1,3

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)

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

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)

(* 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 *)

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.

Sequence in context: A202033 A340742 A029610 * A094246 A169592 A245600

Adjacent sequences: A279964 A279965 A279966 * A279968 A279969 A279970

nonn,tabl

Alec Jones, Dec 24 2016

approved