A229183 -id:A229183 - cp's OEIS frontend

A061989 Number of ways to place 3 nonattacking queens on a 3 X n board.

0, 0, 0, 0, 4, 14, 36, 76, 140, 234, 364, 536, 756, 1030, 1364, 1764, 2236, 2786, 3420, 4144, 4964, 5886, 6916, 8060, 9324, 10714, 12236, 13896, 15700, 17654, 19764, 22036, 24476, 27090, 29884, 32864, 36036, 39406, 42980, 46764, 50764

Offset: 0

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), May 29 2001

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Vaclav Kotesovec, Ways of placing non-attacking queens and kings..., part of "Between chessboard and computer", 1996, pp. 204 - 206.
E. Lucas, Recreations mathematiques I, Albert Blanchard, Paris, 1992, p. 231.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A061990, A117937, A229183.

Essentially the same as A079908.

[0,0,0] cat [(n-3)*(n^2-6*n+12): n in [3..50]]; // G. C. Greubel, Apr 29 2022

A061989 := proc(n)
    if n >= 3 then
        (n-3)*(n^2-6*n+12) ;
    else
        0;
    end if;
end proc:
seq(A061989(n),n=0..30) ; # R. J. Mathar, Aug 16 2019

CoefficientList[Series[2*x^4*(2-x+2*x^2)/(1-x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, May 02 2013 *)

[0,0,0]+[(n-3)*((n-3)^2 +3) for n in (3..50)] # G. C. Greubel, Apr 29 2022

G.f.: 2*x^4*(2-x+2*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n >= 7.
Explicit formula (H. Tarry, 1890): a(n) = (n-3)*(n^2-6*n+12), n >= 3.
(4, 14, 36, ...) is the binomial transform of row 4 of A117937: (4, 10, 12, 6). - Gary W. Adamson, Apr 09 2006
a(n) = 2*A229183(n-3). - R. J. Mathar, Aug 16 2019
E.g.f.: 36 + 14*x + 2*x^2 + (-36 + 22*x - 6*x^2 + x^3)*exp(x). - G. C. Greubel, Apr 29 2022

A270109 a(n) = n^3 + (n+1)*(n+2).

Original entry on oeis.org

2, 7, 20, 47, 94, 167, 272, 415, 602, 839, 1132, 1487, 1910, 2407, 2984, 3647, 4402, 5255, 6212, 7279, 8462, 9767, 11200, 12767, 14474, 16327, 18332, 20495, 22822, 25319, 27992, 30847, 33890, 37127, 40564, 44207, 48062, 52135, 56432, 60959, 65722, 70727, 75980, 81487, 87254

Offset: 0

Bruno Berselli, Mar 11 2016, at the suggestion of Giuseppe Amoruso in BASE Cinque forum

For n>1, many consecutive terms of the sequence are generated by floor(sqrt(n^2 + 2)^3) + n^2 + 2.
It appears that this is a subsequence of A000037 (the nonsquares).
The primes in the sequence belong to A045326.
Inverse binomial transform is 2, 5, 8, 6, 0, 0, 0, ... (0 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
MathsSmart, Number pattern and Puzzle - 7, 20, 47, 94, 167.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Subsequence of A001651, A047212.
Cf. A000037, A045326.
Cf. A027444: numbers of the form n^3+n*(n+1); A085490: numbers of the form n^3+(n-1)*n.
Cf. A008865: numbers of the form n+(n+1)*(n+2); A130883: numbers of the form n^2+(n+1)*(n+2).

Magma
```
[n^3+(n+1)*(n+2): n in [0..50]];
```

Table[n^3 + (n + 1) (n + 2), {n, 0, 50}]

Maxima
```
makelist(n^3+(n+1)*(n+2), n, 0, 50);
```
PARI
```
vector(50, n, n--; n^3+(n+1)*(n+2))
```
Sage
```
[n^3+(n+1)*(n+2) for n in (0..50)]
```

O.g.f.: (2 - x + 4*x^2 + x^3)/(1 - x)^4.
E.g.f.: (2 + x)*(1 + x)^2*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3.
a(n+h) - a(n) + a(n-h) = n^3 + n^2 + (6*h^2+3)*n + (2*h^2+2) for any h. This identity becomes a(n) = n^3 + n^2 + 3*n + 2 if h=0.
a(h*a(n) + n) = (h*a(n))^3 + (3*n+1)*(h*a(n))^2 + (3*n^2+2*n+3)*(h*a(n)) + a(n) for any h, therefore a(h*a(n) + n) is always a multiple of a(n).
a(n) + a(-n) = 2*A059100(n) = A255843(n).
a(n) - a(-n) = 4*A229183(n).

A153976 a(n) = n^3 + (n+2)^3.

Original entry on oeis.org

8, 28, 72, 152, 280, 468, 728, 1072, 1512, 2060, 2728, 3528, 4472, 5572, 6840, 8288, 9928, 11772, 13832, 16120, 18648, 21428, 24472, 27792, 31400, 35308, 39528, 44072, 48952, 54180, 59768, 65728, 72072, 78812, 85960, 93528, 101528, 109972, 118872, 128240

Offset: 0

Vladimir Joseph Stephan Orlovsky, Jan 03 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..765
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000537, A000578, A003215, A005898, A006007, A027602.

[n^3+(n+2)^3: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

f[n_]:=n^3;lst={};Do[AppendTo[lst,(f[n]+f[n+2])],{n,0,6!}];lst
Array[#^3+(#+2)^3&,40,0] (* or *) LinearRecurrence[{4,-6,4,-1},{8,28,72,152},40] (* Harvey P. Dale, Aug 02 2011 *)

def a(n): return n**3 + (n+2)**3
print([a(n) for n in range(40)]) # Michael S. Branicky, Aug 28 2021

For n>3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 02 2011
G.f.: 4*( 2-x+2*x^2 ) / (x-1)^4 . - R. J. Mathar, Apr 11 2016
a(n) = 4*A229183(n+1). - R. J. Mathar, Apr 11 2016

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 26 2011

A271636 a(n) = 4n(4*n^2 + 3).

Original entry on oeis.org

0, 28, 152, 468, 1072, 2060, 3528, 5572, 8288, 11772, 16120, 21428, 27792, 35308, 44072, 54180, 65728, 78812, 93528, 109972, 128240, 148428, 170632, 194948, 221472, 250300, 281528, 315252, 351568, 390572, 432360, 477028, 524672, 575388, 629272, 686420

Offset: 0

Vincenzo Librandi, Apr 11 2016

This is the case h=0 of the identity 4*n*(4*n^2 + 3*(2*h + 1)^2) = (2*n - 2*h - 1)^3 + (2*n + 2*h + 1)^3.

Subsequence of A004999 and, after 0, second bisection of A153976.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A004999, A153976, A229183.

Magma
```
[4*n*(4*n^2+3): n in [0..50]];
```
Mathematica
```
Table[4 n (4 n^2 + 3), {n, 0, 50}]
```

x='x+O('x^99); concat(0, Vec(x*(28+40*x+28*x^2)/(1-x)^4)) \\ Altug Alkan, Apr 11 2016

for n in range(0,1000):print(4*n*(4*n**2+3)) # Soumil Mandal, Apr 11 2016

O.g.f.: 4*x*(7 + 10*x + 7*x^2)/(1 - x)^4.
E.g.f.: 4*x*(7 + 12*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, Apr 11 2016
a(n) = -a(-n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 4*A229183(2*n). - Bruno Berselli, Apr 11 2016

Edit and extended by Bruno Berselli, Apr 12 2016

A342940 Triangle read by rows: T(n, k) is the Skolem number of the parallelogram graph P_{n, k}, with 1 < k <= n.

Original entry on oeis.org

2, 3, 4, 4, 6, 8, 5, 8, 11, 14, 6, 10, 14, 18, 22, 7, 12, 17, 22, 27, 32, 8, 14, 20, 26, 32, 38, 44, 9, 16, 23, 30, 37, 44, 51, 58, 10, 18, 26, 34, 42, 50, 58, 66, 74, 11, 20, 29, 38, 47, 56, 65, 74, 83, 92, 12, 22, 32, 42, 52, 62, 72, 82, 92, 102, 112, 13, 24, 35, 46, 57, 68, 79, 90, 101, 112, 123, 134

Offset: 2

Stefano Spezia, Mar 30 2021

For the meaning of Skolem number of a graph, see Definitions 1.4 and 1.5 in Carrigan and Green.

			The triangle T(n, k) begins:
n\k|   2   3   4   5   6   7
---+------------------------
2  |   2
3  |   3   4
4  |   4   6   8
5  |   5   8  11  14
6  |   6  10  14  18  22
7  |   7  12  17  22  27  32
...

Braxton Carrigan and Garrett Green, Skolem Number of Subgraphs on the Triangular Lattice, Communications on Number Theory and Combinatorial Theory 2 (2021), Article 2.

Cf. A014206, A229183, A342938, A342939.

For n > 1, 3*A002061(n) gives the Skolem number of the hexagonal grid graph H_n.

T[n_,k_]:=k*n-2k-n+4; Table[T[n,k],{n,2,13},{k,2,n}]//Flatten

O.g.f.: (4 - 6*y - x*(5 - 8*y))/((1 - x)^2*(1 - y)^2).
E.g.f.: exp(x+y)*(4 - x*(1 - y) - 2*y).
T(n, k) = k*n - 2*k - n + 4 (see Theorem 3.3 in Carrigan and Green).
Sum_{k=2..n} T(n, k) = A229183(n-1).
T(n, n) = A014206(n-2).

A275740 Sums of the next n consecutive nonsquare integers.

Original entry on oeis.org

0, 2, 8, 21, 46, 83, 136, 210, 306, 426, 575, 758, 972, 1223, 1519, 1855, 2236, 2669, 3156, 3694, 4290, 4956, 5678, 6467, 7332, 8269, 9278, 10368, 11548, 12804, 14148, 15593, 17126, 18753, 20485, 22325, 24262, 26308, 28481, 30756, 33148

Offset: 0

Olivier Gérard, Aug 07 2016

Row sums of nonsquare integers (A000037), seen as a regular triangle:
.
2   | 2,
8   | 3,  5,
21  | 6,  7,  8,
46  | 10, 11, 12, 13,
83  | 14, 15, 17, 18, 19,
136 | 20, 21, 22, 23, 24, 26,
210 | 27, 28, 29, 30, 31, 32, 33,
306 | 34, 35, 37, 38, 39, 40, 41, 42,
...
The equivalent for all integers are A006003 (starting from 1), A229183 (starting from 2) and A027480 (starting from 0).
There are several sequences close to nonsquares whose sum of groups of n terms starts like this sequence, notably A028761, A158276, A167759.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000037, A006003, A229183, A027480.

R:= 0: s:= 1:
for n from 1 to 100 do
  if floor(sqrt(s+n)) = floor(sqrt(s)) then
    R:= R, n*s + n*(n+1)/2; s:= s+n;
  else
    R:= R, n*s + n*(n+1)/2 - floor(sqrt(s+n))^2 + s+n+1; s:= s+n+1;
  fi
od:
R; # Robert Israel, Oct 02 2022

Table[Sum[
  i + Floor[1/2 + Sqrt[i]], {i, n (n - 1)/2 + 1, (n + 1) (n)/2}], {n,
  0, 40}]
Join[{0},Module[{nn=1000,nsi,len},nsi=Select[Range[nn],!IntegerQ[Sqrt[#]]&];len=Floor[ (Sqrt[ 8*Length[nsi]+1]-1)/2];Total/@TakeList[nsi,Range[len]]]] (* Harvey P. Dale, Jan 04 2024 *)

Definition clarified by Harvey P. Dale, Jan 04 2024

A267707 a(n) = A000217(A000217(n)+1).

Original entry on oeis.org

1, 3, 10, 28, 66, 136, 253, 435, 703, 1081, 1596, 2278, 3160, 4278, 5671, 7381, 9453, 11935, 14878, 18336, 22366, 27028, 32385, 38503, 45451, 53301, 62128, 72010, 83028, 95266, 108811, 123753, 140185, 158203, 177906, 199396, 222778, 248160, 275653, 305371

Offset: 0

Waldemar Puszkarz, Jan 19 2016

It is the sequence of triangular numbers (A000217) with progressive gaps that grow as 0,1,2,3, ... (consecutive numbers), by which I mean that the 0,1,2,3, ... consecutive triangular numbers are removed from A000217 to form this sequence. For instance, (1), 6, a triangular number, is missing between 3 and 10, which is the gap with 1 triangular number removed, (2), 15 and 21 (two consecutive triangular numbers) are missing between 10 and 28, which is the gap with 2 triangular numbers removed, and so on.
The differences between the consecutive terms of this sequence can be expressed through the sum of cubes of two numbers separated by 2 as (n^3+(n+2)^3)/4, which is the same as A229183, except for the first term in there.
The same pattern when applied to squares, A000290(A000290(n)+1), gives A082044(n). Triangular numbers are also linked in a similar manner to A027927(n) = A000217(A000217(n)+2)/3.

			For n=0, a(0)=1*2/2=1. For n=2, a(2)=4*5/2=10.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217 (triangular numbers), A229183 (consecutive terms differences), A082044 (related sequence for squares), A027927 (related sequence for triangular numbers).

I:=[1,3,10,28,66]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 22 2016

S[n_] :=n*(n+1)/2; Table[S[S[n]+1], {n, 0, 50}]
Table[(n*(n+1)/2+1)(n*(n+1)/2+2)/2, {n, 0, 50}]
Table[(n^4+2*n^3+7*n^2+6*n+8)/8, {n, 0, 50}]
CoefficientList[Series[(1 - 2 x + 5 x^2 - 2 x^3 + x^4) / (1 - x)^5, {x, 0, 33}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 3, 10, 28, 66}, 50] (* Vincenzo Librandi, Jan 22 2016 *)

for(n=0,50,print1((n^4+2*n^3+7*n^2+6*n+8)/8 ", "))

a(n) = A000217(A000217(n)+1) = (n*(n+1)/2+1)(n*(n+1)/2+2)/2.
a(n) = (n^4+2n^3+7n^2+6n+8)/8  = (n^2+n+2)(n^2+n+4)/8.
G.f.: (1-2*x+5*x^2-2*x^3+x^4)/(1-x)^5. - Vincenzo Librandi, Jan 22 2016
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Vincenzo Librandi, Jan 22 2016

cp's OEIS Frontend

A061989 Number of ways to place 3 nonattacking queens on a 3 X n board.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

A270109 a(n) = n^3 + (n+1)*(n+2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Maxima

PARI

Sage

Formula

A153976 a(n) = n^3 + (n+2)^3.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

Python

Formula

Extensions

A271636 a(n) = 4*n*(4*n^2 + 3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Python

Formula

Extensions

A342940 Triangle read by rows: T(n, k) is the Skolem number of the parallelogram graph P_{n, k}, with 1 < k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A275740 Sums of the next n consecutive nonsquare integers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A267707 a(n) = A000217(A000217(n)+1).

Views

Author

Keywords

Comments

A271636 a(n) = 4n(4*n^2 + 3).