A255312 -id:A255312 - cp's OEIS frontend

A088643 Triangle read by rows: row n >= 1 is obtained as follows. Start with n, next term is always largest number m with 1 <= m < n which has not yet appeared in that row and such that m + previous term in the row is a prime. Stop when no further m can be found.

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

Offset: 1

N. J. A. Sloane, Nov 24 2003

It is conjectured that row n is always a permutation of {1..n}. This has been verified for n <= 400000.
Presumably many of the rows, when read from right to left, match the infinite sequence A055265. [But see a more precise comment that follows. - N. J. A. Sloane, Aug 14 2021]
I conjecture that almost all rows have exactly 7 (but not more) trailing terms in common with the initial terms of A055265 = (1, 2, 3, 4, 7, 6, 5, 8, ...): After row 10 whose reversal matches the first 10 terms of A055265, and rows n = 14, 15 and 16 having the last 2 (but not 3) terms equal to A055265(1..2), all rows up to n = 500 have either (about 25%) exactly 1 or (about 73%) exactly 7 trailing terms equal to the first terms of A055265. Between n = 501 and n = 10000 and beyond, all rows end in (..., 9, 14, 5, 6, 7, 4, 3, 2, 1), so they all have exactly m = 7 but not m = 8 trailing terms equal to A055265(1..m). - M. F. Hasler, Aug 03 2021
In fact, the reversed rows converge to the different sequence A132075, essentially defined by this property. - M. F. Hasler, Aug 04 2021
It seems we do not know of a proof (1) that the sequence of reversed rows of this sequence converges or (2) that A132075 is infinite; or that either statement implies the other. The reversed rows converge to A132075 if both statements are true, as suggested empirically by the early rows of this sequence. - Peter Munn, Nov 19 2021

			For example, the 20th row is 20, 17, 14, 15, 16, 13, 18, 19, 12, 11, 8, 9, 10, 7, 6, 5, 2, 3, 4, 1.
Triangle begins:
  1;
  2, 1;
  3, 2, 1;
  4, 3, 2, 1;
  5, 2, 3, 4, 1;
  6, 5, 2, 3, 4, 1;
  (...)

Reinhard Zumkeller, Rows n = 1..150 of triangle, flattened
Peter Munn, Illustration of the relationship between this sequence, A132075 and A255312.
J. W. Roche, Letter regarding "M. J. Kenney and S. J. Bezuszka, Calendar problem 12, 1997", Mathematics Teacher, 91 (1998), 155.

A088631 and A088861 give second and third columns.
Cf. A049476, A049477, A049478, A346778.
Cf. A055265, A132075, A255312, A255313, A255316.

import Data.List (delete)
a088643_tabl = map a088643_row [1..]
a088643 n k = a088643_row n !! (k-1)
a088643_row n = n : f n [n-1, n-2 .. 1] 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

A088643 := proc(n,k)
    option remember ;
    local m,c;
    if n = 1 then
        1;
    else
        if k = 1 then
            return n;
        else
            for m from n-1 to 1 by -1 do
                if not member(m,[seq(procname(n,c),c=1..k-1)]) then
                    if isprime(m+procname(n,k-1)) then
                        return m;
                    end if ;
                end if;
            end do:
        end if;
    end if;
end proc:
for n from 1 to 10 do
for k from 1 to n do
    printf("%d ",A088643(n,k)) ;
end do:
printf("\n") ;
end do: # R. J. Mathar, Aug 18 2021

t[n_, 1] := n; t[n_, k_] := t[n, k] = For[m = n-1, m >= 1, m--, If[ PrimeQ[m + t[n, k-1] ] && FreeQ[ Table[ t[n, j], {j, 1, k-1} ], m], Return[m] ] ]; Table[ t[n, k], {n, 1, 14}, {k, 1, n} ] // Flatten (* Jean-François Alcover, Apr 03 2013 *)

apply( {A088643_row(n, t=List(-[1-n..-1]))=vector(n,i, i>1 && for(j=1,#t, isprime(n+t[j]) && [n=t[j], listpop(t,j), break]);n)}, [1..20]) \\ M. F. Hasler, Aug 02 2021; improved Aug 03 2021 after PARI below

row(n) = { my(res = vector(n), todo = List([1..n-1])); res[1] = n; for(i = 1, n - 1, forstep(j = #todo, 1, -1, if(isprime(res[i] + todo[j]), res[i+1] = todo[j]; listpop(todo, j); next(2) ) ) ); res } \\ David A. Corneth, Aug 02 2021

A255313(n,k) = T(n,k-1) + T(n,k), n > 0 and 1 <= k <= n. - Reinhard Zumkeller, Feb 22 2015

More terms from David Wasserman, Aug 16 2005

A132075 A conjectured permutation of the positive integers such that for every n, a(n) is the largest number among a(1), a(2), ..., a(n) that when added to a(n+1) gives a prime.

Original entry on oeis.org

1, 2, 3, 4, 7, 6, 5, 14, 9, 10, 13, 16, 15, 8, 11, 20, 17, 12, 19, 24, 23, 18, 25, 22, 21, 26, 27, 34, 33, 28, 31, 30, 29, 32, 35, 36, 37, 46, 43, 40, 39, 44, 45, 38, 41, 42, 47, 50, 59, 54, 55, 58, 51, 62, 65, 48, 61, 52, 57, 56, 53, 60, 49, 64, 63, 68, 69, 70, 67, 72, 77, 80, 71, 66, 73, 78, 79, 84, 83, 90, 89, 74, 75, 76, 81, 82, 85, 88, 93, 86, 95, 104, 107, 92

Offset: 1

Paul Boddington, Oct 30 2007, Mar 06 2010

The terms are defined as follows. Start by choosing the initial terms: 1, 2, 3. Then write the rows of table A088643 backwards but always leave off the last three quarters of the terms. This gives: [], [], [], [1], [1], [1], [1], [1, 2], [1, 2], [1, 2], [1, 4], [1, 4, 3,], [1, 4, 3] etc. Then build the sequence up by repeatedly choosing the first such truncated row that extends the terms already chosen. [Edited by Peter Munn, Aug 19 2021]
It is not until the 26th truncated row - [1, 2, 3, 4, 7, 6] - that the initial list is extended at all. It is unclear whether this process can be continued indefinitely, although I have verified by computer that it generates a sequence of at least 2000 terms. Conjecturally: (1) the sequence is infinite, (2) it is the unique sequence containing infinitely many complete rows of table A088643, and (3) for every n > 0 there exists N > 0 such that the first n terms of this sequence are contained in every row of table A088643 from the N-th onwards.
Maybe the idea could be expressed more concisely by defining this sequence as the limit of the reversed rows of A088643? - M. F. Hasler, Aug 04 2021
It seems we do not know of an existence proof for the limit of the reversed rows of A088643. - Peter Munn, Aug 19 2021

Peter Munn, Illustration of the relationship between A088643, this sequence and A255312.

Cf. A088643.

Name edited by Peter Munn, Aug 19 2021

cp's OEIS Frontend

A088643 Triangle read by rows: row n >= 1 is obtained as follows. Start with n, next term is always largest number m with 1 <= m < n which has not yet appeared in that row and such that m + previous term in the row is a prime. Stop when no further m can be found.

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