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A159920 Sums of the antidiagonals of Sundaram's sieve (A159919). 6

%I #46 Oct 04 2022 08:39:08

%S 4,14,32,60,100,154,224,312,420,550,704,884,1092,1330,1600,1904,2244,

%T 2622,3040,3500,4004,4554,5152,5800,6500,7254,8064,8932,9860,10850,

%U 11904,13024,14212,15470,16800,18204,19684,21242,22880,24600,26404

%N Sums of the antidiagonals of Sundaram's sieve (A159919).

%C For every n >= 2, a(n) is the sum of numbers in the (n-1)-th antidiagonal of the Sundaram sieve. (It is not clear why the offset was set to 2 rather than 1.) Thus, if T(j, k) is the element in row j and column k of the Sundaram sieve, we have a(n) = Sum_{i = 1..n-1} T(i, n-i) = Sum_{i = 1..n-1} (2*i*(n-i) + i + (n-i)) = (n - 1)*n*(n + 4)/3 for the sum of the numbers in the (n-1)-th antidiagonal. - _Petros Hadjicostas_, Jun 19 2019

%H G. C. Greubel, <a href="/A159920/b159920.txt">Table of n, a(n) for n = 2..1000</a>

%H Andrew Baxter, <a href="http://banach.millersville.edu/~bob/math478/History/Sundaram.html">Sundaram's Sieve</a>.

%H Julian Havil, <a href="http://plus.maths.org/issue50/features/havil/index.html">Sundaram's Sieve</a>, Plus Magazine, March 2009.

%H New Zealand Maths, <a href="https://web.archive.org/web/20040309001101/www.nzmaths.co.nz/HelpCentre/Newsletter18.pdf">Newletter 18</a>, October 2002.

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Sieve_of_Sundaram">Sundaram's Sieve</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) = (n - 1)*n*(n + 4)/3.

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

%F a(n) = 2*A005581(n), n > 1.

%F a(n) = Sum_{i=1..n-1} i*(i + 3). - _Wesley Ivan Hurt_, Oct 19 2013

%F From _G. C. Greubel_, Oct 03 2022: (Start)

%F G.f.: 2*x^2*(2 - x)/(1-x)^4.

%F E.g.f.: (1/3)*x^2*(6 + x)*exp(x). (End)

%e For n = 5, (4*5*9)/3 = 60. Indeed, T(1, 4) + T(2, 3) + T(3, 2) + T(4, 1) = 13 + 17 + 17 + 13 = 60 for the sum of the terms in the 4th antidiagonal of the Sundaram sieve.

%p A159920:=n->n*(n-1)*(n+4)/3; seq(A159920(k), k=2..100); # _Wesley Ivan Hurt_, Oct 19 2013

%t Table[(n-1)*n*(n+4)/3,{n,2,60}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 28 2010 *)

%t LinearRecurrence[{4,-6,4,-1},{4,14,32,60},61] (* _Harvey P. Dale_, Apr 23 2011 *)

%o (Magma) [n*(n-1)*(n+4)/3: n in [2..60]]; // _G. C. Greubel_, Oct 03 2022

%o (SageMath) [n*(n-1)*(n+4)/3 for n in range(2,60)] # _G. C. Greubel_, Oct 03 2022

%Y Cf. A005581, A159919.

%K nonn,easy

%O 2,1

%A _Russell Walsmith_, Apr 26 2009

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