login
A132427
Triangle, read by rows of 2n+1 terms, where T(n,k) = T(n,k-1) + T(n-1,k-2) for n>0, 1<k<=2n, with T(n,1)=T(n,0)=T(n-1,2n-2) for n>0 and T(0,0)=1.
1
1, 1, 1, 2, 2, 2, 3, 4, 6, 6, 6, 8, 10, 13, 17, 23, 23, 23, 29, 35, 43, 53, 66, 83, 106, 106, 106, 129, 152, 181, 216, 259, 312, 378, 461, 567, 567, 567, 673, 779, 908, 1060, 1241, 1457, 1716, 2028, 2406, 2867, 3434, 3434, 3434, 4001, 4568, 5241, 6020, 6928, 7988
OFFSET
0,4
COMMENTS
Column 0 equals (essentially) column 1 and the rightmost border.
FORMULA
The g.f. of column 0 (A125273) satisfies: G(x) = 1 + x*G( x/(1-x)^2 ) / (1-x).
The central terms (A132428) are the inverse binomial transform of A125273 (offset 1).
EXAMPLE
Triangle begins:
1;
1, 1, 2;
2, 2, 3, 4, 6;
6, 6, 8, 10, 13, 17, 23;
23, 23, 29, 35, 43, 53, 66, 83, 106;
106, 106, 129, 152, 181, 216, 259, 312, 378, 461, 567;
567, 567, 673, 779, 908, 1060, 1241, 1457, 1716, 2028, 2406, 2867, 3434; ...
MATHEMATICA
t[n_, k_] := t[n, k] = t[n, k-1] + t[n-1, k-2]; t[n_, 0] := t[n, 0] = t[n-1, 2n-2]; t[n_, 1] := t[n, 0]; t[0, 0] = 1; Flatten[ Table[t[n, k], {n, 0, 7}, {k, 0, 2 n}]] (* Jean-François Alcover, Jun 18 2012 *)
PROG
(PARI) T(n, k)=local(A=[1]); if(2*n<k || k<0, 0, if(n==0 && k==0, 1, if(k==0 || k==1, T(n-1, 2*n-2), for(i=1, n, A=Vec(Ser(concat([A[ #A], 0], A))/(1-x))); A[k+1])))
CROSSREFS
Cf. A125273; A132428 (central terms).
Sequence in context: A373785 A089150 A056697 * A176975 A333374 A098523
KEYWORD
nonn,tabf
AUTHOR
Paul D. Hanna, Aug 21 2007
STATUS
approved