A299120 -id:A299120 - cp's OEIS frontend

A299198 a(n) = n^4/6 - 2n^3/3 - n^2/6 + 5n/3 + 1.

2, 1, 0, 5, 26, 77, 176, 345, 610, 1001, 1552, 2301, 3290, 4565, 6176, 8177, 10626, 13585, 17120, 21301, 26202, 31901, 38480, 46025, 54626, 64377, 75376, 87725, 101530, 116901, 133952, 152801, 173570, 196385, 221376, 248677, 278426, 310765, 345840, 383801, 424802, 469001

Offset: 1

Juri-Stepan Gerasimov, Feb 04 2018

			For n=2, a(2) = 1^4/6 - 2*1^3/3 - 1^2/6 + 5*1/3 + 1 = 2.

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

Cf. A001263, A067998, A077415, A299120, A299146.

List([1..50], n -> n^4/6-2*n^3/3-n^2/6+5*n/3+1); # Muniru A Asiru, Feb 04 2018

[div((n-3)*(n+1)*(n^2-2*n-2),6) for n in 1:50] |> println # Bruno Berselli, Apr 11 2018

[n^4/6-2*n^3/3-n^2/6+5*n/3+1: n in [1..50]];

seq(n^4/6-2*n^3/3-n^2/6+5*n/3+1,n=1..50); # Muniru A Asiru, Feb 04 2018

f[n_] := n^4/6 - 2 n^3/3 - n^2/6 + 5 n/3 + 1; Array[f, 50] (* or *)
CoefficientList[ Series[(-2 + 9 x - 15 x^2 + 5 x^3 - x^4)/(-1 + x)^5, {x, 0, 50}], x] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {2, 1, 0, 5, 26}, 50] (* Robert G. Wilson v, Feb 09 2018 *)

Vec(x*(2 - 9*x + 15*x^2 - 5*x^3 + x^4) / (1 - x)^5 + O(x^50)) \\ Colin Barker, Feb 05 2018

a(n) = (n - 3)*(n + 1)*(n^2 - 2*n - 2)/6 = A299120(n-1) + A299120(1-n).
From Colin Barker, Feb 05 2018: (Start)
G.f.: x*(2 - 9*x + 15*x^2 - 5*x^3 + x^4) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
(End)
E.g.f.: exp(x)*(6 + 6*x - 6*x^2 + 2*x^3 + x^4)/6. - Iain Fox, Feb 09 2018
6*a(n) = A067998(n)^2 - 5*A067998(n) + 6. - Bruno Berselli, Apr 11 2018

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45

Offset: 0

Paul Curtz, Mar 05 2023

The main diagonal is A002414.
The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
b(n+2) = A299120(n). b(n+4) = A033275(n). b(n+4) - b(n) = A002492(n).

			The rows are
  0, 0,  1,  3,  6,  10,  15,  21, ...   = A161680
  0, 1,  5, 12, 22,  35,  51,  70, ...   = A000326
  0, 2,  9, 21, 38,  60,  87, 119, ...   = A005476
  0, 3, 13, 30, 54,  85, 123, 168, ...   = A022264
  0, 4, 17, 39, 70, 110, 159, 217, ...   = A022266
  ... .
Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741).
Difference between two consecutive rows: A000290. Hence A143844.
This square array read by antidiagonals leads to the triangle
  0
  0   0
  0   1   1
  0   2   5   3
  0   3   9  12   6
  0   4  13  21  22  10
  0   5  17  30  38  35  15
  ... .

Cf. A000004, A000290, A000326, A001477, A002414, A005476, A016777, A016813, A016873, A017017, A017101, A017197, A017581, A022264, A022266, A026741, A034827, A160378, A161680, A360962.

Cf. A002492, A033275, A143844, A299120.

T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *)

a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023

# Seen as a triangle:
from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    r = Trow(n - 1)
    return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)]
for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023

Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024

cp's OEIS Frontend

A299198 a(n) = n^4/6 - 2n^3/3 - n^2/6 + 5n/3 + 1.

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A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

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cp's OEIS Frontend

A299198 a(n) = n^4/6 - 2*n^3/3 - n^2/6 + 5*n/3 + 1.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

Julia

Magma

Maple

Mathematica

PARI

Formula

A361226 Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A299198 a(n) = n^4/6 - 2n^3/3 - n^2/6 + 5n/3 + 1.

A361226 Square array T(n,k) = k((1+2n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.