A345018 For each n, append to the sequence n^2 consecutive integers, starting from n.

1, 2, 3, 4, 5, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41

Offset

1,2

Comments

Irregular triangle read by rows T(n,k) in which row n lists the integers from n to n + n^2 - 1, with n >= 1.

Links

Paolo Xausa, Table of n, a(n) for n = 1..10416 (rows 1..31 of the triangle, flattened)

Formula

T(n,k) = n + k - 1, with n >= 1 and 1 <= k <= n^2.

Example

Written as an irregular triangle T(n,k) the sequence begins:
  n\k|  1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16 ...
  ---+---------------------------------------------------------------
   1 |  1;
   2 |  2,  3,  4,  5;
   3 |  3,  4,  5,  6,  7,  8,  9, 10, 11;
   4 |  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19;
  ...

Maple

T:= n-> (t-> seq(n+i, i=0..t-1))(n^2):
seq(T(n), n=1..6);  # Alois P. Heinz, Nov 05 2024

Mathematica

Table[Range[n, n^2+n-1], {n, 6}] (* Paolo Xausa, Sep 05 2023 *)

Prog

(PARI) row(n) = vector(n^2, k, n+k-1); \\ Michel Marcus, Jun 08 2021
(Python)
from sympy import integer_nthroot
def A345018(n): return n-1+(k:=(m:=integer_nthroot(3*n, 3)[0])+(6*n>m*(m+1)*((m<<1)+1)))*(k*(3-(k<<1))+5)//6 # Chai Wah Wu, Nov 05 2024

Crossrefs

Column 1: A000027.

Right border: A028387.

Row lengths: A000290.

Row sums: A255499.

Cf. A064866, A074279.

Sequence in context: A279850 A117607 A215092 * A376840 A377488 A309079

Adjacent sequences: A345015 A345016 A345017 * A345019 A345020 A345021

Keyword

nonn,tabf

Author

Paolo Xausa, Jun 05 2021

Status

approved