|
|
A156869
|
|
Triangle read by rows: T(n,k) = number of nondecreasing sequences of n positive integers with reciprocals adding up to k (1 <= k <= n).
|
|
6
|
|
|
1, 1, 1, 3, 1, 1, 14, 4, 1, 1, 147, 17, 4, 1, 1, 3462, 164, 18, 4, 1, 1, 294314, 3627, 167, 18, 4, 1, 1, 159330691, 297976, 3644, 168, 18, 4, 1, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
Conjecture: T(2n + m, n + m) = T(2n, n) ( = A156870(n) ) if and only if m >= 0.
Yes, the diagonals are constant for n <= 2k. Any such sequence must have at least one 1; remove that 1, and you get a sequence for n-1,k-1. - Franklin T. Adams-Watters, Feb 20 2009
|
|
LINKS
|
|
|
EXAMPLE
|
Triangle begins:
n=1: 1
n=2: 1, 1
n=3: 3, 1, 1
n=4: 14, 4, 1, 1
n=5: 147, 17, 4, 1, 1
n=6: 3462, 164, 18, 4, 1, 1
n=7: 294314, 3627, 167, 18, 4, 1, 1
For n = 4 and k = 2, the T(4, 2) = 4 sequences are (1, 2, 3, 6), (1, 2, 4, 4), (1, 3, 3, 3) and (2, 2, 2, 2) because 1/1 + 1/2 + 1/3 + 1/6 = 1/1 + 1/2 + 1/4 + 1/4 = 1/1 + 1/3 + 1/3 + 1/3 = 1/2 + 1/2 + 1/2 + 1/2 = 2.
|
|
PROG
|
(PARI) { A156869(n, k, m=1) = n==1 & return(numerator(k)==1 & denominator(k)>=m); sum( i=max(m, 1\k+1), n\k, A156869(n-1, k-1/i, i)); } \\ M. F. Hasler, Feb 20 2009
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|