OFFSET
1,5
COMMENTS
FORMULA
T(n,k) = Sum_{j=[k/2],k} C(j,k-j)*T(n-1,j) for n>=k>1 with T(n,1)=1 and T(n,k)=0 when k>n or k<1.
Main diagonal equals column 0 of triangle A135080, which transforms diagonals in the table of coefficients of the iterations of x+x^2.
Triangle A135080 also transforms diagonals in this triangle into each other.
Diagonal m of this triangle equals column 0 of the m-th power of triangle A135080, with diagonal m=1 being the main diagonal.
EXAMPLE
Triangle begins:
1;
1, 1;
1, 2, 2;
1, 3, 6, 8;
1, 4, 12, 29, 50;
1, 5, 20, 69, 202, 436;
1, 6, 30, 134, 538, 1880, 4912;
1, 7, 42, 230, 1164, 5404, 22108, 68098;
1, 8, 56, 363, 2210, 12646, 67092, 315784, 1122952;
1, 9, 72, 539, 3830, 25930, 166520, 997581, 5322126, 21488640;
1, 10, 90, 764, 6202, 48386, 362556, 2591010, 17337444, 103541022, 468331252; ...
in which rows can be generated as illustrated below.
Row polynomials R_n(y) begin:
R_1(y) = y;
R_2(y) = y + y^2;
R_3(y) = y + 2*y^2 + 2*y^3;
R_4(y) = y + 3*y^2 + 6*y^3 + 8*y^4;
R_5(y) = y + 4*y^2 + 12*y^3 + 29*y^4 + 50*y^5; ...
where row n = the coefficients of y^k in R_{n-1}(y+y^2) for k=1..n;
this method is illustrated by:
n=3: R_2(y+y^2) = (y + 2*y^2 + 2*y^3) + y^4;
n=4: R_3(y+y^2) = (y + 3*y^2 + 6*y^3 + 8*y^4) + 6*y^5 + 2*y^6;
n=5: R_4(y+y^2) = (y + 4*y^2 + 12*y^3 + 29*y^4 + 50*y^5) + 54*y^6 + 32*y^7 + 8*y^8;
where the n-th row polynomial R_n(y) equals R_{n-1}(y+y^2) truncated to the initial n terms.
...
ALTERNATE GENERATING METHOD.
Let F^n(x) denote the n-th iteration of x+x^2 with F^0(x) = x.
Then row n of this triangle may be generated by the coefficients of x^k in G(F^n(x)), k=1..n, where G(x) is the g.f. of A187009:
G(x) = x - x^2 + 2*x^3 - 6*x^4 + 20*x^5 - 80*x^6 + 348*x^7 - 1778*x^8 + 9892*x^9 - 64392*x^10 + 449596*x^11 + 15449192*x^12 +...
and satisfies: [x^(n+1)] G(F^n(x)) = 0 for n>0.
The table of coefficients in G(F^n(x)) begins:
G(x+x^2) : [1, 0, 0, -1, 2, -14, 44, -348, 1476, -14148, ...];
G(F^2(x)): [1, 1, 0, -1, -2, -10, -24, -231, -654, -9276, ...];
G(F^3(x)): [1, 2, 2, 0, -6, -26, -108, -570, -3216, -22622, ...];
G(F^4(x)): [1, 3, 6, 8, 0, -54, -324, -1776, -10594, -71702, ...];
G(F^5(x)): [1, 4, 12, 29, 50, 0, -616, -4846, -32686, -228926, ...];
G(F^6(x)): [1, 5, 20, 69, 202, 436, 0, -8629, -84140, -680298, ...];
G(F^7(x)): [1, 6, 30, 134, 538, 1880, 4912, 0, -143442, -1672428, ..];
G(F^8(x)): [1, 7, 42, 230, 1164, 5404, 22108, 68098, 0, -2762748, ..];
G(F^9(x)): [1, 8, 56, 363, 2210, 12646, 67092, 315784, 1122952, 0, ..]; ...
of which this triangle forms the lower triangular portion.
...
TRANSFORMATIONS OF DIAGONALS BY TRIANGLE A135080.
Given main diagonal = A135081 = [1,1,2,8,50,436,4912,68098,...],
the diagonals can be generated from each other as illustrated by:
Related triangle A135080 begins:
1;
1, 1;
2, 2, 1;
8, 7, 3, 1;
50, 40, 15, 4, 1;
436, 326, 112, 26, 5, 1;
4912, 3492, 1128, 240, 40, 6, 1; ...
MATHEMATICA
f[p_] := Series[p /. y -> y + y^2, {y, 0, 1 + Exponent[p, y]}] // Normal;
Flatten[ Rest[ CoefficientList[#, y]] & /@ NestList[f, y, 10]][[1 ;; 56]] (* Jean-François Alcover, Jun 09 2011 *)
PROG
(PARI) {T(n, k)=local(Rn=y); for(m=1, n, Rn=subst(truncate(Rn), y, y+y^2+y*O(y^m))); polcoeff(Rn, k, y)}
for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print(""))
(PARI) {T(n, k)=if(k>n||k<1, 0, if(n==1, 1, sum(j=k\2, k, binomial(j, k-j)*T(n-1, j))))}
for(n=1, 10, for(k=1, n, print1(T(n, k), ", ")); print(""))
CROSSREFS
KEYWORD
AUTHOR
Paul D. Hanna, Mar 02 2011
STATUS
approved