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a(n) = n*(5*n - 1)/2.
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%I #84 Aug 02 2024 01:59:10

%S 0,2,9,21,38,60,87,119,156,198,245,297,354,416,483,555,632,714,801,

%T 893,990,1092,1199,1311,1428,1550,1677,1809,1946,2088,2235,2387,2544,

%U 2706,2873,3045,3222,3404,3591

%N a(n) = n*(5*n - 1)/2.

%C a(n) is half the number of ways to divide an n X n square into 3 rectangles whose side-lengths are integers. See Matthew Scroggs link. - _George Witty_, Feb 06 2024

%H G. C. Greubel, <a href="/A005476/b005476.txt">Table of n, a(n) for n = 0..1000</a>

%H Matthew Scroggs, <a href="https://www.mscroggs.co.uk/puzzles/advent2023/23">Advent Calendar 2023 Solutions</a>.

%H Leo Tavares, <a href="/A005476/a005476.jpg">Illustration: Triangulated Diamonds</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = C(5*n,2)/5 for n>=0. - _Zerinvary Lajos_, Jan 02 2007

%F a(n) = A033991(n) - A000326(n). - _Zerinvary Lajos_, Jun 11 2007

%F a(n) = a(n-1) + 5*n - 3 for n>0, a(0)=0. - _Vincenzo Librandi_, Nov 18 2010

%F a(n) = A000217(n) + A000384(n) = A000290(n) + A000326(n). - _Omar E. Pol_, Jan 11 2013

%F a(n) = A130520(5*n+1). - _Philippe Deléham_, Mar 26 2013

%F a(n) = A033994(n) - A033994(n-1). - _J. M. Bergot_, Jun 12 2013

%F From _Bruno Berselli_, Oct 17 2016: (Start)

%F G.f.: x*(2 + 3*x)/(1 - x)^3.

%F a(n) = A000217(3*n-1) - A000217(2*n-1). (End)

%F E.g.f.: x*(4 + 5*x)*exp(x)/2. - _G. C. Greubel_, Jul 30 2019

%F Sum_{n>=1} 1/a(n) = 2 * A294833. - _Amiram Eldar_, Nov 16 2020

%F From _Leo Tavares_, Nov 20 2021: (Start)

%F a(n) = A016754(n) - A133694(n+1). See Triangulated Diamonds illustration.

%F a(n) = A000290(n) + A000217(n) + 2*A000217(n-1)

%F a(n) = 2*A000217(n) + 3*A000217(n-1). (End)

%p [seq(binomial(5*n,2)/5,n=0..40)]; # _Zerinvary Lajos_, Jan 02 2007

%t Table[n(5n-1)/2, {n,0,40}] (* _Vladimir Joseph Stephan Orlovsky_, Oct 25 2008 *)

%o (PARI) a(n)=n*(5*n-1)/2 \\ _Charles R Greathouse IV_, Sep 24 2015

%o (Magma) [Binomial(5*n,2)/5: n in [0..40]]; // _G. C. Greubel_, Jul 30 2019

%o (Sage) [binomial(5*n,2)/5 for n in (0..40)] # _G. C. Greubel_, Jul 30 2019

%o (GAP) List([0..40], n-> Binomial(5*n,2)/5); # _G. C. Greubel_, Jul 30 2019

%Y Cf. A000217, A005475, A033994, A294833.

%Y Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.

%Y Cf. similar sequences listed in A022288.

%Y Cf. A016754, A133694, A000290.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_