A101516 -id:A101516 - cp's OEIS frontend

A101514 Shifts one place left under the square binomial transform (A008459): a(0) = 1, a(n) = Sum_{k=0..n-1} C(n-1,k)^2*a(k).

1, 1, 2, 7, 35, 236, 2037, 21695, 277966, 4198635, 73558135, 1475177880, 33495959399, 853167955357, 24182881926558, 757554068775721, 26068954296880361, 980202973852646786, 40079727064364154465, 1774594774575753650941, 84756211791797266285252

Offset: 0

Paul D. Hanna, Dec 06 2004

nonn

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

			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).

Vaclav Kotesovec, Table of n, a(n) for n = 0..330
John M. Campbell, Semisimple algebras related to immaculate tableaux, arXiv:2507.02539 [math.CO], 2025. See p. 7.

Cf. A008459, A101515, A101516.

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

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 *)

{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),", "))

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

Typo in definition corrected by Philip B. Zhang, Oct 10 2014

A101515 Symmetric square array, read by antidiagonals, such that the inverse binomial transform of row n forms the sequence: {C(n,k)*A101514(k), 0<=k<=n}, where A101514 equals the main diagonal shift right.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 35, 21, 6, 1, 1, 7, 31, 77, 77, 31, 7, 1, 1, 8, 43, 146, 236, 146, 43, 8, 1, 1, 9, 57, 249, 596, 596, 249, 57, 9, 1, 1, 10, 73, 393, 1290, 2037, 1290, 393, 73, 10, 1, 1, 11, 91, 585, 2486, 5772, 5772, 2486, 585

Offset: 0

Paul D. Hanna, Dec 06 2004

The main diagonal equals A101514 shift one place left. The antidiagonal sums form A101516.

			Rows begin:
[_1,1,1,1,1,1,1,1,1,...],
[1,_2,3,4,5,6,7,8,9,...],
[1,3,_7,13,21,31,43,57,73,...],
[1,4,13,_35,77,146,249,393,585,...],
[1,5,21,77,_236,596,1290,2486,4387,...],
[1,6,31,146,596,_2037,5772,13987,29987,...],
[1,7,43,249,1290,5772,_21695,67943,181811,...],
[1,8,57,393,2486,13987,67943,_277966,951051,...],
[1,9,73,585,4387,29987,181811,951051,_4198635,...],...
The inverse binomial transform of the rows of this array are generated
from the products of the main diagonal with rows of Pascal's triangle:
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,...],...

Cf. A101514, A101516.

T(n,k)=if(n<0 || k<0,0,if(n==0 || k==0,1,if(n>k,T(k,n), 1+sum(j=1,k,binomial(k,j)*binomial(n,j)*T(j-1,j-1));)))

cp's OEIS Frontend

A101514 Shifts one place left under the square binomial transform (A008459): a(0) = 1, a(n) = Sum_{k=0..n-1} C(n-1,k)^2*a(k).

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