OFFSET
1,4
COMMENTS
The rank of a partition is the largest part minus the number of parts.
Row lengths give A000041, n >= 1.
Just for n <= 6, row n is antisymmetric due to conjugation of partitions (see links under A105806): a(n, p(n)-(k-1)) = a(n,k), k = 1..floor(p(n)/2). [Comment corrected by Franklin T. Adams-Watters, Jan 17 2006]
First differs from A330368 at a(49) = T(7,5). - Omar E. Pol, Dec 31 2019
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
A. M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 831-2.
Freeman J. Dyson, Problems for solution nr. 4261, Am. Math. Month. 54 (1947) 418.
Wolfdieter Lang, First 16 rows.
FORMULA
EXAMPLE
Triangle begins:
[0];
[1, -1];
[2, 0, -2];
[3, 1, 0, -1, -3];
[4, 2, 1, 0, -1, -2, -4];
[5, 3, 2, 1, 1, 0, -1, -1, -2, -3, -5];
...
Row 3 for partitions of 3 in the mentioned order: 3,(1,2),1^3 with ranks 2,0,-2.
From Wolfdieter Lang, Jul 18 2013: (Start)
Row n = 7 is [6, 4, 3, 2, 2, 1, 0 , 0, 0, -1, -2, -2, -3, -4, -6].
This is also antisymmetric, but by accident, because a(7,7) = 0 for the partition (1,3^2), conjugate to (2^2,3) with a(7,8) = 0, and a(7,9) = 0 for (1^3,4) which is self-conjugate.
Row n=8 (see the link) is no longer antisymmetric. See the Franklin T. Adams-Watters correction above. (End)
MAPLE
# ASPrts is implemented in A119441
A105805 := proc(n, k)
local pi;
pi := ASPrts(n)[k] ;
max(op(pi))-nops(pi) ;
end proc:
for n from 1 do
for k from 1 to A000041(n) do
printf("%d, ", A105805(n, k)) ;
end do:
printf("\n") ;
end do: # R. J. Mathar, Jul 17 2013
CROSSREFS
KEYWORD
sign,easy,tabf
AUTHOR
Wolfdieter Lang, Apr 28 2005
EXTENSIONS
Name clarified by Omar E. Pol, Dec 31 2019
STATUS
approved