A108678 -id:A108678 - cp's OEIS frontend

A213962 T(n,k)=Number of n X n 0..k matrices with row and column i=1..n having sum <= i*k.

2, 3, 10, 4, 42, 138, 5, 120, 3186, 5314, 6, 275, 32060, 1299735, 575122, 7, 546, 199735, 74323976, 2860343784, 176041978, 8, 980, 909900, 1839121115, 1494988224948, 34107691901433, 153143689594, 9, 1632, 3323712, 26346319032, 214090209950060

Offset: 1

R. H. Hardin Jun 28 2012

Table starts
......2..........3.............4...............5.................6
.....10.........42...........120.............275...............546
....138.......3186.........32060..........199735............909900
...5314....1299735......74323976......1839121115.......26346319032
.575122.2860343784.1494988224948.214090209950060.13186531254708756

			Some solutions for n=4 k=4
..0..1..1..0....4..0..0..0....1..3..0..0....3..0..1..0....4..0..0..0
..2..4..1..1....0..4..4..0....0..4..2..1....1..2..0..3....0..2..2..3
..0..1..2..4....0..3..4..2....0..1..0..1....0..3..1..2....0..0..1..2
..0..0..1..3....0..0..1..1....1..0..4..0....0..2..1..0....0..0..3..0

R. H. Hardin, Table of n, a(n) for n = 1..53

Row 2 is A108678

Empirical: rows n=1,2,3,4 are polynomials of degree 1,4,9,16

A098077 a(n) = n^2(n+1)(2*n+1)/3.

Original entry on oeis.org

2, 20, 84, 240, 550, 1092, 1960, 3264, 5130, 7700, 11132, 15600, 21294, 28420, 37200, 47872, 60690, 75924, 93860, 114800, 139062, 166980, 198904, 235200, 276250, 322452, 374220, 431984, 496190, 567300, 645792, 732160, 826914, 930580, 1043700

Offset: 1

Alexander Adamchuk, Oct 24 2004

Sum of all matrix elements M(i,j) = i^2 + j^2 (i,j = 1,...,n).
From Torlach Rush, Jan 05 2020: (Start)
  a(n) = n * A006331(n).
  tr(M(n)) = A006331(n).
The sum of the antidiagonal of M(n) equals tr(M(n)).
  M(n)  = M(n)' (Symmetric).
  M(1,) = M(,1) = A002522(n), n > 0.
  M(2,) = M(,2) = A087475(n), n > 0.
  M(3,) = M(,3) = A189834(n), n > 0.
  M(4,) = M(,4) = A241751(n), n > 0.
(End)
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, p and p+q. - Wesley Ivan Hurt, Apr 15 2018

			a(2) = (1^2 + 1^2) + (1^2 + 2^2) + (2^2 + 1^2) + (2^2 + 2^2) = 2 + 5 + 5 + 8 = 20.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002522, A006331, A087475, A100431, A108678, A189834, A241751.

[n^2*(n+1)*(2*n+1)/3: n in [1..40]]; // G. C. Greubel, Apr 09 2023

Table[ Sum[i^2 + j^2, {i, n}, {j, n}], {n, 35}]
LinearRecurrence[{5, -10, 10, -5, 1}, {2, 20, 84, 240, 550}, 40] (* Vincenzo Librandi, Apr 16 2018 *)

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

[n^2*(n+1)*(2*n+1)/3 for n in range(1,41)] # G. C. Greubel, Apr 09 2023

a(n) = Sum_{j=1..n} Sum_{i=1..n} (i^2 + j^2).
G.f.: 2*x*(1 + 5*x + 2*x^2)/(1-x)^5. - Colin Barker, May 04 2012
E.g.f.: (1/3)*exp(x)*x*(6 + 24*x + 15*x^2 + 2*x^3) . - Stefano Spezia, Jan 06 2020
a(n) = a(n-1) + (8*n^3 - 3*n^2 + n)/3. - Torlach Rush, Jan 07 2020
From Amiram Eldar, May 31 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/2 + 24*log(2) - 21.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/4 - 6*Pi - 6*log(2) + 21. (End)
From G. C. Greubel, Apr 09 2023: (Start)
a(n) = (1/4)*A100431(n-1).
a(n) = 2*A108678(n-1). (End)

More terms from Robert G. Wilson v, Nov 01 2004

New definition from Ralf Stephan, Dec 01 2004

A100431 Bisection of A002417.

Original entry on oeis.org

8, 80, 336, 960, 2200, 4368, 7840, 13056, 20520, 30800, 44528, 62400, 85176, 113680, 148800, 191488, 242760, 303696, 375440, 459200, 556248, 667920, 795616, 940800, 1105000, 1289808, 1496880, 1727936, 1984760, 2269200, 2583168, 2928640, 3307656, 3722320

Offset: 0

N. J. A. Sloane, Nov 20 2004

G. C. Greubel, 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, A002417, A014105, A098077, A108678.

[4*(n+1)^2*(n+2)*(2*n+3)/3: n in [0..60]]; // G. C. Greubel, Apr 09 2023

Table[4*(n+1)^2(n+2)(2n+3)/3, {n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)

[4*(n+1)^2*(n+2)*(2*n+3)/3 for n in range(61)] # G. C. Greubel, Apr 09 2023

a(n) = (4/3)*(2*n^4 + 11*n^3 + 22*n^2 + 19*n + 6). - Ralf Stephan, May 15 2007
G.f.: 8*(1 + 5*x + 2*x^2)/(1 - x)^5. - Ilya Gutkovskiy, Feb 24 2017
From G. C. Greubel, Apr 09 2023: (Start)
a(n) = (8/3)*binomial(n+2, 2)*binomial(2*n+3, 2).
a(n) = (8/3)*A000217(n+1)*A014105(n+1).
a(n) = 8*A108678(n).
a(n) = 4*A098077(n+1).
E.g.f.: (4/3)*(6 + 54*x + 69*x^2 + 23*x^3 + 2*x^4)*exp(x). (End)

More terms from Hugo Pfoertner, Nov 26 2004

A207361 Displacement under constant discrete unit surge.

Original entry on oeis.org

0, 1, 11, 53, 173, 448, 994, 1974, 3606, 6171, 10021, 15587, 23387, 34034, 48244, 66844, 90780, 121125, 159087, 206017, 263417, 332948, 416438, 515890, 633490, 771615, 932841, 1119951, 1335943, 1584038, 1867688

Offset: 0

Jonathan Vos Post, Feb 18 2012

Assume discrete times 0, 1, 2, 3, ...
Assume constant discrete unit surge (= jerk = rate of change of acceleration) starting surge(0) = 0.
Also assume acceleration(0) = velocity(0) = displacement(0) = 0.
So at t = 0, 1, 2, 3, 4, ... the acceleration = 0, 1, 2, 3, 4, ...
Then the velocity v(t) = v(t-1) + a(t)*t.
So the displacement = s(t) = s(t-1) + v(t)*t.
v(0,1,2,3,4,...) = 0, 1, 5, 14, 30, 55, 91, 140, ... = A000330(n).
The subsequence of primes is finite with three terms 11, 53, and 173.

			s(4) = s(3) + v(4)*4 =  53 +  30*4 =  53 + 120 =  173;
s(5) = s(4) + v(5)*5 = 173 +  55*5 = 173 + 275 =  448;
s(6) = s(5) + v(6)*6 = 448 +  91*6 = 448 + 546 =  994;
s(7) = s(6) + v(7)*7 = 994 + 140*7 = 994 + 980 = 1974.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000330, A101856, A108678.

a:=n->sum(sum(i^2*j,j=i..n),i=0..n): seq(a(n),n=0..30); # Robert FERREOL, May 24 2022

a[0] = 0; a[n_] := a[n] = a[n-1] + n^2*(n+1)*(2*n+1)/6; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Oct 22 2015 *)

A207361(n) := block(
        n*(1+n)*(2+n)*(1+11*n+8*n^2)/120
)$ /* R. J. Mathar, Mar 08 2012 */

a(0) = 0; for n>0, a(n) = a(n-1) + n*A000330(n) = a(n-1) + n*(0^2 + 1^2 + 2^2 + ... + n^2) = a(n-1) + n^2*(n+1)*(2*n+1)/6 = n*(1+n)*(2+n)*(1 + 11*n + 8*n^2)/120 = (2*n + 25*n^2 + 50*n^3 + 35*n^4 + 8*n^5)/120.
G.f.: x*(2*x^2+5*x+1) / (x-1)^6. - Colin Barker, May 06 2013
a(n) = Sum_{i=0..n-1} A108678(i). - J. M. Bergot, May 02 2018
a(n) = Sum_{0<=i<=j<=n} i^2*j. - Robert FERREOL, May 24 2022

cp's OEIS Frontend

A213962 T(n,k)=Number of n X n 0..k matrices with row and column i=1..n having sum <= i*k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

A098077 a(n) = n^2*(n+1)*(2*n+1)/3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A100431 Bisection of A002417.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

Extensions

A207361 Displacement under constant discrete unit surge.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Maxima

Formula

A098077 a(n) = n^2(n+1)(2*n+1)/3.