A132163 - OEIS

A132163

Triangle read by rows. For row n, start with 1 but from the second term onwards always choose the largest positive integer between 1 and n inclusive that i) has not already appeared in the row ii) gives a prime when added to the previous term. Stop if no such integer can be found.

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

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OFFSET

1,3

COMMENTS

The following statements are conjectural: 1) The n-th row is always a permutation of 1,...,n. 2) For the even rows, the last term is one less than a prime (so the row gives a solution to the prime circle problem - see A051252). 3) There exists a (unique) sequence b(2), b(3),... with the property that for every n > 1 there is a positive integer N such that every even row of the triangle from the 2N-th onwards ends b(n), ..., b(3), b(2) and every odd row from the (2N - 1)-th onwards ends b(n)+(-1)^n, ..., b(3)-1, b(2)+1. (If the sequence b(n) exists it is probably A132075 without the initial term 1.)

LINKS

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened

MATHEMATICA

t[_, 1] = 1; t[n_, k_] := t[n, k] = For[ j = n, j > 1, j--, If[ PrimeQ[ t[n, k-1] + j] && FreeQ[ Table[ t[n, m], {m, 1, k-1}], j], Return[j] ] ]; Table[ t[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Jean-François Alcover, Apr 02 2013 *)

PROG

(Haskell)

import Data.List (delete)

a132163_tabl = map a132163_row [1..]

a132163 n k = a132163_row n !! (k-1)

a132163_row n = 1 : f 1 [n, n-1 .. 2] where

f u vs = g vs where

g [] = []

g (x:xs) | a010051 (x + u) == 1 = x : f x (delete x vs)

| otherwise = g xs

-- Reinhard Zumkeller, Jan 05 2013

CROSSREFS