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A229183 a(n) = n*(n^2 + 3)/2. 8

%I #65 Sep 08 2022 08:46:05

%S 0,2,7,18,38,70,117,182,268,378,515,682,882,1118,1393,1710,2072,2482,

%T 2943,3458,4030,4662,5357,6118,6948,7850,8827,9882,11018,12238,13545,

%U 14942,16432,18018,19703,21490,23382,25382,27493,29718,32060,34522,37107,39818

%N a(n) = n*(n^2 + 3)/2.

%C Numbers a(n) such that (a(n) + B)^(1/3) + (a(n) - B)^(1/3) = n, where B = sqrt(a(n)^2 + 1).

%C 4*a(n) is the sum of two cubes. In fact: 2*n*(n^2 + 3) = (n-1)^3 + (n+1)^3. - _Bruno Berselli_, Apr 11 2016

%C From _Olivier GĂ©rard_, Aug 07 2016 (Start)

%C Row sums of n consecutive integers, starting at 2, seen as a triangle:

%C .

%C 2 | 2

%C 7 | 3 4

%C 18 | 5 6 7

%C 38 | 8 9 10 11

%C 70 | 12 13 14 15 16

%C 117 | 17 18 19 20 21 22

%C (End)

%C Take a long horizontal strip of paper 1 unit high and mark two points on the top edge, n/2 and n units from the top left corner. Then fold over the top left corner so that the fold line passes through the bottom left corner and the point n units along the top edge. Then draw a line from the bottom left corner of the strip through the new position of the n/2 point. The point at which that shallow diagonal line meets the top edge of the strip of paper will be a(n) from the top left corner. - _Elliott Line_, Jul 09 2018

%H Vincenzo Librandi, <a href="/A229183/b229183.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: x*(2 - x + 2*x^2) / (x-1)^4. - _R. J. Mathar_, Sep 22 2013

%F a(n)^2 + 1 = (n^2 + 1)^2 * ((n/2)^2 + 1). - _Joerg Arndt_, Jan 22 2015

%F E.g.f.: exp(x)*x*(4 + 3*x + x^2)/2. - _Stefano Spezia_, Jul 04 2021

%p A229183 := proc(n)

%p n*(n^2+3) /2;

%p end proc:

%p [seq(A229183(n),n=0..30)] ; # _R. J. Mathar_, Aug 16 2019

%t Table[(n^3 + 3n)/2, {n, 0, 100}] (* _T. D. Noe_, Sep 16 2013 *)

%t CoefficientList[Series[x (2 - x + 2 x^2)/(x - 1)^4, {x, 0, 50}], x] (* _Vincenzo Librandi_, Sep 23 2013 *)

%o (Python)

%o {print((n**3+3*n)/2,end=', ') for n in range(0,100)} # Simplified by _Derek Orr_, Mar 12 2015

%o (PARI) vector(100,n,((n-1)^3+3*n-3)/2) \\ _Derek Orr_, Mar 12 2015

%o (Magma) [n*(n^2 + 3) div 2: n in [0..50]]; // _Vincenzo Librandi_, Sep 23 2013

%Y Cf. A006003 (row sums of integers, starting with 1).

%Y Cf. A027480 (row sums of integers, starting with 0).

%K nonn,easy

%O 0,2

%A _Derek Orr_, Sep 15 2013

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Last modified March 29 02:23 EDT 2024. Contains 371264 sequences. (Running on oeis4.)