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Search: id:A129174
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| A129174 |
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Triangle read by rows: T(n,k) is the number of Dyck paths of semilength n such that the sum of the peak-abscissae is k (0<=k<=n^2). |
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+0
3
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| 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 3, 2, 4, 3, 4, 3, 4, 2, 3, 2, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 2, 2, 4, 3, 5, 5, 7, 6, 9, 7, 9, 8, 9, 7, 9, 6, 7, 5, 5, 3, 4, 2, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0
(list; graph; listen)
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OFFSET
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0,27
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COMMENT
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Row n contains 1+n^2 entries. Row sums are the Catalan numbers (A000108). Column sums yield A129528. T(n,n+k)=T(n,n^2-k) (i.e. rows are palindromic). Alternating row sums are (-1)^n*binom(n,floor(n/2))=A126930(n). Sum(k*T(n,k),k=0..n^2)=n*binom(2n-1,n-1)=A002457(n-1). T(n,k)=A129175(n,n-k) (i.e., except for the initial 0's, rows of A129174 and A129175 are the same).
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REFERENCES
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G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.
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FORMULA
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The generating polynomial for row n is P[n](t) = t^n*binom[2n,n]/[n+1], where [n+1]=1+t+t^2+...+t^n and binom[2n,n] is a Gaussian polynomial (in t).
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EXAMPLE
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T(5,11)=3 because we have (i) UDUDUUUDDD with peak-abscissae 1,3,7, (ii) UUUDDUUDD with peak-abscissae 3,8 and (iii) UUUUDDUDDD with peak-abscissae 4,7; here U=(1,1) and D=(1,-1).
Triangle starts:
1;
0,1;
0,0,1,0,1;
0,0,0,1,0,1,1,1,0,1;
0,0,0,0,1,0,1,1,2,1,2,1,2,1,1,0,1;
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MAPLE
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br:=n->sum(q^i, i=0..n-1): f:=n->product(br(j), j=1..n): cbr:=(n, k)->f(n)/f(k)/f(n-k): P:=n->sort(expand(simplify(q^n*cbr(2*n, n)/br(n+1)))): for n from 0 to 7 do seq(coeff(P(n), q, k), k=0..n^2) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000108, A002457, A126930, A129175, A129528.
Sequence in context: A107393 A047885 A072731 this_sequence A129175 A063053 A063050
Adjacent sequences: A129171 A129172 A129173 this_sequence A129175 A129176 A129177
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 20 2007
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