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Search: id:A156869
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| A156869 |
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Triangle read by rows: T(n,k) = Number of increasing sequences of length n with reciprocals adding up to k (1 <= k <= n) |
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+0
3
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| 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; listen)
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OFFSET
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1,4
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COMMENT
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Conjecture: T(2n + m, n + m) = T(2n, n) ( = A156870(n) ) if and only if m >= 0.
Comment from Franklin T. Adams-Watters, Feb 20 2009: 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.
The next term will be a(37) = A002966(9). [From M. F. Hasler (MHasler(AT)univ-ag.fr), Feb 20 2009]
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EXAMPLE
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Triangle begins:
..........................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
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.
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PROGRAM
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Contribution from M. F. Hasler (MHasler(AT)univ-ag.fr), Feb 20 2009: (Start)
(PARI) c(n, k, m=1)={ n==1 & return(numerator(k)==1 & denominator(k)>=m); sum( i=max(m, 1\k+1), n\k, c(n-1, k-1/i, i))}
for(n=1, 7, for(k=1, n, print1( c(n, k)", "))) (End)
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CROSSREFS
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T(n, 1) = A002966(n)
T(2n, n) = A156870(n)
T(n, 1) + ... + T(n, n) = A156871(n)
Sequence in context: A094573 A055154 A015112 this_sequence A153090 A073483 A006956
Adjacent sequences: A156866 A156867 A156868 this_sequence A156870 A156871 A156872
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KEYWORD
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more,nonn,tabl
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AUTHOR
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Jens Voss (jens(AT)voss-ahrensburg.de), Feb 17 2009
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EXTENSIONS
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a(21)-a(36) from M. F. Hasler (MHasler(AT)univ-ag.fr), Feb 20 2009
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