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%I #29 Sep 08 2022 08:45:49
%S 1,5,25,77,177,341,585,925,1377,1957,2681,3565,4625,5877,7337,9021,
%T 10945,13125,15577,18317,21361,24725,28425,32477,36897,41701,46905,
%U 52525,58577,65077,72041,79485,87425,95877,104857,114381,124465,135125,146377,158237,170721,183845
%N a(n) = 1 + 4*n*(1 + 2*n^2)/3.
%C Binomial transform of quasi-finite sequence 1,4,16,16,0,0,... (0 continued).
%H Vincenzo Librandi, <a href="/A171272/b171272.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F First differences: a(n+1) - a(n) = A108099(n).
%F Second differences: a(n+2) - 2*a(n+1) + a(n) = A008598(n+1).
%F Third differences: a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n) = 16.
%F a(n) = (A168574(n) + A168547(n))/2. - This formula is the link to the Janet table of the PSE.
%F G.f.: ( 1 + x + 11*x^2 + 3*x^3 ) / (x-1)^4. - _R. J. Mathar_, Jul 07 2011
%F E.g.f.: (3 +12*x +24*x^2 +8*x^3)*exp(x)/3. - _G. C. Greubel_, Nov 02 2018
%t LinearRecurrence[{4,-6,4,-1},{1,5,25,77},50] (* _Harvey P. Dale_, Nov 22 2011 *)
%o (PARI) a(n)=4*n*(1+2*n^2)/3+1 \\ _Charles R Greathouse IV_, Jul 07 2011
%o (Magma) [1+4*n*(1+2*n^2)/3: n in [0..40]]; // _Vincenzo Librandi_, Aug 05 2011
%K nonn,easy
%O 0,2
%A _Paul Curtz_, Dec 06 2009
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