cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A011379 a(n) = n^2*(n+1).

Original entry on oeis.org

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
Offset: 0

Views

Author

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

Keywords

Comments

(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

Examples

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

References

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

Crossrefs

Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

Programs

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
a(n) = A245334(n+1,2), n > 0. - Reinhard Zumkeller, Aug 31 2014
Sum_{n>=1} 1/a(n) = A013661-1. - R. J. Mathar, Oct 18 2019 [corrected by Jason Yuen, Aug 04 2024]
Sum_{n>=1} (-1)^(n+1)/a(n) = 1 + Pi^2/12 - 2*log(2). - Amiram Eldar, Jul 04 2020
E.g.f.: exp(x)*x*(2 + 4*x + x^2). - Stefano Spezia, May 20 2021
a(n) = n*A002378(n) = A000578(n) + A000290(n). - J.S. Seneschal, Jun 18 2024

A033579 Four times pentagonal numbers: a(n) = 2*n*(3*n-1).

Original entry on oeis.org

0, 4, 20, 48, 88, 140, 204, 280, 368, 468, 580, 704, 840, 988, 1148, 1320, 1504, 1700, 1908, 2128, 2360, 2604, 2860, 3128, 3408, 3700, 4004, 4320, 4648, 4988, 5340, 5704, 6080, 6468, 6868, 7280, 7704, 8140, 8588, 9048, 9520, 10004, 10500, 11008, 11528, 12060
Offset: 0

Views

Author

Keywords

Comments

Subsequence of A062717: A010052(6*a(n)+1) = 1. - Reinhard Zumkeller, Feb 21 2011
Sequence found by reading the line from 0, in the direction 0, 4, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011

Crossrefs

Programs

Formula

a(n) = 4*n*(3*n-1)/2 = 6*n^2 - 2*n = 4*A000326(n). - Omar E. Pol, Dec 11 2008
a(n) = 2*A049450(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 12*n - 8 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 05 2010
a(n) = A014642(n)/2. - Omar E. Pol, Aug 19 2011
G.f.: x*(4+8*x)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 06 2012
a(n) = A191967(2*n). - Reinhard Zumkeller, Jul 07 2012
a(n) = A181617(n+1) - A181617(n). - J. M. Bergot, Jun 28 2013
a(n) = (A174371(n) - 1)/6. - Miquel Cerda, Jul 28 2016
From Ilya Gutkovskiy, Jul 28 2016: (Start)
E.g.f.: 2*x*(2 + 3*x)*exp(x).
a(n+1) = Sum_{k=0..n} A017569(k).
Sum_{i>0} 1/a(i) = (9*log(3) - sqrt(3)*Pi)/12 = 0.3705093754425278... (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(2*sqrt(3)) - log(2). - Amiram Eldar, Feb 20 2022

Extensions

More terms from Michel Marcus, Mar 04 2014

A035006 Number of possible rook moves on an n X n chessboard.

Original entry on oeis.org

0, 8, 36, 96, 200, 360, 588, 896, 1296, 1800, 2420, 3168, 4056, 5096, 6300, 7680, 9248, 11016, 12996, 15200, 17640, 20328, 23276, 26496, 30000, 33800, 37908, 42336, 47096, 52200, 57660, 63488, 69696, 76296, 83300, 90720, 98568, 106856
Offset: 1

Views

Author

Ulrich Schimke (ulrschimke(AT)aol.com)

Keywords

Comments

Obviously A035005(n) = A002492(n-1) + a(n) since Queen = Bishop + Rook. - Johannes W. Meijer, Feb 04 2010
X values of solutions of the equation: (X-Y)^3-2*X*Y=0. Y values are b(n)=2*n*(n-1)^2 (see A181617). - Mohamed Bouhamida, Jul 06 2023

Examples

			On a 3 X 3-board, rook has 9*4 moves, so a(3)=36.
		

References

  • E. Bonsdorff, K. Fabel and O. Riihimaa, Schach und Zahl (Chess and numbers), Walter Rau Verlag, Dusseldorf, 1966.

Crossrefs

Cf. A033586 (King), A035005 (Queen), A035008 (Knight), A002492 (Bishop) and A049450 (Pawn).

Programs

  • Magma
    [(n-1)*2*n^2: n in [1..40]]; // Vincenzo Librandi, Jun 16 2011
  • Mathematica
    Table[(n-1) 2 n^2,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,8,36,96},40] (* Harvey P. Dale, May 12 2012 *)

Formula

a(n) = (n-1)*2*n^2.
a(n) = Sum_{j=1..n} ((n+j-1)^2 - (n-j+1)^2). - Zerinvary Lajos, Sep 13 2006
1/a(n+1) = Integral_{x=1/(n+1)..1/n} x*h(x) = Integral_{x=1/(n+1)..1/n} x*(1/x - floor(1/x)) = 1/((2*(n^2+2*n+1))*n) and Sum_{n>=1} 1/((2*(n^2+2*n+1))*n) = 1-Zeta(2)/2 where h(x) is the Gauss (continued fraction) map h(x)={x^-1} and {x} is the fractional part of x. - Stephen Crowley, Jul 24 2009
a(n) = 4 * A006002(n-1). - Johannes W. Meijer, Feb 04 2010
G.f.: 4*x^2*(2+x)/(1-x)^4. - Colin Barker, Mar 11 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=0, a(2)=8, a(3)=36, a(4)=96. - Harvey P. Dale, May 12 2012
a(n) = A006566(n) - A006564(n). - Peter M. Chema, Feb 10 2016
E.g.f.: 2*exp(x)*x^2*(2 + x). - Stefano Spezia, May 10 2022
From Amiram Eldar, May 14 2022: (Start)
Sum_{n>=2} 1/a(n) = 1 - Pi^2/12.
Sum_{n>=2} (-1)^n/a(n) = Pi^2/24 + log(2) - 1. (End)

A270205 Number of 2 X 2 planar subsets in an n X n X n cube.

Original entry on oeis.org

0, 0, 6, 36, 108, 240, 450, 756, 1176, 1728, 2430, 3300, 4356, 5616, 7098, 8820, 10800, 13056, 15606, 18468, 21660, 25200, 29106, 33396, 38088, 43200, 48750, 54756, 61236, 68208, 75690, 83700, 92256
Offset: 0

Views

Author

Craig Knecht, Mar 13 2016

Keywords

Comments

William H. Press looked at the hybrid structure of a most-perfect magic square and the Hilbert space filling curve and thought it might be the "most uniform" way of putting the consecutive integers in a 2-d square. He thought a definition of "most uniform" would be useful.
Al Zimmermann suggested this: Start by defining the "non-uniformity of a distribution of integers among the cells of a square [or cube or hypercube]" to be the standard deviation of the sums of the 2 X 2 planar subsets. Then define a "most uniform distribution of integers" to be a distribution with the smallest non-uniformity. For both the most-perfect square and most-perfect cube the non-uniformity is 0 and so each is a most uniform distribution. (Of course, you'd want a better word for "non-uniformity". Skewness?) Perhaps use "2 X 2 planar subset" instead of "2 X 2 partition"?
Comment from Dwane Campbell: For cubes, the definition of compact is that all 2 X 2 X 2 subcubes add to the same sum. That definition also includes wrap around. Your most perfect space cube is compact. It has the additional constraint that each orthogonal plane is also compact. There are 64 2 X 2 X 2 subcubes that add to 260 and 192 2 X 2 subsquares that add to 130 in your cube. I did not think either result was possible. Congratulations!
The most-perfect order 4 cube and the reversible order 4 cube are the new findings to look at in the link section.
Most-perfect magic squares require every 2 X 2 cell block to have the same sum. This sequence looks at that same subset in the cube.
Most-perfect space is defined as a structure where all these 2 X 2 subsets have the same sum.
What structure provides the most uniform distribution of integers in a cube?
a(n+1) is the number of unit faces required to make an n X n X n cubic lattice. Number of unit edges required for the same is A059986(n). - Mohammed Yaseen, Aug 22 2021
a(n-3) is the maximum sigma irregularity over all maximal 3-degenerate graphs with n vertices. The extremal graphs are 3-stars (K_3 joined to n-3 independent vertices). (The sigma irregularity of a graph is the sum of the squares of the differences between the degrees over all edges of the graph.) - Allan Bickle, Jun 14 2023

Examples

			The 2 X 2 X 2 cube labeled with the integers 1 to 8 has the following six 2 X 2 planar subsets each containing 4 cells: 1,2,3,4; 5,6,7,8; 1,2,5,6; 3,4,7,8; 1,4,5,8; 2,3,6,7.
		

Crossrefs

Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

Programs

Formula

a(n) = 3*n^3 - 6*n^2 + 3*n.
From Wesley Ivan Hurt, Mar 13 2016: (Start)
G.f.: 6*x^2*(1+2*x)/(x-1)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
E.g.f.: 3*x^2*(1+x)*exp(x). - G. C. Greubel, May 10 2016
a(n) = 6 * A002411(n-1) for n>=1. - Joerg Arndt, May 11 2016
a(n) = A118659((n-1)^3), n>1. - Mohammed Yaseen, Aug 22 2021
From Amiram Eldar, Jul 02 2023: (Start)
Sum_{n>=2} 1/a(n) = Pi^2/18 - 1/3.
Sum_{n>=2} (-1)^n/a(n) = Pi^2/36 - 2*log(2)/3 + 1/3. (End)

A107661 Array read by antidiagonals: T(n,m) = Sum m^max(k,n-k),k=0..n.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 10, 4, 1, 8, 21, 24, 5, 1, 10, 36, 72, 52, 6, 1, 12, 55, 160, 225, 112, 7, 1, 14, 78, 300, 656, 702, 232, 8, 1, 16, 105, 504, 1525, 2688, 2133, 480, 9, 1, 18, 136, 784, 3060, 7750, 10816, 6480, 976, 10, 1, 20, 171, 1152, 5537, 18576, 38875
Offset: 0

Views

Author

Keywords

Examples

			The array starts in row n=0 and column m=1 as:
    1,    1,    1,    1,    1,    1,...
    2,    4,    6,    8,   10,   12,...
    3,   10,   21,   36,   55,   78,...
    4,   24,   72,  160,  300,  504,...
    5,   52,  225,  656, 1525, 3060,...
    6,  112,  702, 2688, 7750,18576,...
    7,  232, 2133,10816,38875,111672,...
    8,  480, 6480,43520,195000,671328,...
		

Crossrefs

Cf. A107659, A107660, A014105 (3rd row), A181617 (4th row)

Formula

a(2n) = m^n(2(m^(n+1)-1)/(m-1)-1), a(2n+1) = 2m^(n+1)(m^(n+1)-1)/(m-1)
T(5,m) = m^2*(2*m^2+2*m+1). - R. J. Mathar, Aug 16 2013

A363706 a(n) is the sigma irregularity of the n-th power of a path graph of length at least 3*n.

Original entry on oeis.org

2, 14, 52, 140, 310, 602, 1064, 1752, 2730, 4070, 5852, 8164, 11102, 14770, 19280, 24752, 31314, 39102, 48260, 58940, 71302, 85514, 101752, 120200, 141050, 164502, 190764, 220052, 252590, 288610, 328352, 372064, 420002, 472430, 529620, 591852, 659414, 732602, 811720, 897080, 989002
Offset: 1

Views

Author

Allan Bickle, Jun 16 2023

Keywords

Comments

The sigma irregularity of a graph is the sum of the squares of the differences between the degrees over all edges of the graph.

Examples

			A path of length at least 3 has two edges between vertices with degrees 1 and 2. Thus a(1) = 2.
		

Crossrefs

Cf. A006325.
Cf. A011379, A181617, A270205 (sigma irregularities of maximal k-degenerate graphs).

Programs

  • Mathematica
    Table[(n^4 + 2*n^3 + 2*n^2 + n)/3, {n, 1, 40}] (* Amiram Eldar, Jul 28 2023 *)

Formula

a(n) = (n^4 + 2*n^3 + 2*n^2 + n)/3.
a(n) = 2*A006325(n+1).
G.f.: 2*x*(1 + x)^2/(1 - x)^5. - Stefano Spezia, Jul 28 2023
Showing 1-6 of 6 results.