A285441 Expansion of q^(-2/5) * r(q)^2 * (1 + r(q) * r(q^2)^2) in powers of q where r() is the Rogers-Ramanujan continued fraction.

1, -1, 0, 2, -2, -2, 5, -1, -6, 7, 2, -12, 6, 11, -15, -2, 22, -14, -20, 31, 4, -41, 24, 37, -58, -9, 80, -44, -68, 105, 12, -143, 83, 119, -184, -16, 238, -144, -196, 307, 30, -391, 234, 317, -502, -49, 638, -374, -511, 804, 68, -1014, 600, 802, -1254, -99, 1562

Offset: 0

Author

Seiichi Manyama, Apr 19 2017

Keywords

• sign

Comments

G.f. A(q) satisfies: A(q) = q^(-2/5) * r(q)^2 * (1 + k(q)) = q^(-2/5) * r(q^2) * (1 - k(q)), where k(q) = r(q) * r(q^2)^2.

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..5000
• Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction

Crossrefs

Cf. A007325 (q^(-1/5) * r(q)), A055101, A112274 (k(q)), A112803 (1 + k(q)), A124242 (1 - k(q)), A285348, A285349.

Programs

Ruby

def s(k, m, n)
  s = 0
  (1..n).each{|i| s += i if n % i == 0 && i % k == m}
  s
end
def A007325(n)
  ary = [1]
  a = [0] + (1..n).map{|i| s(5, 1, i) + s(5, 4, i) - s(5, 2, i) - s(5, 3, i)}
  (1..n).each{|i| ary << (1..i).inject(0){|s, j| s - a[j] * ary[-j]} / i}
  ary
end
def mul(f_ary, b_ary, m)
  s1, s2 = f_ary.size, b_ary.size
  ary = Array.new(s1 + s2 - 1, 0)
  (0..s1 - 1).each{|i|
    (0..s2 - 1).each{|j|
      ary[i + j] += f_ary[i] * b_ary[j]
    }
  }
  ary[0..m]
end
def A285441(n)
  ary1 = A007325(n)
  ary2 = Array.new(n + 1, 0)
  (0..n / 2).each{|i| ary2[i * 2] = ary1[i]}
  ary = [-1] + mul(ary1, mul(ary2, ary2, n), n)[0..-2]
  mul(ary2, (0..n).map{|i| -ary[i]}, n)
end
p A285441(100)