cp's OEIS Frontend

(1) a(n) = sum of second string of n triangular numbers - sum of first n triangular numbers, or the 2n-th partial sum of triangular numbers (A000217) - the n-th partial sum of triangular numbers (A000217). The same for natural numbers gives squares. (2) a(n) = (n-th triangular number)*(the n-th 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) = (Product_{i=0..n-1} 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^2-1). - Benoit Cloitre, Jan 15 2003

Y values of solutions of the equation: (X-Y)^3-X*Y=0. X values are a(n)=n*(n+1)^2 (see A045991) - Mohamed Bouhamida, May 09 2006

a(2d-1) is the number of self-avoiding walk of length 3 in the d-dimensional 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..n-1), and this is the case d=3 in the identity n^2*(d*n+d-2)/2 - Sum_{k=0..n-1} k*(d*k+d-2)/2 = n*(n+d)*(2*d*n+d-3)/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) is the sum of n-th 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) = (n-k)^2 if n >= k >= 0 and by m(n,k) = (k+n)*(k-n) 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) is the sum of all pairs with repetition drawn from the set of odd numbers 2*n-3. 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

a(n-2) is the maximum sigma irregularity over all trees with n vertices. The extremal graphs are stars. (The sigma irregularity of a graph is the sum of squares of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 14 2023

Nonnegative X values of solutions to the equation X + (X + 3)^2 + (X + 6)^3 = Y^2. To prove that X = n^2 + 6n: Y^2 = X + (X + 3)^2 + (X + 6)^3 = X^3 + 19*X^2 + 115X + 225 = (X + 9)*(X^2 + 10X + 25) = (X + 9)*(X + 5)^2 it means: (X + 9) must be a perfect square, so X = k^2 - 9 with k>=3. we can put: k = n + 3, which gives: X = n^2 + 6n and Y = (n + 3)*(n^2 + 6n + 5). - Mohamed Bouhamida, Nov 12 2007

a(m) where m is a positive integer are the only positive integer values of t for which the Binet-de Moivre Formula of the recurrence b(n)=6*b(n-1)+t*b(n-2) with b(0)=0 and b(1)=1 has a root which is a square. In particular, sqrt(6^2+4*t) is an integer since 6^2+4*t=6^2+4*a(m)=(2*m+6)^2. Thus, the charcteristic roots are k1=6+m and k2=-m. - Felix P. Muga II, Mar 27 2014

Also, numbers k such that k + 9 is a perfect square.

Number of ZnS molecules in cluster of n layers in zinc blende crystal.

(Zinc sulfide crystallizes in two different forms: wurtzite and zinc blende, the latter is also spelled zincblende.) - Jonathan Vos Post, Jan 22 2013

The Kn4 triangle sums of the Connell-Pol triangle A159797 lead to the sequence given above. For the definitions of the Kn4 and other triangle sums see A180662. - Johannes W. Meijer, May 20 2011

If one generated primitive Pythagorean triangles (2n+1, 2n+3) the collective sum of their perimeters for each n is four times the numbers listed in this sequence. - J. M. Bergot, Jul 18 2011

a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and nA000292(n)+A000292(n+1)=n^3. - Clark Kimberling, Jun 04 2012

Degrees of the Hilbert polynomials for B_3 and C_3, per p. 13 of Gashi et al. - Jonathan Vos Post, Dec 14 2013

Number of solutions to a + b = c + d when 0 < a <= k, 0 <= b, c, d <= k, k = 0, 1, 2, 3.... Taken from Step 1 2007 problem #1(i) using 4 digit balanced numbers. - Bobby Milazzo, Mar 09 2013

From J. M. Bergot, Jun 18 2013: (Start)

Consider the lower half, including the main diagonal, of the array in A144216 as a triangle. The rows begin:

0;

1, 2;

3, 4, 6;

6, 7, 9, 12, ...

The sum of the terms in row(n) is a(n). (End)

This sequence is related to A008865 by a(n) = n*A008865(n+1) - Sum_{i=1..n} A008865(i) for n>0. - Bruno Berselli, Aug 06 2015

These are the solutions to the equation x^2 + xy = n where y mod 2 = 0, y is positive and x is any positive integer. - Andrew S. Plewe, Oct 19 2007

Ordered different terms of A120070 = 3, 8, 5, 15, 12, 7, ... (which contains two 15's, two 40's, and two 48's). Complement: A139544. (See A139491.) - Paul Curtz, Sep 01 2009

A024359(a(n)) > 0. - Reinhard Zumkeller, Nov 09 2012

If a(n) mod 6 = 3, n > 1, then a(n) = c^2 - f(a(n))^2 where f(n) = (floor(4*n/3) - 3 - n)/2. For example, 171 = 30^2 - 27^2 and f(171) = 27. - Gary Detlefs, Jul 15 2014

a(m) where m is a positive integer are the only positive integer values of t for which the Binet-de Moivre Formula of the recurrence b(n) = 8*b(n-1) + t*b(n-2) with b(0) = 0 and b(1) = 1 has a root which is a square. In particular, sqrt(8^2 + 4*t) is a positive integer since 8^2 + 4*t = 8^2 + 4*a(m) = (2*m + 8)^2. Thus, the characteristics roots are r1 = 8 + m and r2 = -m. - Felix P. Muga II, Mar 28 2014

These are the only positive integer values of t for which the Binet-de Moivre formula for the recurrence b(n) = 10*b(n-1)+t*b(n-2) with b(0)=0 and b(1)=1 has a root which is a square. In particular, sqrt(10^2+4*t) is a positive integer since 10^2+4*t = 10^2+4*a(m) = (2*m+10)^2. Thus the characteristic roots are r1=10+m and r2 = -m. - Felix P. Muga II, Mar 28 2014

Frequencies or energies of the spectral lines of the hydrogen (H) atom are given, according to quantum theory, by r(m,n)*3.287*PHz (1 Peta Hertz= 10^15 s^{-1}) or r(m,n)*13.599 eV (electron Volts), respectively. The wave lengths are lambda(m,n) = (1/r(m,n))* 91.196 nm (all decimals rounded). See the W. Lang link for more details.

The spectral series for n=1,2,...,7, m>=n+1, are named after Lyman, Balmer, Paschen, Brackett, Pfund, Humphreys, Hansen-Strong, respectively.

The corresponding denominator triangle is A120073.

The rationals are r(m,n):= a(m,n)/A120073(m,n) = A120070(m,n)/(m^2*n^2) = 1/ n^2 - 1/m^2 and they are given in lowest terms.