A279211 -id:A279211 - cp's OEIS frontend

A279212 Fill an array by antidiagonals upwards; in the top left cell enter a(0)=1; thereafter, in the n-th cell, enter the sum of the entries of those earlier cells that can be seen from that cell.

1, 1, 2, 2, 6, 11, 4, 15, 39, 72, 8, 37, 119, 293, 543, 16, 88, 330, 976, 2364, 4403, 32, 204, 870, 2944, 8373, 20072, 37527, 64, 464, 2209, 8334, 26683, 74150, 176609, 331072, 128, 1040, 5454, 22579, 79534, 246035, 673156, 1595909, 2997466, 256, 2304, 13176, 59185, 226106, 762221, 2303159, 6231191, 14721429, 27690124

Offset: 0

N. J. A. Sloane, Dec 24 2016

"That can be seen from" means "that are on the same row, column, diagonal, or antidiagonal as".
Inspired by A279967.
Conjecture: Every column has a finite number of odd entries, and every row and diagonal have an infinite number of odd entries. - Peter Kagey, Mar 28 2020. The conjecture about columns is true, see that attached pdf file from Alec Jones.
The "look" keyword refers to Peter Kagey's bitmap. - N. J. A. Sloane, Mar 29 2020
The number of sequences of queen moves from (1, 1) to (n, k) in the first quadrant moving only up, right, diagonally up-right, or diagonally up-left. - Peter Kagey, Apr 12 2020
Column 0 gives A011782. In the column 1, the only powers of 2 occur at positions A233328(k) with value a(k(k+1)/2 + 1), k >=1 (see A335903). Conjecture: Those are the only multiple occurrences of numbers greater than 1 in this sequence (checked through the first 2000 antidiagonals). - Hartmut F. W. Hoft, Jun 29 2020

			The array begins:
i/j|  0    1    2     3     4      5      6       7       8
-------------------------------------------------------------
0  |  1    2   11    72   543   4403  37527  331072 2997466 ...
1  |  1    6   39   293  2364  20072 176609 1595909 ...
2  |  2   15  119   976  8373  74150 673156 ...
3  |  4   37  330  2944 26683 246035 ...
4  |  8   88  870  8334 79534 ...
5  | 16  204 2209 22579 ...
6  | 32  464 5454 ...
7  | 64 1040 ...
8  |128 ...
  ...
For example, when we get to the antidiagonal that reads 4, 15, 39, ..., the reason for the 39 is that from that cell we can see one cell that has been filled in above it (containing 11), one cell to the northwest (2), two cells to the west (1, 6), and two to the southwest (4, 15), for a total of a(8) = 39.
The next pair of duplicates greater than 2 is 2^20 = 1048576 = a(154) = a(231), located in antidiagonals 17 = A233328(2) and 21, respectively. For additional duplicate numbers in this sequence see A335903.  - _Hartmut F. W. Hoft_, Jun 29 2020

Alois P. Heinz, Antidiagonals n = 0..200, flattened (first 20 antidiagonals from Alec Jones)
Alec Jones, Proof that columns have finitely many odd entries.
Peter Kagey, Bitmap showing parity of first 1024 rows and 512 columns. (Odd values are white; even values are black.)
Peter Kagey, Animated example illustrating the first fifteen terms.

Cf. A064642 is analogous if a cell can only "see" its immediate neighbors.
Cf. A035002, A059450, A132439, A279966, A279967, A279211.
See A280026, A280027 for similar sequences based on a spiral.
Cf. A011782, A335903.

s[0, 0] = 1; s[i_, j_] := s[i, j] = Sum[s[k, j], {k, 0, i-1}] + Sum[s[i, k], {k, 0, j-1}] + Sum[s[i+j-k, k], {k, 0, j-1}] + Sum[s[i-k-1, j-k-1], {k, 0, Min[i, j] - 1}]
aDiag[m_] := Map[s[m-#, #]&, Range[0, m]]
a279212[n_] := Flatten[Map[aDiag, Range[0, n]]]
a279212[9] (* data - 10 antidiagonals;  Hartmut F. W. Hoft, Jun 29 2020 *)

T(0, 0) = 1; T(i, j) = Sum_{k=0..i-1} T(k, j) + Sum_{k=0..j-1} T(i, k) + Sum_{k=0..j-1} T(i+j-k, k) + Sum_{k=0..min(i, j)-1} T(i-k-1, j-k-1), with recursion upwards along antidiagonals. - Hartmut F. W. Hoft, Jun 29 2020

A279967 Square array read by antidiagonals upwards in which each term is the sum of prior elements in the same row, column, diagonal, or antidiagonal that divide n; the array is seeded with an initial value a(1)=1.

Original entry on oeis.org

1, 1, 2, 2, 2, 7, 2, 9, 10, 15, 2, 10, 1, 13, 17, 8, 0, 13, 1, 14, 9, 8, 0, 13, 3, 30, 13, 10, 2, 16, 1, 23, 5, 7, 14, 15, 2, 8, 28, 32, 2, 23, 2, 9, 49, 12, 0, 48, 2, 11, 1, 20, 3, 18, 13, 28, 0, 4, 1, 56, 5, 8, 16, 35, 46, 4, 2, 6, 2, 10

Offset: 1

Alec Jones, Dec 24 2016

From Hartmut F. W. Hoft, Jan 23 2017: (Start)
Shown by induction and direct (modular) computations for
column 1: Every number is even, except for the first two 1's; in addition to row 3, value 2 occurs in rows 4*k and 4*k+1, and every value in rows 4*k+2 and 4*k+3 is divisible by 4, for all k>=1.
column 2: The first four entries, 2, 2, 9 and 10, contain the only odd number; no nonzero entry in row k>3 has 9 as a factor, and value 0 occurs in rows 4*k+1 and 4*k+2, for all k>=1.
Conjecture:
a({1, 6, 8, 9, 10, 15, 26, 45, 48, 84, 96, 112, 115, 252, 336, 343}) =
  {1, 7, 9,10, 15, 17, 30, 49, 48,104,117, 115, 122, 257, 343, 395} are the only numbers in the sequence with the property a(n) >= n (verified through n=500500, i.e., the triangle with 1000 antidiagonals).
This conjecture together with Bouniakowsky's conjecture that certain quadratic integer polynomials generate infinitely many primes (e.g. see A002496 for n^2+1 and A188382 for 2*n^2+n+1) implies that in every column in the triangle infinitely many prime sequence indices occur and therefore infinitely many 0's whenever the column contains no 1's. The proof is based on the fact that for a large enough prime sequence index p in whose prior column no 1 occurs then a(p)=0; therefore infinitely many 0's occur in that column. Obviously, once value 1 occurs in a column no 0 value can occur in a subsequent row.
Conjecture:
Every row in the triangle contains exactly two 1's.
(End)

			After 6 terms, the array looks like:
.
1   2   7
1   2
2
We have a(6) = 7 because a(1) = 1, a(3) = 2, a(4) = 2, and a(5) = 2 divide 6; 1 + 2 + 2 + 2 = 7.
From _Hartmut F. W. Hoft_, Jan 23 2017: (Start)
1   2   7  15  17   9  10  15  49  13   4  31  22
1   2  10  13  14  13  14   9  18  46  12  66
2   9   1   1  30   7   2   3  35  12   3
2  10  13   3   5  23  20  16  14  17
2   0  13  23   2   1   8  11   2
8   0   1  32  11   5   3   6
8  16  28   2  56  42   8
2   8  48   1   2 104
2   0   4  10   1
12   0   2  10
28   6   2
2  42
2
.
Expanded the triangle to the first 13 antidiagonals of the array, i.e. a(1) ... a(91), to show the start of the 2- and 0-value patterns in columns 1 and 2. The first 0 beyond column 2 is a(677) in row 27, column 11 of the triangle.
A188382(n)=2*n^2+n+1 for n>=0 are the alternate sequence indices for column 1 starting in row 1, 2*n^2+n+2 for n>=1 are the alternate sequence indices for column 2 starting in row 2, and 2*n^2+n+11 for n>=5 are the alternate sequence indices for column 11 starting in row 1.
The sequence indices in the triangle for row positions k>=1 in columns 1,..., 5 are given in sequences A000124(k), A152948(k+3), A152950(k+3), A145018(k+4) and A167499(k+4).
(End)

Peter Kagey, Table of n, a(n) for n = 1..5000

Cf. A279966 for the related sequence which counts prior terms.
Cf. A269347 for a one-dimensional version of this sequence.
Cf. also A279211, A279212.
Cf. A000124, A002496, A145018, A152948, A152950, A167499, A188382.

(*  printing of the triangle is commented out of function a279967[]  *)
pCol[{i_, j_}] := Map[{#, j}&, Range[1, i-1]]
pDiag[{i_, j_}] := If[j>=i, Map[{#, j-i+#}&, Range[1, i-1]], Map[{i-j+#, #}&, Range[1, j-1]]]
pRow[{i_, j_}] := Map[{i, #}&, Range[1, j-1]]
pAdiag[{i_, j_}] := Map[{i+j-#, #}&, Range[1, j-1]]
priorPos[{i_, j_}] := Join[pCol[{i, j}], pDiag[{i, j}], pRow[{i, j}], pAdiag[{i, j}]]
seqPos[{i_, j_}] := (i+j-2)(i+j-1)/2+j
antiDiag[k_] := Map[{k+1-#, #}&, Range[1, k]]
upperTriangle[k_] := Flatten[Map[antiDiag, Range[1, k]], 1]
a279967[k_] := Module[{ut=upperTriangle[k], ms=Table[" ", {i, 1, k}, {j, 1, k}], h, pos, val, seqL={1}}, ms[[1, 1]]=1; For[h=2, h<=Length[ut], h++, pos=ut[[h]]; val=Apply[Plus, Select[Map[ms[[Apply[Sequence, #]]]&, priorPos[pos]], #!=0 && Mod[seqPos[pos], #]==0&]]; AppendTo[seqL, val]; ms[[Apply[Sequence, pos]]]=val]; (* Print[TableForm[ms]]; *) seqL]
a279967[13] (* values in first 13 antidiagonals *)
(* Hartmut F. W. Hoft, Jan 23 2017 *)

A280026 Fill an infinite square array by following a spiral around the origin; in the n-th cell, enter the number of earlier cells that can be seen from that cell.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 4, 5, 6, 5, 6, 7, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 11, 12, 9, 10, 11, 12, 13, 10, 11, 12, 13, 14, 15, 11, 12, 13, 14, 15, 16, 12, 13, 14, 15, 16, 17, 18, 13, 14, 15, 16, 17, 18, 19, 14, 15, 16, 17, 18, 19, 20, 21, 15, 16, 17, 18, 19, 20, 21

Offset: 0

N. J. A. Sloane, Dec 24 2016

The spiral track being used here is the same as in A274640, except that the starting cell here is numbered 0 (as in A274641).
"Can be seen from" means "are on the same row, column, diagonal, or antidiagonal as".
The entry in a cell gives the number of earlier cells that are occupied in any of the eight cardinal directions. - Robert G. Wilson v, Dec 25 2016
First occurrence of k = 0,1,2,3,...: 0, 1, 2, 3, 5, 7, 8, 11, 14, 15, 19, 23, 24, 29, 34, 35, 41, 47, 48, 55, 62, ... - Robert G. Wilson v, Dec 25 2016

			The central portion of the spiral is:
.
    7---9---8---7---6
    |               |
    8   3---3---2   7
    |   |       |   |
    9   4   0---1   6
    |   |           |
   10   4---5---6---5
    |
    8---9--10--11--12 ...

Lars Blomberg, Table of n, a(n) for n = 0..10000

See A280027 for an additive version.
Cf. A274640, A274641, A278354.
See A279211, A279212 for versions that follow antidiagonals in just one quadrant.

a[n_] := a[n - 1] + If[ IntegerQ@ Sqrt@ n || IntegerQ@ Sqrt[4n +1], 2 - Select[{Sqrt@ n, (Sqrt[4n +1] -1)/2}, IntegerQ][[1]], 1]; a[0] = 0; Array[a, 76, 0] (* Robert G. Wilson v, Dec 25 2016 *)

Empirically: a(0)=0, a(n+1)=a(n)+d for n>0, when n=k^2 or n=k*(k+1) then d=2-k, else d=1.

Corrected a(23) and more terms from Lars Blomberg, Dec 25 2016

A279968 Square array read by antidiagonals upwards in which each term is the number of prior elements in the same row, column, diagonal, or antidiagonal whose parity is not the same as the parity of n.

Original entry on oeis.org

0, 0, 2, 0, 4, 0, 3, 1, 4, 2, 3, 3, 5, 3, 4, 2, 3, 5, 5, 4, 7, 2, 4, 5, 5, 6, 8, 3, 5, 4, 5, 10, 7, 7, 7, 7, 5, 5, 5, 9, 6, 12, 6, 8, 9, 4, 5, 8, 6, 6, 10, 9, 9, 7, 10, 4, 6, 8, 8, 5, 13, 9, 9, 12, 12, 8, 7, 6, 7, 8, 10, 9, 10, 12, 7, 11, 14, 9

Offset: 1

Alec Jones, Dec 24 2016

			The first six terms of this array are:
.
0 2 0
0 4
0
.
a(8) = 1 because the parity of 8 is different from the parity of a(7) = 1.
a(3) = 2 because the parity of 3 is different from the parity of a(2) = 0 and a(1) = 0.

Peter Kagey, Table of n, a(n) for n = 1..5000

Cf. A279965 for the related sequence which counts same-parity prior elements.

Cf. also A279211, A279212.

a279968 n = genericIndex a279968_list (n - 1)
a279968_list = map count [1..] where
  count n = genericLength $ filter (odd . (n+)) adjacentLabels where
    adjacentLabels = map a279968 (a274080_row n)

cp's OEIS Frontend

A279212 Fill an array by antidiagonals upwards; in the top left cell enter a(0)=1; thereafter, in the n-th cell, enter the sum of the entries of those earlier cells that can be seen from that cell.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A279967 Square array read by antidiagonals upwards in which each term is the sum of prior elements in the same row, column, diagonal, or antidiagonal that divide n; the array is seeded with an initial value a(1)=1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A280026 Fill an infinite square array by following a spiral around the origin; in the n-th cell, enter the number of earlier cells that can be seen from that cell.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A279968 Square array read by antidiagonals upwards in which each term is the number of prior elements in the same row, column, diagonal, or antidiagonal whose parity is not the same as the parity of n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell