A346778 -id:A346778 - cp's OEIS frontend

A049476 Positions of records in A346778.

1, 2, 5, 13, 14, 26, 61, 63, 111, 131, 151, 153, 155, 161, 179, 295, 390, 391, 398, 425, 428, 459, 485, 656, 675, 1142, 1143, 1169, 1243, 1247, 1255, 1263, 1267, 1639, 1643, 1646, 1748, 2690, 2702, 2703, 2728, 2767, 2777, 2786, 2840, 2877

Offset: 1

N. J. A. Sloane

Previous name: Row numbers that set records for initial gap lengths g in the permutations found in A088643.

			For n = 4, when we examine row 13 in A088643, the Roche algorithm produces the initial row values 13, 10, 9, 8, 11, 12. The remaining values are equal to row 7 in A088643, and at no earlier point in row 13 are the remaining values equal to row m, 7 < m < 13. So we calculate the difference between 13 and 7 ("the uncharted length") to be 6, which is longer than the previous record uncharted length (A049478(3) = 4) set by row a(3) = 5. So a(4) = 13. - _Peter Munn_, Aug 03 2021 (based on text supplied by _J. Stauduhar_)

J. W. Roche, Letter regarding "M. J. Kenney and S. J. Bezuszka, Calendar problem 12, 1997", Mathematics Teacher, 91 (1998), 155.

Cf. A049477, A049478, A088643, A346778.

{print1(m=0); for( n=1, oo, my( r=A088643_row(n)); for( g=1, #r-1, if( Set(r[1..g]) == [n-g+1..n] && r[g+1]==n-g, g > m && print1(","n)+ m=g; break)))} \\ M. F. Hasler, Aug 04 2021

Revised by Sean A. Irvine, Aug 03 2021

A049478 Record values in A346778.

Original entry on oeis.org

0, 1, 4, 6, 12, 22, 27, 62, 64, 84, 104, 106, 108, 114, 132, 142, 178, 179, 186, 213, 216, 247, 273, 322, 341, 357, 358, 449, 458, 462, 470, 478, 482, 495, 499, 502, 581, 585, 597, 598, 623, 662, 672, 681, 735, 772, 778, 794, 818, 827, 851, 864

Offset: 1

N. J. A. Sloane

Record values of the least g >= 1 such that the first g terms of row n of A088643 [start with n and always append the largest k < n not yet used such that n+k is a prime] constitute the set {n, n-1, ..., n-g+1}, and the next (i.e., (g+1)-th) term of that row equals n-g. - M. F. Hasler, Aug 04 2021

Cf. A049476, A049477, A088643, A346778.

{print1(m=0); for( n=1, oo, my( r=A088643_row(n)); for( g=1, #r-1, if( Set(r[1..g]) == [n-g+1..n] && r[g+1]==n-g, g > m && print1(", "m=g); break)))} \\ M. F. Hasler, Aug 04 2021

Revised by Sean A. Irvine, Aug 03 2021

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.

Original entry on oeis.org

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

cp's OEIS Frontend

A049476 Positions of records in A346778.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Extensions

A049478 Record values in A346778.

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

Extensions

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

Extensions

A049477 For each record-breaking row, m, of A088643, the sequence gives the length, k, of the longest earlier row contained in row m.

Views

Author

Keywords

Comments

Crossrefs

Extensions