login
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.
14

%I #64 Nov 19 2021 16:44:10

%S 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,

%T 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,

%U 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

%N 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.

%C It is conjectured that row n is always a permutation of {1..n}. This has been verified for n <= 400000.

%C 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]

%C 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

%C In fact, the reversed rows converge to the different sequence A132075, essentially defined by this property. - _M. F. Hasler_, Aug 04 2021

%C 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

%H Reinhard Zumkeller, <a href="/A088643/b088643.txt">Rows n = 1..150 of triangle, flattened</a>

%H Peter Munn, <a href="/A255312/a255312_1.txt">Illustration of the relationship between this sequence, A132075 and A255312</a>.

%H J. W. Roche, <a href="http://www.jstor.org/stable/27970468">Letter regarding "M. J. Kenney and S. J. Bezuszka, Calendar problem 12, 1997"</a>, Mathematics Teacher, 91 (1998), 155.

%F A255313(n,k) = T(n,k-1) + T(n,k), n > 0 and 1 <= k <= n. - _Reinhard Zumkeller_, Feb 22 2015

%e 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.

%e Triangle begins:

%e 1;

%e 2, 1;

%e 3, 2, 1;

%e 4, 3, 2, 1;

%e 5, 2, 3, 4, 1;

%e 6, 5, 2, 3, 4, 1;

%e (...)

%p A088643 := proc(n,k)

%p option remember ;

%p local m,c;

%p if n = 1 then

%p 1;

%p else

%p if k = 1 then

%p return n;

%p else

%p for m from n-1 to 1 by -1 do

%p if not member(m,[seq(procname(n,c),c=1..k-1)]) then

%p if isprime(m+procname(n,k-1)) then

%p return m;

%p end if ;

%p end if;

%p end do:

%p end if;

%p end if;

%p end proc:

%p for n from 1 to 10 do

%p for k from 1 to n do

%p printf("%d ",A088643(n,k)) ;

%p end do:

%p printf("\n") ;

%p end do: # _R. J. Mathar_, Aug 18 2021

%t 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 *)

%o (Haskell)

%o import Data.List (delete)

%o a088643_tabl = map a088643_row [1..]

%o a088643 n k = a088643_row n !! (k-1)

%o a088643_row n = n : f n [n-1, n-2 .. 1] where

%o f u vs = g vs where

%o g [] = []

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

%o | otherwise = g xs

%o -- _Reinhard Zumkeller_, Jan 05 2013

%o (PARI) 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

%o (PARI) 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

%Y A088631 and A088861 give second and third columns.

%Y Cf. A049476, A049477, A049478, A346778.

%Y Cf. A055265, A132075, A255312, A255313, A255316.

%K nonn,tabl,nice,easy

%O 1,2

%A _N. J. A. Sloane_, Nov 24 2003

%E More terms from _David Wasserman_, Aug 16 2005