OEIS/Eta products

Eta product signatures (EPS)

The coefficient sequences for generating functions that are products of the eta function can be computed conveniently by an Euler transform of some periodic integer sequence. Following Michael Somos, we describe an eta product by a matrix resp. a list of pairs (q_i, e_i) of the form

[q1,e1;q2,e2;q3,e3;...]

The q_i are the powers of q and the e_i are the powers of the eta functions. The latter can be negative.

Common generating functions

Partition numbers

A000041 1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231
Euler transform of period 1: [1]

Dedkind eta η (without the q^(1/24) factor)

A010815 1, -1, -1, 0, 0, 1, 0, 1, 0, 0, 0, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1
eta(q)
Euler transform of period 1: [-1]
eps P="[1,1]"

Jacobi theta_2 θ_2

A089800 2, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2 ... = theta_2(q)/q^(1/4)

Jacobi theta_3 θ_3, Ramanujan phi φ

A000122 1, 2, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2
a(0) = 1, for n >= 1: a(n) = 2 if n is a square, otherwise 0.
eta(q^2)^5 / (eta(q)*eta(q^4))^2
Euler transform of period 4: [2,-3,2,-1]
eps P="[2,5;1,-2;4,-2]"

Jacobi theta_4 θ_4

A002448 1, -2, 0, 0, 2, 0, 0, 0, 0, -2, 0, 0, 0, 0, 0, 0, 2
a(0) = 1, for n >= 1: a(n) = 2 * (-1)^sqrt(n) if n is a square, otherwise 0.
eta(q)^2 / eta(q^2)
Euler transform of period 2: [2,-1]
eps P="[1,2;2,-1]"

Ramanujan psi ψ

A010054 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0
a(n) = 1 if n is a triangular number, otherwise 0.
q^(-1/8) * eta(q^2)^2 / eta(q)
Euler transform of period 2: [1,-1]
eps P="[2,2;1,-1]"

Ramanujan chi χ

A000700 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5
q^(1/24) * eta(q^2)^2 /(eta(q) * eta(q^4))
Euler transform of period 4: [1,-1,1,0]
eps P="[2,2;1,-1;4,-1]"

Ramanujan f

A121373 1, 1, -1, 0, 0, -1, 0, -1, 0, 0, 0, 0, -1, 0, 0, 1
a(n) = (-1)^n * A010815(n)
q^(-1/24) * eta(q^2)^3 / (eta(q) * eta(q^4))
Euler transform of period 4: [[1,-2,1,-1]
eps P="[2,3;1,-1;4,-1]"

Leading power of q factor (pqf)

Some of the generating functions have an additional power of q factor, in order to normalize the leading coefficient to q^0.

• Example A226862:

Expansion of q^(-1/6) * eta(q^4) * eta(q^6)^5 / (eta(q^3) * eta(q^12))^2
eps P="[4,1;6,5;3,-2;12,-2]"

For all pairs we multiply the q power with the eta power and add the results, giving 4*1 + 6*5 - 3*2 - 12*2 = 4 + 30 - 6 - 24 = 4. This is the negative numerator of the pqf (expanded for denominator 24).

• Example A286813:

q^(-9/8) * (eta(q^2) * eta(q^16))^2 / (eta(q) * eta(q^8))
eps P="[2,2;16,2;1,-1;8,-1]" gives 2*2 * 16*2  - 1*1 - 8*1 = 36 - 9 = 27 and pqf = q^(-27/24)

• Example A286134 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, -1, -1

q^(-1/2) * eta(q^5) * eta(q^6) * eta(q^7) * eta(q^210)
eps P="[5,1;6,1;7,1;210,1]" gives 18 + 210 = 228 and pqf = q^(-228/24) = -9 -1/2. 

The sequence is shifted right by 9 positions in this case.

Normalization of an EPS

For long lists of generating functions it is convenient to require all EPSs to be normalized, such that they can be tested for equality and can be sorted. We use the following rules:

• The sublist of pairs with positive e_i comes before the sublist with negative e_i.
• Inside a sublist the first components q_i are in descending order.
• Successive pairs with the same q_i are combined into a single pair with the sum of the e_i as second component.

Exponentiation of an EPS

An EPS is taken to a power m by simply multiplying all q_i (the first components in the pairs) by m. The exponent may be 1/2 for the square root, or any other fraction m_n/md as long as the q_i are all divisible by md.

Multiplication of two EPSs

Two EPSs could be multiplied by simply concatenating their list of pairs. It is convenient, howver, to combine all occurrences of specific q_i by adding the corresponding eta exponents e.