = |A078038(n)|. a(2n+1)= A028495(n). - _R. J. Mathar_, Jan 13 2023
%e a(6) = 7 because the palindromic compositions of 6 with parts in {1,2,4,6,8,...} are 6, 141, 222, 2112, 1221, 11211, and 111111.
%p g := (1+z^2)*(1+z-z^3)/(1-z^2-2*z^4+z^6): gser:= series(g,z=0,55): seq(coeff(gser,z,n),n=0..50);
%t CoefficientList[Series[(1 + x^2) (1 + x - x^3)/(1 - x^2 - 2 x^4 + x^6), {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 18 2016 *)
%t LinearRecurrence[{0,1,0,2,0,-1},{1,1,2,1,4,2},50] (* _Harvey P. Dale_, Jul 03 2021 *)
%Y Cf. A028495, A276053,
%K nonn,easy
%O 0,3
%A _Emeric Deutsch_, Aug 17 2016
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