OFFSET
0,2
COMMENTS
Number of 321-avoiding permutations in which the length of the longest increasing subsequence is n. Example: a(2)=9 because we have 12, 132, 312, 213, 231, 3142, 3412, 2143 and 2413. Column sums of triangle in A126217 (n >= 1). - Emeric Deutsch, Sep 07 2007
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..830
FORMULA
a(n) = Sum_{k=0..n} (C(2*n-k+1,n-k)*(k+1)/(2*n-k+1))^2.
EXAMPLE
The dot product of Catalan row 4 with itself equals
a(4) = [14,14,9,4,1]*[14,14,9,4,1] = 490
which is equivalent to obtaining the final term in these repeated partial sums of Catalan row 4:
14, 14, 9, 4, 1
28, 37, 41, 42
65, 106, 148
171, 319
490
MAPLE
a:=proc(k) options operator, arrow: sum((2*k-n+1)^2*binomial(n+1, k+1)^2/(n+1)^2, n=k..2*k) end proc: 1, seq(a(k), k=1..17); # Emeric Deutsch, Sep 07 2007
MATHEMATICA
Table[Sum[(Binomial[2*n-j+1, n-j]*(j+1)/(2*n-j+1))^2, {j, 0, n}], {n, 0, 30}] (* G. C. Greubel, May 12 2019 *)
PROG
(PARI) a(n)=sum(k=0, n, ((k+1)*binomial(2*n-k+1, n-k)/(2*n-k+1))^2)
(Magma) [(&+[(Binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1))^2: j in [0..n]]): n in [0..30]]; // G. C. Greubel, May 12 2019
(Sage) [sum(( binomial(2*n-j+1, n-j)*(j+1)/(2*n-j+1) )^2 for j in (0..n)) for n in (0..30)] # G. C. Greubel, May 12 2019
(GAP) List([0..30], n-> Sum([0..n], j-> (Binomial(2*n-j+1, n-j)* (j+1)/(2*n-j+1))^2 )) # G. C. Greubel, May 12 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 04 2006
STATUS
approved