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A036799 a(n) = 2 + 2^(n+1)*(n-1).

%I A036799 #91 Jul 15 2023 06:24:55
%S A036799 0,2,10,34,98,258,642,1538,3586,8194,18434,40962,90114,196610,425986,
%T A036799 917506,1966082,4194306,8912898,18874370,39845890,83886082,176160770,
%U A036799 369098754,771751938,1610612738,3355443202,6979321858,14495514626,30064771074,62277025794
%N A036799 a(n) = 2 + 2^(n+1)*(n-1).
%C A036799 This sequence is a part of a class of sequences of the type: a(n) = Sum_{i=0..n} (C^i)*(i^k). This sequence has C=2, k=1. Sequence A036800 has C=2, k=2. Suppose C >= 2, k >= 1 are integers. What is the general closed form for a(n)? - _Ctibor O. Zizka_, Feb 07 2008
%C A036799 Partial sums of A036289. - _Vladimir Joseph Stephan Orlovsky_, Jul 09 2011
%C A036799 a(n) is the number of swaps needed in the worst case, when successively inserting 2^(n+1) - 1 keys into an initially empty binary heap (thus creating a tree with n+1 full levels). - _Rudy van Vliet_, Nov 09 2015
%C A036799 a(n) is also the total path length of the complete binary tree of height n, with nodes at depths 0,...,n. Total path length is defined to be the sum of depths over all nodes. - _F. Skerman_, Jul 02 2017
%C A036799 For n >= 1, every number greater than or equal to a(n-1) can be written as a sum of (not necessarily distinct) numbers of the form 2^n - 2^k with 0 <= k < n. However, a(n-1) - 1 cannot be written in this way. See problem N1 from the 2014 International Mathematics Olympiad Shortlist. - _Dylan Nelson_, Jun 02 2023
%D A036799 M. Petkovsek et al., A=B, Peters, 1996, p. 97.
%H A036799 Reinhard Zumkeller, <a href="/A036799/b036799.txt">Table of n, a(n) for n = 0..1000</a>
%H A036799 R. K. Guy, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL3/GUY/catwalks.html">Catwalks, sandsteps and Pascal pyramids</a>, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6.
%H A036799 Jia Huang, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL26/Huang/huang8.html">Partially Palindromic Compositions</a>, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 3, 4, 9.
%H A036799 Problem Selection Committee for the 2014 IMO, <a href="https://www.imo-official.org/problems/IMO2014SL.pdf">Shortlisted Problems with Solutions for the 55th International Mathematical Olympiad</a>. See pp. 9, 68-70.
%H A036799 Stanislav Sykora, <a href="http://dx.doi.org/10.3247/SL1Math06.002">Finite and Infinite Sums of the Power Series (k^p)(x^k)</a>, DOI 10.3247/SL1Math06.002, Section V.
%H A036799 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-8,4).
%F A036799 a(n) = (n-1) * 2^(n+1) + 2.
%F A036799 a(n) = 2 * A000337(n).
%F A036799 a(n) = Sum_{k=1..n} k*2^k. - _Benoit Cloitre_, Oct 25 2002
%F A036799 G.f.: 2*x/((1-x)*(1-2*x)^2). - _Colin Barker_, Apr 30 2012
%F A036799 a(n) = 5*a(n-1) - 8*a(n-2) + 4*a(n-3) for n > 2. - _Wesley Ivan Hurt_, Nov 12 2015
%F A036799 a(n) = Sum_{k=0..n} Sum_{i=0..n} k * binomial(k,i). - _Wesley Ivan Hurt_, Sep 21 2017
%F A036799 E.g.f.: 2*exp(x) - 2*(1-2*x)*exp(2*x). - _G. C. Greubel_, Mar 29 2021
%p A036799 A036799:=n->2+2^(n+1)*(n-1): seq(A036799(n), n=0..40); # _Wesley Ivan Hurt_, Nov 12 2015
%t A036799 Accumulate[Table[n*2^n, {n, 0, 40}]] (* _Vladimir Joseph Stephan Orlovsky_, Jul 09 2011 *)
%o A036799 (Haskell) a036799 n = (n-1)*2^(n+1) + 2  -- _Reinhard Zumkeller_, May 24 2012
%o A036799 (PARI) a(n)=2+(n-1)<<(n+1) \\ _Charles R Greathouse IV_, Sep 28 2015
%o A036799 (PARI) concat(0, Vec(2*x/((1-x)*(1-2*x)^2) + O(x^40))) \\ _Altug Alkan_, Nov 09 2015
%o A036799 (Magma) [2+2^(n+1)*(n-1) : n in [0..40]]; // _Wesley Ivan Hurt_, Nov 12 2015
%o A036799 (Sage) [2^(n+1)*(n-1) +2 for n in (0..40)] # _G. C. Greubel_, Mar 29 2021
%Y A036799 Cf. A000337, A036289, A036800, A232599, A232600, A232601, A232602.
%K A036799 nonn,easy
%O A036799 0,2
%A A036799 _N. J. A. Sloane_