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Antidiagonal sums in A101321.
3

%I #46 Jan 11 2019 11:40:45

%S 1,2,4,9,20,41,77,134,219,340,506,727,1014,1379,1835,2396,3077,3894,

%T 4864,6005,7336,8877,10649,12674,14975,17576,20502,23779,27434,31495,

%U 35991,40952,46409,52394,58940,66081,73852,82289,91429,101310,111971

%N Antidiagonal sums in A101321.

%C Equals binomial transform of [1, 1, 1, 2, 1, 0, 0, 0, ...]. Example: a(5) = 20 = [1, 1, 1, 2, 1] dot [1, 4, 6, 4, 1] = (1 + 4 + 6 + 8 + 1). - _Gary W. Adamson_, Aug 25 2010

%H Vincenzo Librandi, <a href="/A101338/b101338.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5, -10, 10, -5, 1).

%F a(n) = n^4/24 + n^3/12 - n^2/24 + 11*n/12 + 1.

%F G.f.: (1-3*x+4*x^2-x^3)/(1-x)^5. - _Colin Barker_, Mar 22 2012

%F a(0)=1, a(1)=2, a(2)=4, a(3)=9, a(4)=20, a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - _Harvey P. Dale_, May 21 2013

%F a(n) = A000127(n+1) + A000292(n-2). - _Bruce J. Nicholson_, Jan 06 2019

%t CoefficientList[Series[(1-3*x+4*x^2-x^3)/(1-x)^5,{x,0,40}],x] (* _Vincenzo Librandi_, Mar 24 2012 *)

%t LinearRecurrence[{5,-10,10,-5,1},{1,2,4,9,20},50] (* _Harvey P. Dale_, May 21 2013 *)

%Y Cf. A000127, A000292.

%K nonn,easy

%O 0,2

%A Eugene McDonnell (eemcd(AT)mac.com), Dec 24 2004