A343446 Coefficients of the series S(p, q) for which -(p^(1/3))*S converges to the largest real root of x^4 - p*x + q, where 0 < p and 0 < q < 3*(p/4)^(4/3).

-1, 1, 4, 40, 648, 14560, 418880, 14696640, 608608000, 29056867840, 1571364748800, 94937979136000, 6337884013260800, 463301182536192000, 36806315255277568000, 3157533815406530560000, 290912372128665391104000, 28648563542097847828480000

Offset: 0

Author

Dixon J. Jones, May 26 2021

Keywords

• sign
• easy

Comments

Based on formulas for series solutions of trinomials given in Eagle article.

S(p, q) = Sum_{n>=0} (a(n)*q^n)/((3^n)*(p^(4n/3))*n!)

In general, given m > 1, p > 0 and 0 < q < m*(p/(m + 1))^((m + 1)/m), the series S(m, p, q) for which (-p^(1/m))*S converges to the largest real root of x^(m + 1) - p*x + q has coefficients c(n) = m^(n - 1)*((n + m - 1)/m)(n - 1), where (x)_k is the Pochhammer symbol for Gamma(x + k)/Gamma(k), and S(m, p, q) = Sum{n>=0}(c(n)*q^n)/((m^n)*(p^(n*(m + 1)/m)*n!).

Links

• Albert Eagle, Series for all the roots of a trinomial equation, Am. Math. Monthly, vol. 46, no. 7 (Aug. - Sep., 1939), pp. 422 - 425.

Crossrefs

A343445 relates similarly to the largest real root of x^3 - p*x + q.

A206300 relates similarly to the largest real root of x^3 - 3*u*x + 4*u, u >= 4.

Programs

Maple

a := proc(n) option remember; if n < 3 then [-1, 1, 4][n+1] else 4*(4*n - 7)*(4*n - 10)*(4*n - 13)*a(n-3) fi; end:
seq(a(n), n = 0..20); # Peter Bala, Jul 23 2024

Mathematica

Clear[a]; a=Table[3^(n - 1) Pochhammer[(n + 2)/3, n - 1], {n, 0, 20}]
(* In general, for the series S(m, p, q) for which (-p^(1/m))*S converges to the largest real root of x^(m + 1) - p*x + q, the first n + 1 coefficients are: *)
Clear[c]; c[m_,n_] := Table[m^(k - 1) Pochhammer[(k + m - 1)/m, k - 1], {k, 0, n}](* and S(m, p, q) to n + 1 terms is given by *)
Clear[s]; s[m_,p_,q_,n_]:= Sum[c[m,n][[k + 1]]*q^k/((m^k)*(p^(k (m + 1)/m))*k!), {k, 0, n}]

Formula

a(n) = 3^(n - 1)*((n + 2)/3)_(n - 1), where (x)_k is the Pochhammer symbol for Gamma(x + k)/Gamma(x).

a(n) = 4*(4*n - 7)*(4*n - 10)*(4*n - 13)*a(n-3) with a(0) = -1, a(1) = 1 and a(2) = 4. - Peter Bala, Jul 23 2024