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A166911 a(n) = (9 + 14*n + 12*n^2 + 4*n^3)/3.

%I A166911 #18 Sep 08 2022 08:45:48
%S A166911 3,13,39,89,171,293,463,689,979,1341,1783,2313,2939,3669,4511,5473,
%T A166911 6563,7789,9159,10681,12363,14213,16239,18449,20851,23453,26263,29289,
%U A166911 32539,36021,39743,43713,47939,52429,57191,62233,67563,73189,79119,85361,91923
%N A166911 a(n) = (9 + 14*n + 12*n^2 + 4*n^3)/3.
%C A166911 The inverse binomial transform yields the quasi-finite sequence 3,10,16,8,0,.. (0 continued).
%C A166911 These are the bottom-left numbers in the blocks (each with 2 rows) shown in A172002, the
%C A166911 atomic number of the leftmost element in the 2nd, 4th, 6th etc. row of the Janet table.
%D A166911 Charles Janet, La structure du noyau de l'atome .., Nov 1927, page 15.
%H A166911 Vincenzo Librandi, <a href="/A166911/b166911.txt">Table of n, a(n) for n = 0..10000</a>
%H A166911 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F A166911 First differences: a(n)-a(n-1) = 2+4*n+4*n^2 = 1+(1+2n)^2 = 1 + A016754(n+1) = A069894(n+1).
%F A166911 Second differences: a(n) - 2*a(n-1) + a(n-2) = 8*n = A008590(n+2).
%F A166911 Third differences: a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 8.
%F A166911 a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F A166911 G.f.: (3 + x + 5*x^2 - x^3)/(1-x)^4.
%F A166911 a(n) = A166464(n) + 2*(n+1)^2 = A166464(n) + A001105(n+1).
%F A166911 E.g.f.: (1/3)*(9 + 30*x + 24*x^2 + 4*x^3)*exp(x). - _G. C. Greubel_, May 28 2016
%t A166911 LinearRecurrence[{4,-6,4,-1}, {3, 13, 39, 89}, 100] (* _G. C. Greubel_, May 28 2016 *)
%o A166911 (Magma) [(9+14*n+12*n^2+4*n^3)/3: n in [0..40] ]; // _Vincenzo Librandi_, Aug 06 2011
%o A166911 (PARI) a(n)=n*(4*n^2+12*n+14)/3+3 \\ _Charles R Greathouse IV_, Dec 21 2011
%K A166911 nonn,easy
%O A166911 0,1
%A A166911 _Paul Curtz_, Oct 23 2009
%E A166911 Edited and extended by _R. J. Mathar_, Mar 02 2010