

A011379


a(n) = n^2*(n+1).


47



0, 2, 12, 36, 80, 150, 252, 392, 576, 810, 1100, 1452, 1872, 2366, 2940, 3600, 4352, 5202, 6156, 7220, 8400, 9702, 11132, 12696, 14400, 16250, 18252, 20412, 22736, 25230, 27900, 30752, 33792, 37026, 40460, 44100, 47952, 52022, 56316, 60840
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OFFSET

0,2


COMMENTS

(1) a(n) = sum of second string of n triangular numbers  sum of first n triangular numbers, or the 2nth partial sum of triangular numbers (A000217 )  the nth partial sum of triangular numbers(A000217 ). The same for natural numbers gives squares. (2) a(n) = (nth triangular number)*(the nth even number) = n(n+1)/2 * (2n)  Amarnath Murthy, Nov 05 2002
Let M(n) be the n X n matrix m(i,j)=1/(i+j+x), let P(n,x) = (Prod_{i=0..n1} i!^2)/det(M(n)). Then P(n,x) is a polynomial with integer coefficients of degree n^2 and a(n) is the coefficient of x^(n^21).  Benoit Cloitre, Jan 15 2003
Y values of solutions of the equation: (XY)^3X*Y=0. X values are a(n)=n*(n+1)^2 (see A045991)  Mohamed Bouhamida, May 09 2006
a(2d1) is the number of selfavoiding walk of length 3 in the ddimensional hypercubic lattice.  Michael Somos, Sep 06 2006
a(n) mod 10 is periodic 5: repeat [0, 2, 2, 6, 0].  Mohamed Bouhamida, Sep 05 2009
This sequence is related to A005449 by a(n) = n*A005449(n)sum(A005449(i), i=0..n1), and this is the case d=3 in the identity n^2*(d*n+d2)/2  Sum_{k=0..n1} k*(d*k+d2)/2 = n*(n+d)*(2*d*n+d3)/6.  Bruno Berselli, Nov 18 2010
Using (n, n+1) to generate a primitive Pythagorean triangle, the sides will be 2*n+1, 2*(n^2+n), and 2*n^2+2*n+1. Inscribing the largest rectangle with integral sides will have sides of length n and n^2+n. Side n is collinear to side 2*n+1 of the triangle and side n^2+n is collinear to side 2*(n^2+n) of the triangle. The areas of theses rectangles are a(n). [J. M. Bergot, Sep 22 2011]
a(n+1) = sum of nth row of the triangle in A195437.  Reinhard Zumkeller, Nov 23 2011
Partial sums of A049450.  Omar E. Pol, Jan 12 2013
From Jon Perry, May 11 2013: (Start)
Define a 'stable brick triangle' as:

 c 

 a   b 

with a, b, c > 0 and c <= a + b. This can be visualized as two bricks with a third brick on top. The third brick can only be as strong as a+b, otherwise the wall collapses  for example, (1,2,4) is unstable.
a(n) gives the number of stable brick triangles that can be formed if the two supporting bricks are 1<=a<=n and 1<=b<=n: a(n) = sum_{a=1..n} sum_{b=1..n} sum_c 1 = n^3+n^2 as given in the Adamchuk formula.
So for i=j=n=2 we have 4:
1 2 3 4
2 2 2 2 2 2 2 2
For example, n=2 gives 2 from [a=1,b=1], 3 from both [a=1,b=2] and [a=2,b=1] and 4 from [a=2,b=2] so a(2) = 2 + 3 + 3 + 4 = 12. (end)
Define the infinite square array m(n,k) by m(n,k) = (nk)^2 if n>=k>=0 and by m(n,k) = (k+n)*(kn) if 0<=n<=k. This contains A120070 below the diagonal. Then a(n) = Sum_{k=0..n} m(n,k) + Sum_{r=0..n} m(r,n), the "hook sum" of the terms to the left of m(n,n) and above m(n,n) with irrelevant (vanishing) terms on the diagonal.  J. M. Bergot, Aug 16 2013
a(n) = A245334(n+1,2), n > 0.  Reinhard Zumkeller, Aug 31 2014
a(n) is the sum of all pairs with repetition drawn from the set of odd numbers 2*n3. This is similar to A027480 but using the odd integers instead. Example using n=3 gives the odd numbers 1,3,5: 1+1, 1+3, 1+5, 3+3, 3+5,5+5 having a total of 36=a(3).  J. M. Bergot, Apr 05 2016
a(n) is the first Zagreb index of the complete graph K[n+1]. The first Zagreb index of a simple connected graph is the sum of the squared degrees of its vertices. Alternately, it is the sum of the degree sums d(i)+d(j) over all edges ij of the graph.  Emeric Deutsch, Nov 07 2016


REFERENCES

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, pp. 50, 64.


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
I. Gutman and K. C. Das, The first Zagreb index 30 years after, MATCH Commun. Math. Comput. Chem. 50, 2004, 8392.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian).
Index entries for linear recurrences with constant coefficients, signature (4,6,4,1).


FORMULA

a(n) = 2*A002411(n).
a(n) = Sum_{j=1..n} (Sum_{i=1..n} (i+j)), row sums of A126890 skipping numbers in the first column.  Alexander Adamchuk, Oct 12 2004
Sum_{n>0} 1/a(n) = (Pi^2  6)/6 = 0.6449340.. [Jolley eq 272]  Gary W. Adamson, Dec 22 2006
a(n) = 2*n*binomial(n+1,2) = 2*n*A000217(n).  Arkadiusz Wesolowski, Feb 10 2012
G.f.: 2*x*(1 + 2*x)/(1  x)^4.  Arkadiusz Wesolowski, Feb 11 2012
a(n) = A000330(n) + A002412(n) = A000292(n) + A002413(n).  Omar E. Pol, Jan 11 2013
Sum_{n>=1) 1/a(n) = (A0136611).  R. J. Mathar, Oct 18 2019
Sum_{n>=1} (1)^(n+1)/a(n) = 1 + Pi^2/12  2*log(2).  Amiram Eldar, Jul 04 2020


EXAMPLE

a(3) = 3^2+3^3 = 36.


MAPLE

A011379:=n>n^2*(n+1); seq(A011379(n), n=0..40); # Wesley Ivan Hurt, Feb 25 2014


MATHEMATICA

Table[n^3+n^2, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Jan 03 2009, modified by G. C. Greubel, Aug 10 2019 *)
LinearRecurrence[{4, 6, 4, 1}, {0, 2, 12, 36}, 40] (* Harvey P. Dale, Sep 13 2018 *)


PROG

(MAGMA) [n^2+n^3: n in [0..40]]; // Vincenzo Librandi, May 02 2011
(Haskell)
a011379 n = a000290 n + a000578 n  Reinhard Zumkeller, Apr 28 2013
(PARI) a(n)=n^3+n^2 \\ Charles R Greathouse IV, Apr 06 2016
(Sage) [n^2*(n+1) for n in (0..40)] # G. C. Greubel, Aug 10 2019
(GAP) List([0..40], n> n^2*(n+1) ); # G. C. Greubel, Aug 10 2019


CROSSREFS

Cf. A000290, A000578, A005449, A022549, A045991, A245334.
Sequence in context: A200543 A246426 A176583 * A338610 A073404 A141208
Adjacent sequences: A011376 A011377 A011378 * A011380 A011381 A011382


KEYWORD

nonn,easy


AUTHOR

Glen Burch (gburch(AT)erols.com); Felice Russo


STATUS

approved



