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A106597
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Triangle T(n,k) = T(n-1, k-1) + T(n-1, k) + Sum_{i >= 1} T(n-2*i, k-i), with T(n, 0) = T(n, n) = 1, read by rows.
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3
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1, 1, 1, 1, 3, 1, 1, 5, 5, 1, 1, 7, 14, 7, 1, 1, 9, 27, 27, 9, 1, 1, 11, 44, 72, 44, 11, 1, 1, 13, 65, 149, 149, 65, 13, 1, 1, 15, 90, 266, 388, 266, 90, 15, 1, 1, 17, 119, 431, 836, 836, 431, 119, 17, 1, 1, 19, 152, 652, 1585, 2150, 1585, 652, 152, 19, 1, 1, 21, 189, 937, 2743, 4753, 4753, 2743, 937, 189, 21, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,5
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COMMENTS
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Next term is sum of two terms above you in previous row (as in Pascal's triangle A007318) plus sum of terms directly above you on a vertical line.
T(n,n-k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0), (0,1), and (s,s) for s>=1. - Joerg Arndt, Jul 01 2011
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins:
1;
1, 1;
1, 3, 1;
1, 5, 5, 1;
1, 7, 14, 7, 1;
1, 9, 27, 27, 9, 1;
1, 11, 44, 72, 44, 11, 1;
1, 13, 65, 149, 149, 65, 13, 1;
1, 15, 90, 266, 388, 266, 90, 15, 1;
1, 17, 119, 431, 836, 836, 431, 119, 17, 1;
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MATHEMATICA
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PROG
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(PARI) /* same as in A092566, but last line (output) replaced by the following */
/* show as triangle T(n-k, k): */
{ for(n=0, N-1, for(k=0, n, print1(T(n-k, k), ", "); ); print(); ); }
(Sage)
@CachedFunction
def T(n, k):
if (k<0): return 0
elif (k==0 or k==n): return 1
else: return + T(n-1, k-1) + T(n-1, k) + sum( T(n-2*j, k-j) for j in (1..min(k, n//2, n-k)))
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 08 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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