A276055 Number of palindromic compositions of n with parts in {1,2,4,6,8,10,...}.

1, 1, 2, 1, 4, 2, 7, 3, 13, 6, 23, 10, 42, 19, 75, 33, 136, 61, 244, 108, 441, 197, 793, 352, 1431, 638, 2576, 1145, 4645, 2069, 8366, 3721, 15080, 6714, 27167, 12087, 48961, 21794, 88215, 39254, 158970, 70755, 286439, 127469, 516164, 229725, 930072

Offset: 0

Author

Emeric Deutsch, Aug 17 2016

Keywords

• nonn
• easy

Examples

			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.
		

References

• S. Heubach and T. Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.

Links

• V. E. Hoggatt, Jr. and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.
• Index entries for linear recurrences with constant coefficients, signature (0,1,0,2,0,-1).

Crossrefs

Cf. A028495, A276053,

Programs

Maple

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);

Mathematica

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 *)
LinearRecurrence[{0,1,0,2,0,-1},{1,1,2,1,4,2},50] (* Harvey P. Dale, Jul 03 2021 *)

Formula

G.f.: g(z) = (1+z^2)*(1+z-z^3)/(1-z^2-2*z^4+z^6). In the more general situation of compositions into a[1]=1} z^(a[j]), we have g(z) = (1+F(z))/(1-F(z^2)) (see Theorem 1.2 in the Hoggatt et al. reference).

a(2*n) = |A078038(n)|. a(2*n+1) = A028495(n). - R. J. Mathar, Jan 13 2023