A144006 Triangle, read by rows of coefficients of x^n*y^k for k=0..n(n-1)/2 for n>=0, defined by e.g.f.: A(x,y) = 1 + Series_Reversion( Integral A(-x*y,y) dx ), with leading zeros in each row suppressed.
1, 1, 1, 3, -1, 15, -10, 3, -1, 105, -105, 55, -30, 10, -3, 1, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1, 135135, -270270, 294525, -275275, 228375
Offset: 0
Examples
Triangle begins (without suppressing leading zeros): 1; 1; 0, 1; 0,0, 3, -1; 0,0,0, 15, -10, 3, -1; 0,0,0,0, 105, -105, 55, -30, 10, -3, 1; 0,0,0,0,0, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1; 0,0,0,0,0,0, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1; 0,0,0,0,0,0,0, 135135, -270270, 294525, -275275, 228375, -172200, 120960, -78519, 48006, -28336, 16065, -8609, 4461, -2166, 1018, -470, 196, -76, 30, -10, 3, -1; ...
Crossrefs
Programs
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PARI
{T(n,k)=local(A=1+x*O(x^n)); for(i=0,n,A=1+serreverse(intformal(subst(A,x,-x*y))));n!*polcoeff(polcoeff(A,n,x),k,y)} -
Python
#This is only correct if the observation in the comment from 2024/08/20 is true. def T(n,k): if 0 <= n <= 1: return 1 if k == 0 else 0 c = {(-1,):1} #Polynomial in infinitely many variables (function iterates) for _ in range(n-1): cnext = {} for key, value in c.items(): key += (0,) for i, ni in enumerate(key): term = tuple(nj-2 if j==i else nj-1 if j<=i+1 else nj for j,nj in enumerate(key)) cnext[term] = cnext.get(term,0) + value*ni if cnext[term] == 0: del cnext[term] c = cnext pairs = {} #Reduction to single variable (evaluation at fixpoint) for key, value in c.items(): s = -sum(key) pairs[s] = pairs.get(s,0) + value _, row = zip(*sorted(pairs.items())) #Coefficients if 0 <= k-n+1 < len(row): #Correcting number of leading 0s return (-1)**(n+k+1)*abs(row[k-n+1]) #Correcting signs else: return 0 # Lucas Larsen, Aug 22 2024
Formula
E.g.f. satisfies: A(x,y) = 1 + Series_Reversion[Integral A(-x*y,y) dx].
T(n,k) = [x^n*y^k] n!*A(x,y) for k=0..n(n-1)/2, n>=0.
Row sums equal A144005.
A067146(n) = Sum_{k=0..n(n-1)/2} (-1)^k*T(n,k).
This is a signed version of table A014621 because setting f((1+x)/y):=A(-x*y,y)/y for fixed y>0 implies f(f(x))*f'(x)=-1 and f(1/y)=1/y, as in the second formula of A014621. Therefore, the row sums form A014623 and the unsigned row sums form A014622. - Roland Miyamoto, Jun 03 2024
Comments