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A010018 a(0) = 1, a(n) = 28*n^2 + 2 for n>0.

%I A010018 #50 May 07 2024 06:05:55
%S A010018 1,30,114,254,450,702,1010,1374,1794,2270,2802,3390,4034,4734,5490,
%T A010018 6302,7170,8094,9074,10110,11202,12350,13554,14814,16130,17502,18930,
%U A010018 20414,21954,23550,25202,26910,28674,30494,32370,34302,36290,38334,40434,42590,44802
%N A010018 a(0) = 1, a(n) = 28*n^2 + 2 for n>0.
%C A010018 First bisection of A005919. - _Bruno Berselli_, Feb 07 2012
%C A010018 a(n) = the second level of difference between the sum of the terms in the n+1 X n+1 matrices and those in the n X n matrices starting at n=1 for an array constructed by using the terms in A016813 as the antidiagonals; the first few antidiagonals are 1; 5,9; 13,17,21; 25,29,33,37. - _J. M. Bergot_, Jul 05 2013
%C A010018 [More formally: (sum[m(n+1),j {j=0..n+1}]+sum[m(i,n+1) {i=0..n}]) - (sum[m(n,j) {j=0...n}] + sum[m(i,n) {i=0..n-1}])=a(n)]
%C A010018 [The first five rows begin: 1,9,21,37,57; 5,17,33,53,77; 13,29,49,73,101;25,45,69,97,129; 41,65,93,125,161]
%H A010018 Bruno Berselli, <a href="/A010018/b010018.txt">Table of n, a(n) for n = 0..1000</a>
%H A010018 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A010018 G.f.: (1+x)*(1+26*x+x^2)/(1-x)^3. - _Bruno Berselli_, Feb 07 2012
%F A010018 E.g.f.: (x*(x+1)*28+2)*e^x-1. - _Gopinath A. R._, Feb 14 2012
%F A010018 Sum_{n>=0} 1/a(n) = 3/4 + sqrt(14)/56*Pi*coth(Pi/sqrt 14) = 1.05615979263340... - _R. J. Mathar_, May 07 2024
%F A010018 a(n) = 2*A158482(n), n>0. - _R. J. Mathar_, May 07 2024
%F A010018 a(n) = A195314(n)+A195314(n+1). - _R. J. Mathar_, May 07 2024
%t A010018 Join[{1}, 28 Range[40]^2 + 2] (* _Bruno Berselli_, Feb 07 2012 *)
%t A010018 LinearRecurrence[{3, -3, 1}, {1, 30, 114, 254}, 40] (* _Robert G. Wilson v_, Jul 06 2013 *)
%Y A010018 Cf. A206399.
%K A010018 nonn,easy
%O A010018 0,2
%A A010018 _N. J. A. Sloane_