A180920 -id:A180920 - cp's OEIS frontend

A180921 a(n) is the square root of the sum of the cubes of the b(n) consecutive integers starting from b(n), where b(n) = A180920.

1, 2079, 7876385, 30254180671, 116236127290689, 446579144331338591, 1715756954644453458529, 6591937773063166150358655, 25326223208345427203876398721, 97303342974524967600723097592479, 373839418381901692962342398114034081

Offset: 1

Vladimir Pletser, Sep 24 2010

Colin Barker's linear recurrence conjecture confirmed, see A180920. - Ray Chandler, Jan 12 2024

			a(3) = 2017*(31*(2079/33) + 8*sqrt(15*((2079/33)^2) + 1)).

Colin Barker, Table of n, a(n) for n = 1..279
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (3904,-238206,3904,-1).

Cf. A180920, A240137.

LinearRecurrence[{3904,-238206,3904,-1},{1,2079,7876385,30254180671},20] (* Harvey P. Dale, Apr 27 2025 *)

default(realprecision, 1000);
b=vector(20, n, if(n==1, t=1, t=round(31*t-14+8*((3*t-1)*(5*t-3))^(1/2))));
vector(#b, n, if(n==1, t=1, t=round(b[n]*(31*(t/b[n-1])+8*(15*((t/b[n-1])^2)+1)^(1/2))))) \\ Colin Barker, Feb 19 2015

a(n) = b(n)*(31*(a(n-1)/b(n-1)) + 8*sqrt(15*((a(n-1)/b(n-1))^2) + 1)) where b(n) = A180920(n).
From Colin Barker, Feb 19 2015: (Start)
a(n) = 3904*a(n-1) - 238206*a(n-2) + 3904*a(n-3) - a(4).
G.f.: x*(x+1)*(x^2-1826*x+1) / ((x^2-3842*x+1)*(x^2-62*x+1)). (End)
a(n) = Sqrt(A240137(A180920(n))). - Ray Chandler, Jan 12 2024

Name clarified by Jon E. Schoenfield, Mar 11 2022

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

Original entry on oeis.org

0, 1, 35, 216, 748, 1925, 4131, 7840, 13616, 22113, 34075, 50336, 71820, 99541, 134603, 178200, 231616, 296225, 373491, 464968, 572300, 697221, 841555, 1007216, 1196208, 1410625, 1652651, 1924560, 2228716, 2567573, 2943675, 3359656, 3818240, 4322241, 4874563

Offset: 0

Bruno Berselli, Apr 02 2014

Sum_{i>=1} 1/a(i) = 1.0356568858420883122567711052556541...
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
A180920 lists the numbers k such that a(k) is a square. - Jon E. Schoenfield, Mar 13 2022

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

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Formula for the constant 1.035656885842088312...
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Subsequence of A217843.
Cf. A116149: sum of n consecutive cubes after n^3.
Cf. A050410: sum of n consecutive squares starting from n^2.
Cf. A000326 (pentagonal numbers): sum of n consecutive integers starting from n.
Cf. A126274: n-th triangular number (A000217) * n-th pentagonal number (A000326).

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

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

Table[n^2 (3 n - 1) (5 n - 3)/4, {n, 0, 40}]
CoefficientList[Series[x (1 + 30 x + 51 x^2 + 8 x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)

a(n)=n^2*(3*n-1)*(5*n-3)/4 \\ Charles R Greathouse IV, Oct 07 2015

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

G.f.: x*(1 + 30*x + 51*x^2 + 8*x^3)/(1 - x)^5.
a(n) = n^2*(3*n - 1)*(5*n - 3)/4 = A000326(n)*A000566(n).
a(n) = A116149(-n), with A116149(0)=0.
a(n) = Sum_{j=n..2n-1} j^3. - Jon E. Schoenfield, Mar 13 2022

cp's OEIS Frontend

A180921 a(n) is the square root of the sum of the cubes of the b(n) consecutive integers starting from b(n), where b(n) = A180920.

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