A254371 -id:A254371 - cp's OEIS frontend

A163102 a(n) = n^2*(n+1)^2/2.

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418

Offset: 0

Omar E. Pol, Jul 24 2009

Row sums of triangle A163282.
Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010
If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014
Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018
Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019
Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv preprint arXiv:1609.00853 [math.CO], 2016-2020.
A. Umar, Combinatorial Results for Semigroups of Orientation-Preserving Partial Transformations, J. Int. Seq., Vol. 14 (2011), Article 11.7.5.
Eric Weisstein's World of Mathematics, Rook Complement Graph.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Column k=2 of A144084.
Cf. A000217, A000537, A000982, A035287.
Cf. A006002, A099903, A163274, A163275, A163276, A163277, A163282.
Cf. A060300, A254371.

List([0..40],n->(n*(n+1))^2/2); # Muniru A Asiru, Aug 02 2018

[n^2*(n+1)^2/2: n in [0..40]]; // Vincenzo Librandi, Mar 26 2012

CoefficientList[Series[2*x*(1+4*x+x^2)/(1-x)^5,{x,0,40}],x] (* Vincenzo Librandi, Mar 26 2012 *)

a(n)=n^2*(n+1)^2/2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011
G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011
Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012
a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A254371(n)/4 = A060300(n)/8. (End)

A060300 a(n) = (2n(n+1))^2.

Original entry on oeis.org

0, 16, 144, 576, 1600, 3600, 7056, 12544, 20736, 32400, 48400, 69696, 97344, 132496, 176400, 230400, 295936, 374544, 467856, 577600, 705600, 853776, 1024144, 1218816, 1440000, 1690000, 1971216, 2286144, 2637376, 3027600, 3459600, 3936256, 4460544, 5035536, 5664400

Offset: 0

Jason Earls, Mar 25 2001

Arises from middle column 4^2, 12^2, 24^2, ... of following triangle: :
                        3^2 +  4^2 = 5^2
                10^2 + 11^2 + 12^2 = 13^2 + 14^2
         21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2
  36^2 + 37^2 + 38^2 + 39^2 + 40^2 = 41^2 + 42^2 + 43^2 + 44^2
...

C. Stanley Ogilvy and John T. Anderson, Excursions in Number Theory, Oxford University Press, NY, 1966, pp. 90-92.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000537, A002378, A035287, A046092, A163102, A254371.

[(2*n*(n+1))^2: n in [0..30]]; // Vincenzo Librandi, Nov 18 2016

CoefficientList[Series[16 x (1 + 4 x + x^2) / (1 - x)^5, {x, 0, 33}], x] (* Vincenzo Librandi, Nov 18 2016 *)
Table[(2n(n+1))^2,{n,0,30}] (* Harvey P. Dale, Jan 19 2019 *)

a(n) = { (2*n*(n + 1))^2 } \\ Harry J. Smith, Jul 03 2009

G.f.: 16*x*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Apr 22 2012
a(n) = 4*A035287(n+1) = 4*A002378(n)^2. - Michel Marcus, May 24 2016
a(n) = 16*A000537(n) = 16*(n*(n+1)/2)^2 = 16*A000217(n)^2 = A046092(n)^2. - Bruce J. Nicholson, Jun 05 2017
a(n) = Integral_{x=1..2*n+1} (x^3-x) dx. - César Aguilera, Jun 27 2020
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: 4*x*(2 + x)*(2 + 6*x + x^2)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 2*A254371(n) = 8*A163102(n). (End)

Name corrected by Harry J. Smith, Jul 03 2009

A064716 Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different).

Original entry on oeis.org

72, 232, 288, 520, 584, 800, 808, 1096, 1152, 1224, 1312, 1600, 1664, 1744, 1800, 1872, 1960, 2248, 2312, 2384, 2592, 2600, 2824, 3328, 3392, 3528, 3600, 4112, 4176, 4328, 4624, 5120, 5328, 5408, 5904, 6056, 6120, 6272, 6352, 6408, 6568, 6920, 8080

Offset: 1

Robert G. Wilson v, Oct 13 2001

nonn

a(n) == 0 (modulo 4) since no integer == 3 (modulo 4) can be represented as the sum of two squares.
This sequence has as a subsequence 72, 288, 800, 1800, ... which is 8 * (triangular numbers)^2. Proof: If x = 8*(n(n+1)/2)^2 then x = (n(n+1))^2 + (n(n+1))^2, x+1 = ((n-1)(n+1))^2 + (n(n+2))^2 and x+2 = (n^2+n-1)^2 + (n^2+n+1)^2. See A254371 - Joshua Zucker, Nov 01 2002
From Altug Alkan, Apr 13 2016: (Start)
If n is in this sequence, so is n*(n+2). Proof:
If n is in this sequence, then n = a^2 + b^2, n+1 = c^2 + d^2, n+2 = e^2 + f^2 for a, b, c, d, e, f being nonzero integers.
So, n*(n+2) = (a^2 + b^2)*(e^2 + f^2) = (a*e + b*f)^2 + (a*f - b*e)^2. Note that a*f cannot be equal to b*e because of their definitions.
n*(n+2) + 1 = n^2 + 2*n + 1 = (n+1)^2. Since we know that n mod 4 = 0, then n+1 cannot be of the form 2*k^2, that is, c and d must be different. So (n+1)^2 is the sum of two nonzero squares because n+1 = c^2 + d^2.
n*(n+2) + 2 = (n+1)^2 + 1, that is obviously the sum of two nonzero squares.
So if n is in this sequence, then n*(n+2), n*(n+2) + 1 and n*(n+2) + 2 are the sums of two nonzero squares, that is n*(n+2) must also be member of this sequence.
Note that it can be produced by repeating of this result and n*(n+2)*(n*(n+2)+2)*(n*(n+2)*(n*(n+2)+2)+2)... is always a member, if n is a member. (End)
For k > 0, 25*k^2*(10*k+2)^2 and 8*A001080(k)^2 are terms. - Jinyuan Wang, Feb 23 2019

			72 = 6^2 + 6^2, 73 = 3^2 + 8^2, 74 = 5^2 + 7^2.

Robert Israel, Table of n, a(n) for n = 1..10000
W. Allen Whitworth, Problem 356, The Mathematical Gazette, Vol. 1, No. 20 (Mar. 1900), p. 338.

Cf. A000404, A001080.

Cf. A254371 \ {0, 8} (a subsequence).

N:= 10000: # to get all terms <= N
S:= {seq(seq(a^2+b^2, b=1..floor(sqrt(N+2-a^2))),a=1..floor(sqrt(N+2)))}:
sort(convert(S intersect map(`-`,S,1) intersect map(`-`,S,2),list)); # Robert Israel, Apr 14 2016

a = Table[n^2, {n, 1, 100}]; c = {}; Do[ c = Append[c, a[[i]] + a[[j]]], {i, 1, 100}, {j, 1, i} ]; c = Union[c]; c[[ Select[ Range[ Length[c] - 2], c[[ # ]] + 2 == c[[ # + 2 ]] & ]]]
Select[Range@ 8080, AllTrue[# + {0, 1, 2}, Length[ PowersRepresentations[#, 2, 2] /. {0, } -> Nothing] > 0 &] &] (* _Michael De Vlieger, Apr 13 2016, Version 10 *)

is(n)= for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2));
lista(nn) = {for(n=1,nn,if(is(n)==1&&is(n+1)==1&&is(n+2)==1,print1(n,", ")))}; \\ Jinyuan Wang, Feb 23 2019

A369585 Table read by rows. T(n, k) = [z^k] h(n, 1, z) where h(n, v, z) are the modified Lommel polynomials (A369117).

Original entry on oeis.org

1, 0, 2, -1, 0, 8, 0, -8, 0, 48, 1, 0, -72, 0, 384, 0, 18, 0, -768, 0, 3840, -1, 0, 288, 0, -9600, 0, 46080, 0, -32, 0, 4800, 0, -138240, 0, 645120, 1, 0, -800, 0, 86400, 0, -2257920, 0, 10321920, 0, 50, 0, -19200, 0, 1693440, 0, -41287680, 0, 185794560

Offset: 0

Peter Luschny, Jan 30 2024

			The list of coefficients starts:
  [0]  1
  [1]  0,   2
  [2] -1,   0,    8
  [3]  0,  -8,    0,   48
  [4]  1,   0,  -72,    0,   384
  [5]  0,  18,    0, -768,     0,    3840
  [6] -1,   0,  288,    0, -9600,       0,    46080
  [7]  0, -32,    0, 4800,     0, -138240,        0, 645120
  [8]  1,   0, -800,    0, 86400,       0, -2257920,      0, 10321920

David Dickinson, On Lommel and Bessel polynomials, Proc. AMS 5 (1954) 946-956.
Eric Weisstein's World of Mathematics, Lommel Polynomial.

Diagonals include: A000165 (main diagonal), A014479, A286725.
Columns include bisections of: A001105, A254371.
Cf. A093985 (row sums), A036243 (abs row sums), A369117.

p := proc(n,  x) option remember; if n = -1 then 0 elif n = 0 then 1 else
2*n*z*p(n - 1, z) - p(n - 2, z) fi end:
seq(seq(coeff(p(n, z), z, k), k = 0..n), n = 0..9);

Table[CoefficientList[Expand[ResourceFunction["LommelR"][n, 1, 1/z]], z], {n, 0, 8}] // MatrixForm

T(n, k) = [z^k] 2*n*z*p(n-1, z) - p(n-2, z) where p(-1, z) = 0 and p(0, z) = 1.

T(n, k) = (-1)^k * [z^k] h(n, -n, z) where h(n, v, z) are the modified Lommel polynomials (A369117).

cp's OEIS Frontend

A163102 a(n) = n^2*(n+1)^2/2.

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A060300 a(n) = (2*n*(n+1))^2.

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A064716 Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different).

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PARI

A369585 Table read by rows. T(n, k) = [z^k] h(n, 1, z) where h(n, v, z) are the modified Lommel polynomials (A369117).

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Author

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Examples

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Crossrefs

Programs

Maple

Mathematica

Formula

A060300 a(n) = (2n(n+1))^2.