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%I #26 Sep 08 2022 08:45:47
%S 3,11,27,54,95,153,231,332,459,615,803,1026,1287,1589,1935,2328,2771,
%T 3267,3819,4430,5103,5841,6647,7524,8475,9503,10611,11802,13079,14445,
%U 15903,17456,19107,20859,22715,24678,26751,28937,31239,33660,36203,38871
%N a(n) = (6 + 10*n + 5*n^2 + n^3)/2.
%C Row sums of the triangle defined by non-interrupted runs in A080036.
%C If the sequence of integers is split at positions defined by A000124 we obtain A080036. Its runs of consecutive integers can be placed into rows of a triangle:
%C 3;
%C 5, 6;
%C 8, 9, 10;
%C 12, 13, 14, 15;
%C 17, 18, 19, 20, 21;
%C ...
%C The a(n) are the row sums of this triangle.
%C The a(n) are also the binomial transform of the quasi-finite sequence 3, 8, 8, 3, 0 (0 continued).
%C An associated integer sequence could be defined by a(n)/A026741(n+1) = 3, 11, 9, 27, ...
%H Vincenzo Librandi, <a href="/A164845/b164845.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) = A162607(n+3) + n.
%F First differences: a(n+1) - a(n) = A104249(n+2), i.e., a(n) = a(n-1) + 3*n^2/2 + 7*n/2 +3.
%F Second differences: a(n+2) - 2*a(n+1) + a(n) = A016789(n+2).
%F a(n) = 2*a(n-1) - a(n-2) + 3*n + 5, n>1.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3, n>2.
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
%F G.f.: (3-x+x^2)/(x-1)^4.
%F E.g.f.: (6 + 16*x + 8*x^2 + x^3)*exp(x)/2. - _G. C. Greubel_, Apr 21 2018
%t Table[(6 + 10*n + 5*n^2 + n^3)/2, {n,0,50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {3, 11, 27, 54}, 50] (* _G. C. Greubel_, Apr 21 2018 *)
%o (Magma) [3+5*n+5*n^2/2+n^3/2: n in [0..50]]; // _Vincenzo Librandi_, Aug 07 2011
%o (PARI) for(n=0, 50, print1((6+10*n+5*n^2+n^3)/2, ", ")) \\ _G. C. Greubel_, Apr 21 2018
%Y Cf. A135278.
%K nonn,easy
%O 0,1
%A _Paul Curtz_, Aug 28 2009
%E Edited and extended by _R. J. Mathar_, Aug 31 2009
%E Corrected typo in recurrence, observed by _Paul Curtz_ - _R. J. Mathar_, Sep 25 2009
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