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%I #71 Feb 15 2022 09:56:27
%S 1,9,39,119,294,630,1218,2178,3663,5863,9009,13377,19292,27132,37332,
%T 50388,66861,87381,112651,143451,180642,225170,278070,340470,413595,
%U 498771,597429,711109,841464,990264,1159400,1350888,1566873,1809633,2081583,2385279
%N a(n) = binomial(n+4,4)*(4*n+5)/5.
%C Kekulé numbers for certain benzenoids. - _Emeric Deutsch_, Nov 18 2005
%C 5-dimensional form of hexagonal-based pyramid numbers. - Ben Creech (mathroxmysox(AT)yahoo.com), Nov 17 2005
%C Convolution of triangular numbers (A000217) and hexagonal numbers (A000384). - _Bruno Berselli_, Jun 27 2013
%D Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
%D Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.
%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 167-169, Table 10.5/II/4).
%H Vincenzo Librandi, <a href="/A034263/b034263.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%H <a href="/index/Ps#pyramidal_numbers">Index to sequences related to pyramidal numbers</a>.
%F a(n) = A093561(n+5, 5).
%F a(n) = A034261(n+1, 3).
%F G.f.: (1+3*x)/(1-x)^6.
%F a(n) = (n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120. - _Emeric Deutsch_ and Ben Creech (mathroxmysox(AT)yahoo.com), Nov 17 2005, corrected by _Eric Rowland_, Aug 15 2017
%F a(-n-4) = -A059599(n). - _Bruno Berselli_, Aug 23 2011
%F a(n) = Sum_{i=1..n+1} i*A000292(i). - _Bruno Berselli_, Jan 23 2015
%F Sum_{n>=0} 1/a(n) = 28300/231 - 1280*Pi/77 - 7680*log(2)/77. - _Amiram Eldar_, Feb 15 2022
%e By the third comment: A000217(1..6) and A000384(1..6) give the term a(5) = 1*21+5*15+12*10+22*6+35*3+51*1 = 630. - _Bruno Berselli_, Jun 27 2013
%p a:=n->(n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120: seq(a(n),n=0..35); # _Emeric Deutsch_, Nov 18 2005
%t Table[Binomial[n+4, 4]*(4*n+5)/5, {n,0,35}] (* _Vladimir Joseph Stephan Orlovsky_, Jan 26 2012 *)
%t a[n_] := (1+n)(2+n)(3+n)(4+n)(4n+5)/120; Array[a, 36, 0] (* or *)
%t LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 9, 39, 119, 294, 630}, 36] (* or *)
%t CoefficientList[ Series[(1+3*x)/(1-x)^6, {x, 0, 35}], x] (* _Robert G. Wilson v_, Feb 26 2015 *)
%o (PARI) a(n)=(n+1)*(n+2)*(n+3)*(n+4)*(4*n+5)/120 \\ _Charles R Greathouse IV_, Sep 24 2015, corrected by _Altug Alkan_, Aug 15 2017
%o (Magma) [(4*n+5)*Binomial(n+4,4)/5: n in [0..35]]; // _G. C. Greubel_, Aug 28 2019
%o (Sage) [(4*n+5)*binomial(n+4,4)/5 for n in (0..35)] # _G. C. Greubel_, Aug 28 2019
%o (GAP) List([0..35], n-> (4*n+5)*Binomial(n+4,4)/5); # _G. C. Greubel_, Aug 28 2019
%Y Partial sums of A002417.
%Y Cf. A000217, A000384, A000292, A034261, A059599, A093561.
%Y Cf. similar sequences listed in A254142.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, _Barry E. Williams_, Dec 13 1999
%E Corrected and extended by _N. J. A. Sloane_, Apr 21 2000
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