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While solving a problem, I reached a recursive solution of the form:

$$ f(x, N) = 2f(x-1,N+1) - f(x-2,N). $$

I stumbled upon an approach to obtain the individual terms without a long recursive operation. I did not prove this, but it looked somehow logical and supported by Monte Carlo simulations.

$$ f(x, N) = \frac{1}{2}\sum_{i=0}^{\lceil x/2 \rceil-1}(-1)^{i} 2^{x-2i}\binom{x-i-1}{i} f(1, N+x-1-2i). $$

For example, for $x=6$, we have

$$ f(6,N) = 32f(1,N+5)-32f(1,N+3)+6f(1,N+1). $$

I do this because $f(1,N)$ has an easy solution while $f(x,N)$ does not.

Is there a polynomial function that generates such coefficients? I assume this kind of a recursive relation should be common in math so that there might be more literature in it.

Here is a list of the values for small $x$:

  • $x=3 \rightarrow [4,-1]$
  • $x=4 \rightarrow [8,-4]$
  • $x=5 \rightarrow [16,-12,1]$
  • $x=6 \rightarrow [32,-32,6]$
  • $x=7 \rightarrow [64,-80,24,-1]$
  • $x=8 \rightarrow [128,-192,80, -8]$
  • $x=9 \rightarrow [256,-448,240,-40,1]$
ck1987pd
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See OEIS A113156 with generating function $ \dfrac 1{1 - 2tx + t^2}$ i.e. "Chebyshev polynomials of the second kind".

Raymond Manzoni
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  • Is it common knowledge that Chebyshev Polynomials appear in recursive relations or was it just somehow luck? – ck1987pd Apr 19 '23 at 12:24
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    Two questions here : many classical polynomials may be obtained this way, Bernoulli, Stirling, Gegenbauer (generalizing Chebyshev), Legendre and so on (classical orthogonal polynomials) $$-$$ How did I get this? Well typing 4,-1,8,-4,16 and so on at OEIS (works fine for triangles too). Excellent continuation, – Raymond Manzoni Apr 19 '23 at 12:38
  • This was fantastically helpful, not just for the question at hand but also for other problems I might have. Thank you! – ck1987pd Apr 19 '23 at 12:40
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    Glad it helped @C.Koca! OEIS is a great resource as well as free software like pari/gp (fast number theory) and others (Mathematica remains the best for symbolic integration but pari is faster!). – Raymond Manzoni Apr 19 '23 at 12:45
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    This was not directly related but somehow even more useful advise :) Thanks a lot! – ck1987pd Apr 19 '23 at 14:22