A191747 -id:A191747 - cp's OEIS frontend

A191748 Sequence of all m in {1,2,3,...} such that A191747(m) = 1.

1, 2, 5, 6, 10, 14, 15, 20, 25, 30, 31, 37, 43, 49, 55, 56, 63, 70, 77, 84, 91, 92, 100, 108, 116, 124, 132, 140, 141, 150, 159, 168, 177, 186, 195, 204, 205, 215, 225, 235, 245, 255, 265, 275, 285, 286, 297, 308, 319, 330, 341, 352, 363, 374, 385, 386, 398

Offset: 0

L. Edson Jeffery, Jun 29 2011

nonn

Note that A191747={1,1,0,0,1,1,0,0,0,1,0,0,0,1,...} is the sequence formed by concatenation of the row entries of successive N X N identity matrices, N=1,2,....
This sequence is read from the antidiagonals of the table
T(n,k)=
   1,   5,  14,   30,   55,  ..
   2,  10,  25,   49,   84,  ..
   6,  20,  43,   77,  124,  ..
  15,  37,  70,  116,  177,  ..
  31,  63, 100,  168,  245,  ..
  ...
  in which the n-th row is found from the n-th generating function (-n+(2*n+1)*x-(n-1)*x^2)/(1-x)^4, n in {0,1,2,...}, by taking the (n+1)-th term on, and, similarly, the k-th column is found from the k-th generating function  (2*k+1-(5*k+2)*x+4*(k+1)*x^2-(k+1)*x^3)/(1-x)^4, k in {0,1,2,...}, by taking the k-th term on. For the first three rows, n=0 gives the core sequence A000330, n=1 gives essentially A058373, ignoring the two initial zeros, and n=2 gives -A058372. The first column, for k=0, is A056520, where it is known that A056520(m)=A000330(m)+1. Thus a trivial relation, A191748(m,j)=A056520(m)+j*(m+2)=A000330(m)+j*(m+2)+1, j in {0,...,m}, m>0, with A191748(0,0)=1, gives the triangle
   1
   2,  5,
   6, 10, 14,
   ...
However, the j-th row R_j of the table is given by R_j(n)=(n+1)*(2*n^2+n-6*j)/6, n=j+1,j+2,j+3,..., and the k-th column C_k by C_k(n)=(n+2)*(2*n^2-n+6*k+3)/6, n=k,k+1,k+2,..., with j,k in {0,1,...}. Substituting n+k for n in the second formula (to account for varying offsets) gives the formula for T(n,k) below.

Cf. A000330, A056520, A058372, A058373, A191747.

For the table: T(n,k) = (n+k+2)*(2*(n+k)^2-n+5*k+3)/6, n,k=0,1,2,....

A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

Original entry on oeis.org

1, 2, 0, 0, 2, 3, 0, 0, 0, 3, 0, 0, 0, 3, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 7, 0

Offset: 1

Julia Zimmerman, Nov 10 2020

			For every identity matrix of size n starting with n=1, append n*(each entry of each row of the matrix), e.g., n=1 -> 1, n=2 -> 2,0,0,2, so the first 5 terms of the sequence are 1,2,0,0,2.

Cf. A191747 (with 1's), A191748 (locations of nonzero terms), A056520 (start location of each matrix).

import numpy as np
def n_id_sequence(iterations):
    sequence = []
    for i in range(1,iterations+1):
        matrix = i*(np.identity(i))
        for row in matrix:
            for entry in row:
                sequence.append(int(entry))
    return sequence

cp's OEIS Frontend

A191748 Sequence of all m in {1,2,3,...} such that A191747(m) = 1.

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A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

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