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A240137 Sum of n consecutive cubes starting from n^3. 7

%I

%S 0,1,35,216,748,1925,4131,7840,13616,22113,34075,50336,71820,99541,

%T 134603,178200,231616,296225,373491,464968,572300,697221,841555,

%U 1007216,1196208,1410625,1652651,1924560,2228716,2567573,2943675,3359656,3818240,4322241,4874563

%N Sum of n consecutive cubes starting from n^3.

%C Sum_{i>=1} 1/a(i) = 1.0356568858420883122567711052556541...

%C Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of cubes with side length q. - _Wesley Ivan Hurt_, Apr 15 2018

%H Bruno Berselli, <a href="/A240137/b240137.txt">Table of n, a(n) for n = 0..1000</a>

%H Bruno Berselli, <a href="/A240137/a240137_1.jpg">Formula for the constant 1.035656885842088312...</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: x*(1 + 30*x + 51*x^2 + 8*x^3)/(1 - x)^5.

%F a(n) = n^2*(3*n - 1)*(5*n - 3)/4 = A000326(n)*A000566(n).

%F a(n) = A116149(-n), with A116149(0)=0.

%e a(3) = 216 because 216 = 3^3 + 4^3 + 5^3.

%p A240137:=n->n^2*(3*n-1)*(5*n-3)/4; seq(A240137(n), n=0..40); # _Wesley Ivan Hurt_, May 09 2014

%t Table[n^2 (3 n - 1) (5 n - 3)/4, {n, 0, 40}]

%t CoefficientList[Series[x (1 + 30 x + 51 x^2 + 8 x^3)/(1 - x)^5, {x, 0, 40}], x] (* _Vincenzo Librandi_, May 09 2014 *)

%o (Sage) [n^2*(3*n-1)*(5*n-3)/4 for n in [0..40]]

%o (MAGMA) [n^2*(3*n-1)*(5*n-3)/4: n in [0..40]];

%o (PARI) a(n)=n^2*(3*n-1)*(5*n-3)/4 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Subsequence of A217843.

%Y Cf. A116149: sum of n consecutive cubes after n^3.

%Y Cf. A050410: sum of n consecutive squares starting from n^2.

%Y Cf. A000326 (pentagonal numbers): sum of n consecutive integers starting from n.

%Y Cf. A126274: n-th triangular number (A000217) * n-th pentagonal number (A000326).

%K nonn,easy

%O 0,3

%A _Bruno Berselli_, Apr 02 2014

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Last modified November 12 19:25 EST 2019. Contains 329078 sequences. (Running on oeis4.)