

A207831


Square array such that T[n+1,k] =  T[n,k] + T[n,k+1] , filling antidiagonals with the smallest possible positive integers not occurring earlier.


2



1, 2, 3, 4, 6, 9, 7, 11, 5, 14, 8, 15, 26, 21, 35, 10, 18, 33, 59, 38, 73, 13, 23, 41, 74, 133, 95, 22, 12, 25, 48, 89, 163, 30, 65, 43, 16, 28, 53, 101, 190, 27, 57, 122, 79, 20, 36, 64, 117, 218, 408, 381, 324, 202, 123, 19, 39, 75, 139, 256, 474, 66, 315, 639, 437, 314, 32, 51, 90, 165, 304, 560, 86, 152, 467, 172, 265, 49
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OFFSET

1,2


COMMENTS

The first 50 antidiagonals of this table were computed by A. Groeneveld.
The table A207826 is a variant obtained by discarding a candidate for T[1,k] as soon as the "greedy way" of filling the antidiagonal (choose absolute difference or sum if the former is already used, but don't trace back to reconsider earlier choices) does not work.
The present version is computed by considering all possibilities in order to have the smallest possible T[1,k], cf. the example.


LINKS

Table of n, a(n) for n=1..78.
E. Angelini, Tableau avec soustractions/additions, Feb 19 2012
E. Angelini, Tableau avec soustractions/additions [Cached copy, with permission]


EXAMPLE

From M. F. Hasler, Feb 23 2012: (Start)
The triangle (upper left of the square array) starts
row 1: 1 2
row 2: 3 (because 12=1 already occurred).
The next antidiagonal reads (4, 4+2=6, 6+3=9) (since 42=2 and 63=3 are already used earlier), and so on.
The 25th antidiagonal can start with (83, 130, 223, 443, 393, 174, 569, 302, 890, 279, 181, 155, 398, 255, 102, 1029, 1679, 1256, 840, 116, ...). Then one could put 597116=481. However, at the next step, both 481+88 and 48188 occurred earlier (here in the very same antidiagonal). Therefore, we revise the earlier choice and change it to 597+116=713. Then the subsequent values in this antidiagonal are (..., 713+88=801, 801505=296, 296126=170, 911170=741). The table A207826 is obtained if we do not reconsider the earlier choice of "" vs. "+", but discard the whole antidiagonal once the greedy method cannot be continued down to the bottom, and start over with the next possible element in the first row. This would yield (91, 138, 231, 451, 401, 968, 573, ..., 549) for the 25th antidiagonal.
(End)


PROG

(PARI) [Contribution by M. F. Hasler, Feb 20 2012] (Start)
/* assumes that the first line A207829 is already computed. NOTE: This code does not yield the correct values T[n, k] beyond n+k=25 */
{T=matrix(#A=A207829, #A); u=Set(T[1, ]=A); for(j=2, #T, for(i=2, j, setsearch(u, T[i, j]=abs(T[i1, j1]T[i1, j]))&T[i, j]=T[i1, j1]+T[i1, j]; u=setunion(u, Set(T[i, j]))))}
for(j=1, 12, for(i=1, j, print1(T[i, j]", "))) \\ (End)


CROSSREFS

Sequence in context: A159849 A098168 A306441 * A207826 A035312 A056230
Adjacent sequences: A207828 A207829 A207830 * A207832 A207833 A207834


KEYWORD

nonn,tabl


AUTHOR

Eric Angelini, Feb 20 2012


STATUS

approved



