A126274 -id:A126274 - cp's OEIS frontend

A006011 a(n) = n^2*(n^2 - 1)/4.

0, 0, 3, 18, 60, 150, 315, 588, 1008, 1620, 2475, 3630, 5148, 7098, 9555, 12600, 16320, 20808, 26163, 32490, 39900, 48510, 58443, 69828, 82800, 97500, 114075, 132678, 153468, 176610, 202275, 230640, 261888, 296208, 333795, 374850, 419580, 468198

Offset: 0

N. J. A. Sloane, Simon Plouffe

Products of two consecutive triangular numbers (A000217).
a(n) is the number of Lyndon words of length 4 on an n-letter alphabet. A Lyndon word is a primitive word that is lexicographically earliest in its cyclic rotation class. For example, a(2)=3 counts 1112, 1122, 1222. - David Callan, Nov 29 2007
For n >= 2 this is the second rightmost column of A163932. - Johannes W. Meijer, Oct 16 2009
Partial sums of A059270. - J. M. Bergot, Jun 27 2013
Using the integers, triangular numbers, and squares plot the points (A001477(n),A001477(n+1)), (A000217(n), A000217(n+1)), and (A000290(n),A000290(n+1)) to create the vertices of a triangle. One-half the area of this triangle = a(n). - J. M. Bergot, Aug 01 2013
a(n) is the Wiener index of the triangular graph T(n+1). - Emeric Deutsch, Aug 26 2013

			From _Bruno Berselli_, Aug 29 2014: (Start)
After the zeros, the sequence is provided by the row sums of the triangle:
   3;
   4, 14;
   5, 16, 39;
   6, 18, 42,  84;
   7, 20, 45,  88, 155;
   8, 22, 48,  92, 160, 258;
   9, 24, 51,  96, 165, 264, 399;
  10, 26, 54, 100, 170, 270, 406, 584;
  11, 28, 57, 104, 175, 276, 413, 592, 819;
  12, 30, 60, 108, 180, 282, 420, 600, 828, 1110; etc.,
where T(r,c) = c*(c^2+r+1), with r = row index, c = column index, r >= c > 0. (End)

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Miguel Azaola and Francisco Santos, The number of triangulations of the cyclic polytope C(n,n-4), Discrete Comput. Geom., Vol. 27 (2002), pp. 29-48 (see Prop. 4.2(a)).
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
Eric Weisstein's World of Mathematics, Triangular Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Thrice A002415. Row 4 of A074650.
Cf. A000217, A000290, A000537, A001477, A002415, A008911, A047928, A059270, A083374, A126274, A163932, A228317
A column of A124428.

[n^2*(n^2-1)/4: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

A006011 := proc(n)
    n^2*(n^2-1)/4 ;
end proc: # R. J. Mathar, Nov 29 2015

Table[n^2 (n^2 - 1)/4, {n, 0, 38}]
Binomial[Range[20]^2, 2]/2 (* Eric W. Weisstein, Sep 08 2017 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 3, 18, 60, 150}, 20] (* Eric W. Weisstein, Sep 08 2017 *)
CoefficientList[Series[-3 x (1 + x)/(-1 + x)^5, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 08 2017 *)
Join[{0},Times@@@Partition[Accumulate[Range[0,40]],2,1]] (* Harvey P. Dale, Aug 08 2025 *)

a(n)=binomial(n^2,2)/2 \\ Charles R Greathouse IV, Jun 27 2013

G.f.: 3*(1 + x) / (1 - x)^5.
a(n) = (n-1)*n/2 * n*(n+1)/2 = A000217(n-1)*A000217(n) = 1/2*(n^2-1)*n^2/2 = 1/2*A000217(n^2-1). - Alexander Adamchuk, Apr 13 2006
a(n) = 3*A002415(n) = A047928(n-1)/4 = A083374(n-1)/2 = A008911(n)*3/2. - Zerinvary Lajos, May 09 2007
a(n) = (A126274(n) - A000537(n+1))/2. - Enrique Pérez Herrero, Mar 11 2013
Ceiling(sqrt(a(n)) + sqrt(a(n-1)))/2 = A000217(n). - Richard R. Forberg, Aug 14 2013
a(n) = Sum_{i=1..n-1} i*(i^2+n) for n > 1 (see Example section). - Bruno Berselli, Aug 29 2014
Sum_{n>=2} 1/a(n) = 7 - 2*Pi^2/3 = 0.42026373260709425411... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n^2+n) - A000217(n)*A000217(n+1). - Charlie Marion, Feb 15 2020
Sum_{n>=2} (-1)^n/a(n) = Pi^2/3 - 3. - Amiram Eldar, Nov 02 2021
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/4. - Stefano Spezia, Mar 12 2024

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

A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

Original entry on oeis.org

1, 15, 84, 300, 825, 1911, 3920, 7344, 12825, 21175, 33396, 50700, 74529, 106575, 148800, 203456, 273105, 360639, 469300, 602700, 764841, 960135, 1193424, 1470000, 1795625, 2176551, 2619540, 3131884, 3721425, 4396575, 5166336, 6040320, 7028769, 8142575

Offset: 0

Emeric Deutsch, Jun 17 2005

Kekulé numbers for certain benzenoids.
This is the case P(3,n) of the family of sequences defined in A132458. - Ottavio D'Antona (dantona(AT)dico.unimi.it), Oct 31 2007
Using the triangular numbers 0, 1, 3, ..., create a sequence of advancing sums of k-tuples with k=n*(n+1)/2 of the odd numbers: 0, 1, 15, 84, 300, 825, 1911, 3920, ... . This begins 0, then 1, then 3+5+7=15, then 9+11+13+15+17+19=84, then 21+23+...+39=300 and so on. - J. M. Bergot, Dec 08 2014
Partial sums of A008354. - J. M. Bergot, Dec 19 2014
Coefficients in the terminating series identity 1 - 15*n/(n + 4) + 84*n*(n - 1)/((n + 4)*(n + 5)) - 300*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 2,3,4,.... Cf. A000330. - Peter Bala, Feb 12 2019

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 231, # 33).

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000537, A126274.

Cf. A000217, A000330, A008354 (first diffs.), A132458.

List([0..40],k->(k+1)^2*(k+2)^2*(2*k+3)/12); # Muniru A Asiru, Feb 18 2019

A108674 :=n->(n+1)^2*(n+2)^2*(2*n+3)/12;
seq(A108674(n),n=0..35);

Table[(n^2+4*n^3+5*n^4+2*n^5)/12, {n, 40}] (* Enrique Pérez Herrero, Feb 27 2013 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,15,84,300,825,1911},40] (* Harvey P. Dale, Apr 04 2024 *)

a(n)=(n+1)^2*(n+2)^2*(2*n+3)/12 \\ Charles R Greathouse IV, Feb 27 2013

G.f.: (1+z)*(1+8*z+z^2)/(1-z)^6.
a(n) = Sum_{j=1..n+1} j^2 Sum_{i=1..n+1} i. - Alexander Adamchuk, Jun 25 2006
a(n) = A000330(n+1) * A000217(n+1). - Daniel Suteu, Nov 26 2020
E.g.f.: exp(x)*(12 + 168*x + 330*x^2 + 184*x^3 + 35*x^4 + 2*x^5)/12. - Stefano Spezia, Mar 02 2022
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) 192*log(2) - 132.
Sum_{n>=0} (-1)^n/a(n) = 2*Pi^2 - 48*Pi + 132. (End)

A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

Original entry on oeis.org

0, 1, 0, 8, 7, 0, 27, 26, 19, 0, 64, 63, 56, 37, 0, 125, 124, 117, 98, 61, 0, 216, 215, 208, 189, 152, 91, 0, 343, 342, 335, 316, 279, 218, 127, 0, 512, 511, 504, 485, 448, 387, 296, 169, 0, 729, 728, 721, 702, 665, 604, 513, 386, 217, 0, 1000, 999, 992, 973, 936, 875, 784, 657, 488, 271, 0

Offset: 0

Wolfdieter Lang, May 12 2015

See the comments in A025581 and A079904 on a problem by François Viète (Vieta)(1593). Regarding that problem, note the simple identity: n^3 - k^3 = (n - k)^3 + 3*n*k*(n - k), for n > = k >= 0.
Row sums give A126274(n-1) for n >= 1, and 0 for n=0.
Alternating row sums are ars(2*n) = ars(2*n-1) = (4*n-3)*n^2 = A103532(n-1), for n >= 1, and ars(0) = 0.

			The triangle T(n, k) begins:
  n\k    0   1   2   3   4   5   6   7   8   9  10
   0:    0
   1:    1   0
   2:    8   7   0
   3:   27  26  19   0
   4:   64  63  56  37   0
   5:  125 124 117  98  61   0
   6:  216 215 208 189 152  91   0
   7:  343 342 335 316 279 218 127   0
   8:  512 511 504 485 448 387 296 169   0
   9:  729 728 721 702 665 604 513 386 217   0
  10: 1000 999 992 973 936 875 784 657 488 271   0
  ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A126274, A103532, A025581, A079904, A335821.

for n from 0 to 10 do seq(n^3-k^3,k=0..n) od; # Robert Israel, May 10 2018

Table[n^3-k^3,{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Jan 02 2021 *)

T(n, k) =  A025581(n, k)*(A025581(n, k)^2 + 3* A079904(n, k)) (see the identity mentioned in a comment).
Columns (with one leading zero and offset 0): k=0: l^3 = A000578(l), k=1: (l+1)^3 - 1 = A068601(l+1), k=2: l*(l^2 + 6*l + 12), k=3: l*(l^2 + 9*l + 27), k=4: l*(l^2 + 12*l + 48), k=5: l*(l^2 + 15*l + 75), ...
G.f. for T(n,k): (1+4*x+4*x*y+x^2-14*x^2*y+x^2*y^2-2*x^3*y-2*x^3*y^2+7*x^4*y^2)*x/((1-x*y)^3*(1-x)^4). - Robert Israel, May 10 2018

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A006011 a(n) = n^2*(n^2 - 1)/4.

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

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A108674 a(n) = (n+1)^2 * (n+2)^2 * (2*n+3) / 12.

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A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

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