OFFSET
0,3
COMMENTS
Equals the main diagonal of symmetric square array A101515 shift right.
Empirical: a(n) = sum((number of standard immaculate tableaux of shape m)^2, m|=n), where this sum is over all compositions m of n > 0. - John M. Campbell, Jul 07 2017
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..330
FORMULA
E.g.f. satisfies: B(x)/A(x) = Sum_{n>=0} x^n/n!^2 where A(x) = Sum_{n>=0} a(n)*x^n/n!^2 and B(x) = Sum_{n>=0} a(n+1)*x^n/n!^2. - Paul D. Hanna, Oct 10 2014
EXAMPLE
The binomial transform of the rows of the term-wise product of this sequence with the rows of Pascal's triangle produces the symmetric square array A101515, in which the main diagonal equals this sequence shift left:
BINOMIAL[1*1] = [(1),1,1,1,1,1,1,1,1,...],
BINOMIAL[1*1,1*1] = [1,(2),3,4,5,6,7,8,9,...],
BINOMIAL[1*1,1*2,2*1] = [1,3,(7),13,21,31,43,57,73,...],
BINOMIAL[1*1,1*3,2*3,7*1] = [1,4,13,(35),77,146,249,393,...],
BINOMIAL[1*1,1*4,2*6,7*4,35*1] = [1,5,21,77,(236),596,1290,...],
BINOMIAL[1*1,1*5,2*10,7*10,35*5,236*1] = [1,6,31,146,596,(2037),...],...
Thus the square binomial transform shifts this sequence one place left:
a(5) = 236 = 1^2*(1) + 4^2*(1) + 6^2*(2) + 4^2*(7) + 1^2*(35),
a(6) = 2037 = 1^2*(1) + 5^2*(1) + 10^2*(2) + 10^2*(7) + 5^2*(35) + 1^2*(236).
MAPLE
a:= proc(n) option remember; if n<=0 then 1 else
add(binomial(n-1, k)^2 *a(k), k=0..n-1) fi
end:
seq(a(n), n=0..25); # Alois P. Heinz, Sep 05 2008
MATHEMATICA
a[0] = 1;
a[n_] := Sum[Binomial[n - 1, k]^2 a[k], {k, 0, n - 1}];
Table[a[i], {i, 0, 20}] (* Philip B. Zhang, Oct 10 2014 *)
PROG
(PARI) {a(n)=if(n==0, 1, sum(k=0, n-1, binomial(n-1, k)^2*a(k)))}
for(n=0, 20, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 06 2004
EXTENSIONS
Typo in definition corrected by Philip B. Zhang, Oct 10 2014
STATUS
approved