A167991 -id:A167991 - cp's OEIS frontend

A167381 The numbers read down the left-center column of an arrangement of the natural numbers in square blocks.

1, 3, 6, 10, 14, 18, 23, 29, 35, 41, 47, 53, 60, 68, 76, 84, 92, 100, 108, 116, 125, 135, 145, 155, 165, 175, 185, 195, 205, 215, 226, 238, 250, 262, 274, 286, 298, 310, 322, 334, 346, 358, 371, 385, 399, 413, 427, 441, 455, 469, 483, 497, 511, 525, 539, 553

Offset: 1

Paul Curtz, Nov 02 2009

The natural numbers are filled into square blocks of edge length 2, 4, 6, 8, ...
by taking A016742(n+1) = 4, 16, 36, ... at a time:
.......1..2......
.......3..4......
....5..6..7..8...
....9.10.11.12...
...13.14.15.16...
...17.18.19.20...
21.22.23.24.25.26
27.28.29.30.31.32
33.34.35.36.37.38
39.40.41.42.43.44
Reading down the column just left from the center yields a(n).
The length of the rows is given by A001670.
The number of elements in each square block, 4, 16, 36, etc., are the first differences of A166464:
A016742(n) = A166464(n)-A166464(n-1).
Reading the blocks from right to left, row by row, we obtain a permutation of the integers, which starts similar to A166133.

Harvey P. Dale, Table of n, a(n) for n = 1..10000

Cf. A113127, A167991 (first differences).

r[1] = Range[4]; r[n_] := r[n] = Range[r[n-1][[-1]]+1, r[n-1][[-1]]+(2n)^2 ];
s[n_] := Partition[r[n], Sqrt[Length[r[n]]]][[All, n]];
A167381 = Table[s[n], {n, 1, 7}] // Flatten (* Jean-François Alcover, Mar 26 2017 *)
Module[{nn=7,c},c=TakeList[Range[(2/3)*nn(nn+1)(2*nn+1)],(2*Range[ nn])^2]; Table[Take[c[[n]],{n,-1,2*n}],{n,nn}]]//Flatten (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 18 2018 *)

Edited by R. J. Mathar, Aug 29 2010

More terms from Jean-François Alcover, Mar 26 2017

A284359 Double triangle (2n+2 terms by row). Every row is 2n + 1 followed by 2n + 1 times 2n + 2.

Original entry on oeis.org

1, 2, 3, 4, 4, 4, 5, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8, 8, 8, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17

Offset: 0

Paul Curtz, Mar 25 2017

In essence the same as A167991. - R. J. Mathar, Mar 27 2017

			1,  2,
3,  4,  4,  4,
5,  6,  6,  6,  6,  6,
7,  8,  8,  8,  8,  8,  8,  8,
9, 10, 10, 10, 10, 10, 10, 10, 10, 10,
... .
The row sum is A000466(n+1).

Cf. A000466, A005408, A103517 (main diagonal), A167381.

Table[2 n + 2 - Boole[k == 1], {n, 0, 8}, {k, 2 n + 2}] // Flatten (* Michael De Vlieger, Mar 25 2017 *)

for(n=0, 10, for(k=1, 2*n + 2, print1(2*n + 2 - (k==1), ", ");); print();) \\ Indranil Ghosh, Mar 26 2017, translated from Mathematica code

for n in range(0, 11):
    print([2*n + 2 -(k==1) for k in range(1, 2*n + 3)])
# Indranil Ghosh, Mar 26 2017

a(n) = A167381(n+1) - A167381(n).

cp's OEIS Frontend

A167381 The numbers read down the left-center column of an arrangement of the natural numbers in square blocks.

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A284359 Double triangle (2n+2 terms by row). Every row is 2n + 1 followed by 2n + 1 times 2n + 2.

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cp's OEIS Frontend

A167381 The numbers read down the left-center column of an arrangement of the natural numbers in square blocks.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Extensions

A284359 Double triangle (2*n+2 terms by row). Every row is 2*n + 1 followed by 2*n + 1 times 2*n + 2.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A284359 Double triangle (2n+2 terms by row). Every row is 2n + 1 followed by 2n + 1 times 2n + 2.