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%I #23 Jul 28 2020 11:10:44
%S 1,3,11,39,131,421,1309,3971,11823,34691,100611,289033,823801,2332419,
%T 6566291,18394911,51310979,142587181,394905493,1090444931,3002921271,
%U 8249479163,22612505091,61857842449,168903452401,460409998851
%N Expansion of (1 - 4*x + 7*x^2 - 4*x^3 + x^4)/(1 - 7*x + 17*x^2 - 17*x^3 + 7*x^4 - x^5).
%C The diagonal sums of number triangle A166335 are 1, 0, 3, 0, 11, 0, ...
%H Vincenzo Librandi, <a href="/A166336/b166336.txt">Table of n, a(n) for n = 0..1000</a>
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (7,-17,17,-7,1).
%F G.f.: (1 - 4*x + 7*x^2 - 4*x^3 + x^4)/((1 - x)*(1 - 3*x + x^2)^2);
%F a(n) = 1 + 2*Sum{k=0..n} k*C(n + k, 2*k) = 1 + 2*Sum{k=0..n} (n-k)*C(2*n - k, k) = 1 + 2*A001870(n).
%F a(0) = 1, a(1) = 3, a(2) = 11, a(3) = 39, a(4) = 131, and a(n) = -17*a(n-1) + 17*a(n-2) - 7*a(n-3) + a(n-4) for n >= 4. - _Harvey P. Dale_, Jul 05 2014
%t CoefficientList[Series[(1-4x+7x^2-4x^3+x^4)/(1-7x+17x^2-17x^3+7x^4-x^5),{x,0,30}],x] (* or *) LinearRecurrence[{7,-17,17,-7,1},{1,3,11,39,131},30] (* _Harvey P. Dale_, Jul 05 2014 *)
%Y Cf. A001870, A166335.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Oct 12 2009