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User: Lucas Larsen

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A375720 Irregular triangle, read by rows: Coefficients of the polynomials P_n, n>=2 such that the series f(x) = c + c(x-c) + Sum_{n>=2} P_n(c)/c^((n-1)*(n+2)/2+1) (x-c)^n/n! satisfies f(c) = c and f'(f(x)) = x near the fixed point c in (0,oo).

Original entry on oeis.org

1, -1, 1, 3, -1, -3, -10, -15, 1, 3, 10, 30, 55, 105, 105, -1, -3, -10, -30, -76, -168, -350, -630, -910, -1260, -945, 1, 3, 10, 30, 76, 196, 434, 910, 1806, 3381, 5789, 9135, 12880, 15750, 17325, 10395, -1, -3, -10, -30, -76, -196, -470, -1018, -2166, -4461, -8609, -16065, -28336, -48006, -78519, -120960, -172200, -228375, -275275, -294525, -270270, -135135
Offset: 2

Author

Lucas Larsen, Aug 26 2024

Keywords

Comments

The indices in each row range from 0 to (n-3)*(n-2)/2
When c = phi = (1+sqrt(5))/2 the series becomes the Taylor expansion of f(x) = phi^(-1/phi)*x^phi centered at phi, in particular the radius of convergence is positive for at least this choice of c.

Examples

			Triangle begins:
  1;
  -1;
  1, 3;
  -1, -3, -10, -15;
  1, 3, 10, 30, 55, 105, 105;
  -1, -3, -10, -30, -76, -168, -350, -630, -910, -1260, -945;
  ...
Polynomials are:
  P_2(c) = 1
  P_3(c) = -1
  P_4(c) = 1 + 3c
  P_5(c) = -1 - 3c - 10c^2 - 15c^3
  etc.
Hence the series begins
f(x) = c + c*(x-c) + c^(-1)(x-c)^2/2 - c^(-4)(x-c)^3/6 + (3c^(-7) + c^(-8))(x-c)^4/24 + ...
		

Crossrefs

Cf. A144006.

Programs

  • Python
    def T(n,k):
        c = {(-1,):1} #Polynomial in infinitely many variables (function iterates)
        for _ in range(n-2):
            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
        return pairs.get(1+k-(n-1)*(n+2)//2,0)