%I #36 Sep 08 2022 08:45:17

%S 1,0,1,1,2,2,3,4,6,8,11,15,21,29,40,55,76,105,145,200,276,381,526,726,

%T 1002,1383,1909,2635,3637,5020,6929,9564,13201,18221,25150,34714,

%U 47915,66136,91286,126000,173915,240051,331337,457337,631252,871303,1202640

%N Expansion of (1 - x + x^2)/(1 - x - x^4).

%C Diagonal sums of A103631.

%C The Kn11 sums, see A180662, of triangle A065941 equal the terms of this sequence without a(0) and a(1). - _Johannes W. Meijer_, Aug 11 2011

%C For n >= 2, a(n) is the number of palindromic compositions of n-2 with parts in {2,1,3,5,7,9,...}. The generating function follows easily from Theorem 1,2 of the Hoggatt et al. reference. Example: a(9) = 8 because we have 7, 151, 11311, 232, 313, 12121, 21112, and 1111111. - _Emeric Deutsch_, Aug 17 2016

%C Essentially the same as A003411. - _Georg Fischer_, Oct 07 2018

%H Muniru A Asiru, <a href="/A103632/b103632.txt">Table of n, a(n) for n = 0..1000</a>

%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_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1).

%F G.f.: (1 - x + x^2)/(1 - x - x^4).

%F a(n) = a(n-1) + a(n-4) with a(0)=1, a(1)=0, a(2)=1 and a(3)=1.

%F a(n) = Sum_{k=0..floor(n/2)} binomial(floor((2*n-3*k-1)/2), n-2*k).

%F a(n) = A003269(n+1) - A003269(n-4), n > 4.

%p A103632 := proc(n): add( binomial(floor((2*n-3*k-1)/2), n-2*k), k=0..floor(n/2)) end: seq(A103632(n), n=0..46); # _Johannes W. Meijer_, Aug 11 2011

%t LinearRecurrence[{1,0,0,1}, {1,0,1,1}, 50] (* _G. C. Greubel_, Mar 10 2019 *)

%o (GAP) a:=[1,0,1,1];; for n in [5..50] do a[n]:=a[n-1]+a[n-4]; od; a; # _Muniru A Asiru_, Oct 07 2018

%o (PARI) my(x='x+O('x^50)); Vec((1-x+x^2)/(1-x-x^4)) \\ _G. C. Greubel_, Mar 10 2019

%o (Magma) I:=[1,0,1,1]; [n le 4 select I[n] else Self(n-1) + Self(n-4): n in [1..50]]; // _G. C. Greubel_, Mar 10 2019

%o (Sage) ((1-x+x^2)/(1-x-x^4)).series(x, 50).coefficients(x, sparse=False) # _G. C. Greubel_, Mar 10 2019

%Y Cf. A275446.

%K easy,nonn

%O 0,5

%A _Paul Barry_, Feb 11 2005

%E Formula corrected by _Johannes W. Meijer_, Aug 11 2011