A348845 -id:A348845 - cp's OEIS frontend

A350051 Part three of the trisection of A017101: a(n) = 19 + 24*n.

19, 43, 67, 91, 115, 139, 163, 187, 211, 235, 259, 283, 307, 331, 355, 379, 403, 427, 451, 475, 499, 523, 547, 571, 595, 619, 643, 667, 691, 715, 739, 763, 787, 811, 835, 859, 883, 907, 931, 955, 979, 1003, 1027, 1051, 1075

Offset: 0

Wolfdieter Lang, Dec 11 2021

The trisection of A017101 = {3 + 8*k}A017077%20=%20%7B3*(1%20+%2012*n)%7D">{k>=0} gives 3*A017077 = {3*(1 + 12*n)}{n>=0}, {A348845(n)}{n >= 0} and {a(n)}{n>=0}. These three sequences are congruent to 3 modulo 8 and to 3, 5, and 1 modulo 6, respectively.

Winston de Greef, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008606, 3*A017077, A017101, A348845.

24 * Range[0, 44] + 19 (* Amiram Eldar, Dec 18 2021 *)

a(n) = 19 + 24*n \\ Winston de Greef, Jan 28 2024

a(n) = 19 + 24*n = 19 + A008606(n), for n >= 0
a(n) = 2*a(n-1) - a(n-2), for n >= 1, with a(-1) = -5, a(0) = 19.
G.f.: (19 + 5*x)/(1-x)^2.
E.g.f.: (19 + 24*x)*exp(x).

A360962 Square array T(n,k) = k((3+6n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 1, 0, 4, 5, 0, 7, 17, 12, 0, 10, 29, 39, 22, 0, 13, 41, 66, 70, 35, 0, 16, 53, 93, 118, 110, 51, 0, 19, 65, 120, 166, 185, 159, 70, 0, 22, 77, 147, 214, 260, 267, 217, 92, 0, 25, 89, 174, 262, 335, 375, 364, 284, 117, 0, 28, 101, 201, 310, 410, 483, 511, 476, 360, 145

Offset: 0

Paul Curtz, Feb 27 2023

The main diagonal is A024394.

The antidiagonals sums are A000537.

			The rows are:
  0  1  5  12  22  35  51  70 ... = A000326
  0  4 17  39  70 110 159 217 ... = A022266
  0  7 29  66 118 185 267 364 ... = A022272
  0 10 41  93 166 260 375 511 ... = A022278
  0 13 53 120 214 335 483 658 ... = A022284
  ... .
Columns: A000004, A016777, A017581, A154266=3*A017209, 2*A348845, 5*A161447, 3*A158057(n+1), ... (coefficients from A026741).
Difference between two consecutive rows are: A033428.
This square array read by antidiagonals leads to the triangle
  0
  0  1
  0  4  5
  0  7 17 12
  0 10 29 39  22
  0 13 41 66  70  35
  0 16 53 93 118 110 51
  ... .

Cf. A000326, A000537, A022266, A022272, A022278, A022284, A024394.
Cf. A033428.
Cf. A000004, A016777, A017209, A017581, A026741, A154266, A158057, A161447, A348845.

T:= (n,k)-> k*(k*(3+6*n)-1)/2:
seq(seq(T(d-k,k), k=0..d), d=0..10);  # Alois P. Heinz, Feb 28 2023

T[n_, k_] := ((6*n + 3)*k - 1)*k/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Feb 27 2023 *)

T(n,k) = k*((3+6*n)*k-1)/2; \\ Michel Marcus, Feb 27 2023

Take successively sequences n*(3*n-1)/2, n*(9*n-1)/2, n*(15*n-1)/2, n*(21*n-1)/2, ... listed in the EXAMPLE section.
From Stefano Spezia, Feb 21 2024: (Start)
G.f.: y*(1 + 2*y + x*(2 + y))/((1 - x)^2*(1 - y)^3).
E.g.f.: exp(x+y)*y*(2 + 3*y + 6*x*(1 + y))/2. (End)

cp's OEIS Frontend

A350051 Part three of the trisection of A017101: a(n) = 19 + 24*n.

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A360962 Square array T(n,k) = k((3+6n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

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cp's OEIS Frontend

A350051 Part three of the trisection of A017101: a(n) = 19 + 24*n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A360962 Square array T(n,k) = k*((3+6*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A360962 Square array T(n,k) = k((3+6n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.