A000914 -id:A000914 - cp's OEIS frontend

A036464 Number of ways to place two nonattacking queens on an n X n board.

0, 0, 8, 44, 140, 340, 700, 1288, 2184, 3480, 5280, 7700, 10868, 14924, 20020, 26320, 34000, 43248, 54264, 67260, 82460, 100100, 120428, 143704, 170200, 200200, 234000, 271908, 314244, 361340, 413540, 471200, 534688, 604384

Offset: 1

Robert G. Wilson v, Raymond Bush (c17h21no4(AT)hotmail.com), Kirk Conely, N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-queens problem I. General theory, January 26, 2013. - _N. J. A. Sloane_, Feb 16 2013
S. Chaiken, C. R. H. Hanusa and T. Zaslavsky, A q-Queens Problem. IV. Queens, Bishops, Nightriders (and Rooks), arXiv:1609.00853 [math.CO], Sep 03 2016.
V. Kotesovec, Non-attacking chess pieces
I. Rivin, I. Vardi and P. Zimmermann, The n-queens problem, Amer. Math. Monthly, 101 (1994), 629-639.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A047659, A061994, A108792, A176186, A178721.

Column k=2 of A348129.

f:=n->n^4/2 - 5*n^3/3 + 3*n^2/2 - n/3; [seq(f(n),n=1..200)]; # N. J. A. Sloane, Feb 16 2013

f[k_] := 2 k; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[2, t[n]]
Table[a[n], {n, 2, 50}]   (* A036464 *)
Table[a[n]/4, {n, 2, 50}] (* A000914 *)
(* Clark Kimberling, Dec 31 2011 *)
CoefficientList[Series[4 x^2 (2 + x) / (1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 02 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{0,0,8,44,140},50] (* Harvey P. Dale, Mar 26 2015 *)

a(n) = C(n, 3)*(3*n-1).
G.f.: 4*x^3*(2+x)/(1-x)^5. - Colin Barker, May 02 2012
a(n) = 2*sum_{i=1..n-2} i(i + 1)^2. - Wesley Ivan Hurt, Mar 18 2014
E.g.f.: (exp(x) * x^3 * (8 + 3*x))/6. - Vaclav Kotesovec, Feb 15 2015
For n>0, a(n) = A163102(n-1) - A006331(n-1). - Antal Pinter, Sep 20 2015

A145142 Triangle T(n,k), n>=1, 0<=k<=n-1, read by rows: T(n,k)/(n-1)! is the coefficient of x^k in polynomial p_n for the n-th row sequence of A145153.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 24, 6, 11, 6, 1, 120, 144, 50, 35, 10, 1, 720, 1200, 634, 225, 85, 15, 1, 5040, 9960, 6804, 2464, 735, 175, 21, 1, 80640, 89040, 71868, 29932, 8449, 1960, 322, 28, 1, 1088640, 1231776, 789984, 375164, 112644, 25473, 4536, 546, 36, 1

Offset: 1

Alois P. Heinz, Oct 03 2008

			Triangle begins:
    1;
    0,   1;
    0,   1,   1;
    0,   2,   3,   1;
   24,   6,  11,   6,   1;
  120, 144,  50,  35,  10,  1;

Alois P. Heinz, Rows n = 1..45, flattened

T(n,k)/(n-1)! gives: A145140 / A145141.
Columns 0-9 give: A052581, A145143, A145144, A145145, A145146, A145147, A145148, A145149, A145150.
Diagonal and lower diagonals 1-3 give: A000012, A000217, A000914, A001303.
Cf. A145153, A001477, A000292, A145126, A145127, A145128, A145129, A145130.
Row sums are in A052593.

row:= proc(n) option remember; local f,i,x; f:= unapply(simplify(sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); unapply(subs(solve({seq(f(i+1)= coeftayl(x/ (1-x-x^4)/ (1-x)^i, x=0, n), i=0..n-1)}, {seq(cat(a||i), i=0..n-1)}), sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); end: T:= (n,k)-> `if`(k<0 or k>=n,0, coeff(row(n)(x),x,k)*(n-1)!): seq(seq(T(n,k), k=0..n-1), n=1..12);

row[n_] := Module[{f, eq}, f = Function[x, Sum[a[k]*x^k, {k, 0, n-1}]]; eq = Table[f[k+1] == SeriesCoefficient[x/(1-x-x^4)/(1-x)^k, {x, 0, n}], {k, 0, n-1}]; Table[a[k], {k, 0, n-1}] /. Solve[eq] // First]; Table[row[n]*(n-1)!, {n, 1, 12}] // Flatten (* Jean-François Alcover, Feb 04 2014, after Alois P. Heinz *)

See program.

A112486 Coefficient triangle for polynomials used for e.g.f.s for unsigned Stirling1 diagonals.

Original entry on oeis.org

1, 1, 1, 2, 5, 3, 6, 26, 35, 15, 24, 154, 340, 315, 105, 120, 1044, 3304, 4900, 3465, 945, 720, 8028, 33740, 70532, 78750, 45045, 10395, 5040, 69264, 367884, 1008980, 1571570, 1406790, 675675, 135135, 40320, 663696, 4302216, 14777620, 29957620

Offset: 0

Wolfdieter Lang, Sep 12 2005

The k-th diagonal of |A008275| appears as the k-th column in |A008276| with k-1 leading zeros.
The recurrence, given below, is derived from (d/dx)g1(k,x) - g1(k,x)= x*(d/dx)g1(k-1,x) + g1(k-1,x), k >= 1, with input g(-1,x):=0 and initial condition g1(k,0)=1, k >= 0. This differential recurrence for the e.g.f. g1(k,x) follows from the one for unsigned Stirling1 numbers.
The column sequences start with A000142 (factorials), A001705, A112487- A112491, for m=0,...,5.
The main diagonal gives (2*k-1)!! = A001147(k), k >= 1.
This computation was inspired by the Bender article (see links), where the Stirling polynomials are discussed.
The e.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle |A008275| with k-1 leading zeros is g1(k-1,x) = exp(x)*Sum_{m=0..k-1} a(k,m)*(x^(k-1+m))/(k-1+m)!.
a(k,n) = number of lists with entries from [n] such that (i) each element of [n] occurs at least once and at most twice, (ii) for each i that occurs twice, all entries between the two occurrences of i are > i, and (iii) exactly k elements of [n] occur twice. Example: a(1,2)=5 counts 112, 121, 122, 211, 221, and a(2,2)=3 counts 1122,1221,2211. - David Callan, Nov 21 2011

			Triangle begins:
    1;
    1,    1;
    2,    5,     3;
    6,   26,    35,    15;
   24,  154,   340,   315,   105;
  120, 1044,  3304,  4900,  3465,   945;
  720, 8028, 33740, 70532, 78750, 45045, 10395;
k=3 column of |A008276| is [0,0,2,11,35,85,175,...] (see A000914), its e.g.f. exp(x)*(2*x^2/2! + 5* x^3/3! + 3*x^4/4!).

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
C. M. Bender, D. C. Brody and B. K. Meister, Bernoulli-like polynomials associated with Stirling Numbers, arXiv:math-ph/0509008 [math-ph], 2005.
Wolfdieter Lang, First 10 rows.

Cf. A112007 (triangle for o.g.f.s for unsigned Stirling1 diagonals). A112487 (row sums).

A112486 := proc(n,k)
    if n < 0 or k<0 or  k> n then
        0 ;
    elif n = 0 then
        1 ;
    else
        (n+k)*procname(n-1,k)+(n+k-1)*procname(n-1,k-1) ;
    end if;
end proc: # R. J. Mathar, Dec 19 2013

A112486 [n_, k_] := A112486[n, k] = Which[n<0 || k<0 || k>n, 0, n == 0, 1, True, (n+k)*A112486[n-1, k]+(n+k-1)*A112486[n-1, k-1]]; Table[A112486[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 05 2014, after R. J. Mathar *)

a(k, m) = (k+m)*a(k-1, m) + (k+m-1)*a(k-1, m-1) for k >= m >= 0, a(0, 0)=1, a(k, -1):=0, a(k, m)=0 if k < m.
From Tom Copeland, Oct 05 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = -(1 + t)
P(3,t) = 2 + 5 t + 3 t^2
P(4,t) = -( 6 + 26 t + 35 t^2 + 15 t^3)
P(5,t) = 24 + 154 t +340 t^2 + 315 t^3 + 105 t^4
Apparently, P(n,t) = (-1)^(n+1) PW[n,-(1+t)] where PW are the Ward polynomials A134991. If so, an e.g.f. for the polynomials is
  A(x,t) = -(x+t+1)/t - LW{-((t+1)/t) exp[-(x+t+1)/t]}, where LW(x) is a suitable branch of the Lambert W Fct. (e.g., see A135338). The comp. inverse in x (about x = 0) is B(x) = x + (t+1) [exp(x) - x - 1]. See A112487 for special case t = 1. These results are a special case of A134685 with u(x) = B(x), i.e., u_1=1 and (u_n)=(1+t) for n>0.
Let h(x,t) = 1/(dB(x)/dx) = 1/[1+(1+t)*(exp(x)-1)], an e.g.f. in x for row polynomials in t of signed A028246 , then P(n,t), is given by
(h(x,t)*d/dx)^n x, evaluated at x=0, i.e., A(x,t)=exp(x*h(u,t)*d/du) u, evaluated at u=0. Also, dA(x,t)/dx = h(A(x,t),t).
The e.g.f. A(x,t) = -v * Sum_{j>=1} D(j-1,u) (-z)^j / j! where u=-(x+t+1)/t, v=1+u, z=(1+t*v)/(t*v^2) and D(j-1,u) are the polynomials of A042977. dA/dx = -1/[t*(v-A)].(End)
A133314 applied to the derivative of A(x,t) implies (a.+b.)^n = 0^n, for (b_n)=P(n+1,t) and (a_0)=1, (a_1)=t+1, and (a_n)=t*P(n,t) otherwise. E.g., umbrally, (a.+b.)^2 = a_2*b_0 + 2 a_1*b_1 + a_0*b_2 =0. - Tom Copeland, Oct 08 2011
The row polynomials R(n,x) may be calculated using R(n,x) = 1/x^(n+1)*D^n(x), where D is the operator (x^2+x^3)*d/dx. - Peter Bala, Jul 23 2012
For n>0, Sum_{k=0..n} a(n,k)*(-1/(1+W(t)))^(n+k+1) = (t d/dt)^(n+1) W(t), where W(t) is Lambert W function. For t=-x, this gives Sum_{k>=1} k^(k+n)*x^k/k! = - Sum_{k=0..n} a(n,k)*(-1/(1+W(-x)))^(n+k+1). - Max Alekseyev, Nov 21 2019
Conjecture: row polynomials are R(n,x) = Sum_{i=0..n} Sum_{j=0..i} Sum_{k=0..j} (n+i)!*Stirling2(n+j-k,j-k)*x^k*(x+1)^(j-k)*(-1)^(n+j+k)/((n+j-k)!*(i-j)!*k!). - Mikhail Kurkov, Apr 21 2025

A344855 Number T(n,k) of permutations of [n] having k cycles of the form (c1, c2, ..., c_m) where c1 = min_{i>=1} c_i and c_j = min_{i>=j} c_i or c_j = max_{i>=j} c_i; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 4, 11, 6, 1, 0, 8, 40, 35, 10, 1, 0, 16, 148, 195, 85, 15, 1, 0, 32, 560, 1078, 665, 175, 21, 1, 0, 64, 2160, 5992, 5033, 1820, 322, 28, 1, 0, 128, 8448, 33632, 37632, 17913, 4284, 546, 36, 1, 0, 256, 33344, 190800, 280760, 171465, 52941, 9030, 870, 45, 1

Offset: 0

Alois P. Heinz, May 30 2021

The sequence of column k satisfies a linear recurrence with constant coefficients of order k*(k+1)/2 = A000217(k).

			T(4,1) = 4: (1234), (1243), (1423), (1432).
Triangle T(n,k) begins:
  1;
  0,  1;
  0,  1,    1;
  0,  2,    3,    1;
  0,  4,   11,    6,    1;
  0,  8,   40,   35,   10,    1;
  0, 16,  148,  195,   85,   15,   1;
  0, 32,  560, 1078,  665,  175,  21,  1;
  0, 64, 2160, 5992, 5033, 1820, 322, 28, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Permutation

Columns k=0-10 give: A000007, A166444, A346317, A346318, A346319, A346320, A346321, A346322, A346323, A346324, A346325.
Row sums give A187251.
Main diagonal gives A000012, lower diagonal gives A000217, second lower diagonal gives A000914.
T(n+1,n) gives A000217.
T(n+2,n) gives A000914.
T(2n,n) gives A345342.
Cf. A060281, A132393, A186366, A345341.

b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*
      b(n-j)*binomial(n-1, j-1)*ceil(2^(j-2))), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);

b[n_] := b[n] = If[n == 0, 1, Sum[Expand[x*b[n-j]*
     Binomial[n-1, j-1]*Ceiling[2^(j-2)]], {j, n}]];
T[n_] := CoefficientList[b[n], x];
Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 23 2021, after Alois P. Heinz *)

Sum_{k=1..n} k * T(n,k) = A345341(n).

For fixed k, T(n,k) ~ (2*k)^n / (4^k * k!). - Vaclav Kotesovec, Jul 15 2021

A241765 a(n) = n(n + 1)(n + 2)(3n + 17)/24.

Original entry on oeis.org

0, 5, 23, 65, 145, 280, 490, 798, 1230, 1815, 2585, 3575, 4823, 6370, 8260, 10540, 13260, 16473, 20235, 24605, 29645, 35420, 41998, 49450, 57850, 67275, 77805, 89523, 102515, 116870, 132680, 150040, 169048, 189805, 212415, 236985, 263625, 292448

Offset: 0

Bruno Berselli, Apr 28 2014

Equivalently, Sum_{i=0..n} (i+4)*A000217(i).
Sequences of the type Sum_{i=0..n} (i+k)*A000217(i):
k = 0,  A001296: 0,  1,  7, 25,  65, 140, 266, 462, ...
k = 1,  A000914: 0,  2, 11, 35,  85, 175, 322, 546, ...
k = 2,  A050534: 0,  3, 15, 45, 105, 210, 378, 630, ... (deleting two 0)
k = 3,  A215862: 0,  4, 19, 55, 125, 245, 434, 714, ...
k = 4,     a(n): 0,  5, 23, 65, 145, 280, 490, 798, ...
k = 5,  A239568: 0,  6, 27, 75, 165, 315, 546, 882, ...
Antidiagonal sums (without 0) give A034263: 1, 9, 39, 119, 294, ...
Diagonal: 1, 11, 45, 125, 280, 546, ... is A051740.
Also: k = -1 gives A050534 deleting a 0; k = -2 gives 0 followed by A059302.
After 0, partial sums of A212343 and third column of A118788.
This sequence is even related to A005286 by a(n) = n*A005286(n) - Sum_{i=0..n-1} A005286(i).

			a(7) = 4*0 + 5*1 + 6*3 + 7*6 + 8*10 + 9*15 + 10*21 + 11*28 = 798.

Bruno Berselli, 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, A005286, A118788, A212343, A227342.

Cf. similar sequences A000914, A001296, A050534, A059302, A215862, A239568 (see table in Comments lines).

/* By first comment: */ k:=4; A000217:=func; [&+[(i+k)*A000217(i): i in [0..n]]: n in [0..40]];

Maple

A241765:=n->n*(n + 1)*(n + 2)*(3*n + 17)/24; seq(A241765(n), n=0..40); # Wesley Ivan Hurt, May 09 2014

Mathematica

Table[n (n + 1) (n + 2) (3 n + 17)/24, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 5, 23, 65, 145}, 40] CoefficientList[Series[x (5 - 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)

Maxima

makelist(coeff(taylor(x*(5-2*x)/(1-x)^5, x, 0, n), x, n), n, 0, 40);

PARI

a(n)=n*(n+1)*(n+2)*(3*n+17)/24 \\ Charles R Greathouse IV, Oct 07 2015

PARI

x='x+O('x^99); concat(0, Vec(x*(5-2*x)/(1-x)^5)) \\ Altug Alkan, Apr 10 2016

Sage

[n*(n+1)*(n+2)*(3*n+17)/24 for n in (0..40)]

Formula

G.f.: x*(5 - 2*x)/(1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = A227342(A055998(n+1)).

a(n) = Sum_{j=0..n+2} (-1)^(n-j)*binomial(-j,-n-2)*S1(j,n), S1 Stirling cycle numbers A132393. - Peter Luschny, Apr 10 2016

A112742 a(n) = n^2*(n^2 - 1)/3.

Original entry on oeis.org

0, 0, 4, 24, 80, 200, 420, 784, 1344, 2160, 3300, 4840, 6864, 9464, 12740, 16800, 21760, 27744, 34884, 43320, 53200, 64680, 77924, 93104, 110400, 130000, 152100, 176904, 204624, 235480, 269700, 307520, 349184, 394944, 445060, 499800, 559440

Offset: 0

Matthew T. Cornick (maruth(AT)gmail.com), Sep 16 2005

Second derivative of the n-th Chebyshev polynomial (of the first kind) evaluated at x=1.
The second derivative at x=-1 is just (-1)^n * a(n).
The difference between two consecutive terms generates the sequence a(n+1) - a(n) = A002492(n).
Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, |q-p| and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

			a(4)=80 because
C_4(x) = 1 - 8x^2 + 8x^4,
C'_4(x) = -16x + 32x^3,
C''_4(x) = -16 + 96x^2,
C''_4(1) = -16 + 96 = 80.

Eric Weisstein's World of Mathematics, Chebyshev polynomials of the first kind
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000914, A002415, A002492.

Table[D[ChebyshevT[n, x], {x, 2}], {n, 0, 100}] /. x -> 1

a(n)=n^2*(n^2-1)/3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = (n-1)*n^2*(n+1)/3 = 4*A002415(n).
a(n) = 2*( A000914(n-1) + C(n+1,4) ). - David Scambler, Nov 27 2006
From Colin Barker, Jan 26 2012: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: 4*x^2*(1+x)/(1-x)^5. (End)
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/3. - Stefano Spezia, Dec 11 2021
a(n) = A053126(n+2) - A006324(n-1). - Yasser Arath Chavez Reyes, Feb 22 2024

A259748 a(n) = (Sum_{0

Original entry on oeis.org

0, 0, 2, 3, 0, 1, 0, 2, 6, 0, 0, 5, 0, 7, 10, 4, 0, 12, 0, 15, 14, 11, 0, 22, 0, 0, 18, 21, 0, 5, 0, 8, 22, 0, 0, 15, 0, 19, 26, 10, 0, 28, 0, 33, 30, 23, 0, 44, 0, 0, 34, 39, 0, 9, 0, 14, 38, 0, 0, 25, 0, 31, 42, 16, 0, 44, 0, 51, 46, 35, 0, 66, 0, 0, 50

Offset: 1

José María Grau Ribas, Jul 04 2015

{a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.
From Danny Rorabaugh, Oct 22 2015: (Start)
a(n)/n = 0     iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);
a(n)/n = 1/6   iff n mod 24 = 6                         (A259752);
a(n)/n = 1/4   iff n mod 24 = 8,16                      (A259751);
a(n)/n = 5/12  iff n mod 24 = 12                        (A073762);
a(n)/n = 1/2   iff n mod 24 = 14,22                     (A259750);
a(n)/n = 2/3   iff n mod 24 = 3,9,15,18,21              (A259754);
a(n)/n = 3/4   iff n mod 24 = 4,20                      (A259755);
a(n)/n = 11/12 iff n mod 24 = 0                         (A008606).
(End)

Danny Rorabaugh, Table of n, a(n) for n = 1..24000
Danny Rorabaugh, Proof of a(n)/n values for A259748

Cf. A000914,
    A259749 (n such that   a(n)=0),
    A259750 (n such that n/a(n)=2),
    A259751 (n such that n/a(n)=4),
    A259752 (n such that n/a(n)=6),
    A073762 (n such that n/a(n)=12/5),
    A259754 (n such that n/a(n)=3/2),
    A259755 (n such that n/a(n)=4/3),
    A008606 (n such that n/a(n)=12/11).

A[n_]:=Sum[a b,{a,1,n},{b,a+1,n}];Table[Mod[A[n],n],{n,1,122}]

vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015

a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.

a(24k) = 22k; a(24k+1) = 0; a(24k+2) = 0; a(24k+3) = 16k+2; a(24k+4) = 18k+3; a(24k+5) = 0; a(24k+6) = 4k+1, a(24k+7) = 0; a(24k+8) = 6k+2; a(24k+9) = 16k+6; a(24k+10) = 0; a(24k+11) = 0; a(24k+12) = 10k+5; a(24k+13) = 0; a(24k+14) = 12k+7; a(24k+15) = 16k+10; a(24k+16) = 6k+4; a(24k+17) = 0; a(24k+18) = 16k+12; a(24k+19) = 0; a(24k+20) = 18k+15; a(24k+21) = 16k+14; a(24k+22) = 12k+11; a(24k+23) = 0. - Danny Rorabaugh, Oct 22 2015

A052149 Number of nonsquare rectangles on an n X n board.

Original entry on oeis.org

0, 4, 22, 70, 170, 350, 644, 1092, 1740, 2640, 3850, 5434, 7462, 10010, 13160, 17000, 21624, 27132, 33630, 41230, 50050, 60214, 71852, 85100, 100100, 117000, 135954, 157122, 180670, 206770, 235600, 267344, 302192, 340340, 381990, 427350, 476634

Offset: 1

Ronald Arms (ron.arms(AT)stanfordalumni.org), Jan 23 2000

Partial sums of A045991 (n^3-n^2). - Jeremy Gardiner, Jun 30 2013

			a(10) = 10 * 9 * 11 * 32 / 12 = 2640.
a(5) = 170 and the sum from 1 to 5 is 15, giving 1*(15-1)=14, 2*(15-2)=26, 2*(15-3)=36, 4*(15-4)=44 and 5*(15-5)=50; adding 14+26+36+44+50=170. Do the same for each n and get a(n). - _J. M. Bergot_, Oct 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Project Euler, Sum square difference: Problem 6.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A035291, A045991, A033487, A007290, A000537, A000330, A000914.

I:=[0, 4, 22, 70, 170]; [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..45]]; // Vincenzo Librandi, Apr 28 2012

a:=n->sum(j^3-j^2, j=0..n): seq(a(n), n=1..37); # Zerinvary Lajos, May 08 2008

CoefficientList[Series[2*x*(2+x)/(1-5*x+10*x^2-10*x^3+ 5*x^4-x^5), {x,0,50}], x] (* Vincenzo Librandi, Apr 28 2012 *)
LinearRecurrence[{5,-10,10,-5,1},{0,4,22,70,170},40] (* Harvey P. Dale, Jul 30 2019 *)

a(n) = sum(k=1,n,(k-1)*k^2) \\ Michel Marcus, Nov 09 2012

a(n) = n*(n-1)*(n+1)*(3*n+2)/12.
G.f.: 2*x^2*(2+x)/(1-5*x+10*x^2-10*x^3+5*x^4-x^5). - Colin Barker, Jan 04 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 28 2012
a(n) = A033487(n-1) - A007290(n+1) starting at n=1. - J. M. Bergot, Jun 04 2012
a(n) = Sum_{k=1..n} (k-1)*k^2. - Michel Marcus, Nov 09 2012
a(n) = A000537(n) - A000330(n) = 2*A000914(n-1). - Luciano Ancora, Mar 16 2015
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=2} 1/a(n) = 81*log(3)/5 - 9*sqrt(3)*Pi/5 - 192/25.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 - 48*log(2)/5 - 318/25. (End)

A259750 Numbers that are congruent to {14, 22} mod 24.

Original entry on oeis.org

14, 22, 38, 46, 62, 70, 86, 94, 110, 118, 134, 142, 158, 166, 182, 190, 206, 214, 230, 238, 254, 262, 278, 286, 302, 310, 326, 334, 350, 358, 374, 382, 398, 406, 422, 430, 446, 454, 470, 478, 494, 502, 518, 526, 542, 550, 566, 574, 590, 598, 614, 622, 638

Offset: 1

José María Grau Ribas, Jul 04 2015

Original name: Numbers n such that n/A259748(n) = 2.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A168489, A259748.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/2 & ]

vector(100, n, 2*(6*n-(-1)^n)) \\ Altug Alkan, Oct 23 2015

Vec(2*x*(7+4*x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

A259748(a(n))/a(n) = 1/2.
a(n) = 2*A168489(n) - Danny Rorabaugh, Oct 22 2015
From Colin Barker, Aug 26 2016: (Start)
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: 2*x*(7+4*x+x^2) / ((1-x)^2*(1+x)).
(End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24 - log(2+sqrt(3))/(4*sqrt(3)). - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(1 + 6*x*exp(x) - exp(-x)). - David Lovler, Sep 06 2022

Better name from Danny Rorabaugh, Oct 22 2015

A259755 Numbers that are congruent to {4, 20} mod 24.

Original entry on oeis.org

4, 20, 28, 44, 52, 68, 76, 92, 100, 116, 124, 140, 148, 164, 172, 188, 196, 212, 220, 236, 244, 260, 268, 284, 292, 308, 316, 332, 340, 356, 364, 380, 388, 404, 412, 428, 436, 452, 460, 476, 484, 500, 508, 524, 532, 548, 556, 572, 580, 596, 604, 620, 628

Offset: 1

José María Grau Ribas, Jul 18 2015

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A007310.

Other sequences of numbers k such that A259748(k)/k equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259754.

[2*(6*n+(-1)^n-3): n in [1..60]]; // Vincenzo Librandi, Aug 27 2015

A[n_] := A[n] = Sum[a b, {a, 1,n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 3/4 &]
Table[2 (6 n + (-1)^n - 3), {n, 1, 60}] (* Bruno Berselli, Oct 23 2015 *)
LinearRecurrence[{1,1,-1},{4,20,28},60] (* Harvey P. Dale, Jul 19 2016 *)

vector(100, n, 2*(6*n+(-1)^n-3)) \\ Altug Alkan, Oct 23 2015

a(n) = 2*(6*n + (-1)^n - 3).
A259748(a(n))/a(n) = 3/4.
a(n) = 4*A007310(n). - Michel Marcus, Sep 22 2015
G.f.: 4*x*(1 + 4*x + x^2) / ((1 + x)*(1 - x)^2). - Bruno Berselli, Oct 23 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(2 + (6*x - 3)*exp(x) + exp(-x)). - David Lovler, Sep 05 2022

Better name from Danny Rorabaugh, Oct 22 2015

cp's OEIS Frontend

A036464 Number of ways to place two nonattacking queens on an n X n board.

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A145142 Triangle T(n,k), n>=1, 0<=k<=n-1, read by rows: T(n,k)/(n-1)! is the coefficient of x^k in polynomial p_n for the n-th row sequence of A145153.

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A112486 Coefficient triangle for polynomials used for e.g.f.s for unsigned Stirling1 diagonals.

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A344855 Number T(n,k) of permutations of [n] having k cycles of the form (c1, c2, ..., c_m) where c1 = min_{i>=1} c_i and c_j = min_{i>=j} c_i or c_j = max_{i>=j} c_i; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A241765 a(n) = n*(n + 1)*(n + 2)*(3*n + 17)/24.

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A112742 a(n) = n^2*(n^2 - 1)/3.

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A259748 a(n) = (Sum_{0

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A241765 a(n) = n(n + 1)(n + 2)(3n + 17)/24.