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A088643
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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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14
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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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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EXAMPLE
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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;
(...)
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MAPLE
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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
end do:
printf("\n") ;
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MATHEMATICA
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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 *)
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PROG
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(Haskell)
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
(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
(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
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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