The following deformations of usual numbers are well-known: $$ [n]_q=\frac{q^n-q^{-n}}{q-q^{-1}}, $$ and $$ [n]_{pq}=\frac{p^n-q^{-n}}{p-q^{-1}}. $$

Question. Are there any meaningful further deformations $[n]_{pqr...}?$

  • $\begingroup$ What's your source for this definition of $[n]_q$? What I'm familiar with is $[n]_q=\frac{q^n-1}{q-1}$, i.e. $[n]_q^\text{Leox}=q^{1-n}[n]_{q^2}^\text{Ernst}$ viz. Eq. (2.1) here. $\endgroup$ – J.G. Apr 13 at 16:46
  • $\begingroup$ see for example the paper QUANTUM GROUPS AND THEIR APPLICATIONS IN NUCLEAR PHYSICS at arxiv $\endgroup$ – Leox Apr 13 at 16:59
  • $\begingroup$ In other words, Eqs. (2.1) & (2.11) here. Thanks! It looks a good read. $\endgroup$ – J.G. Apr 13 at 17:00
  • 1
    $\begingroup$ Writing $q=e^\eta$ shows that $[n]_q$ is in some sense trigonometric. There are also 'elliptic deformations' of the integers, see e.g. (3.7) in arxiv.org/abs/1512.01720 . These are good (cf Somos' answer) in that there are again versions of the binomial theorem etc, and are related to elliptic quantum groups. $\endgroup$ – Jules Lamers Apr 14 at 7:10
  • $\begingroup$ You might, or might not get something out of the examples displayed here. $\endgroup$ – Cosmas Zachos May 4 at 15:16

The problem is to find a good meaning for "deformation" of the integers. Your second definition $$ [n]_{pq}=\frac{p^n-q^{-n}}{p-q^{-1}} \tag{1}$$ is essentially the same as the definition of the Lucas sequence function $$ U_n(P,Q) := \frac{\alpha^n-\beta^n}{\alpha-\beta} \tag{2}$$ where $\ \alpha,\beta\ $ are the roots of $\ x^2-Px+Q = 0.\ $ An essential property of these deformed integers is that they become integers in some suitable limit. For example, for the Lucas sequence $$ \lim_{\beta\to\alpha} \frac{\alpha^n-\beta^n}{\alpha-\beta} = n\alpha^{n-1} = n(P/2)^{n-1} \tag{3}$$ and so when $\ P=2\ $ and $\ Q=1\ $ we get the definition $\ U_n(2,1) := n,\ $ as it should be.

What is beyond Lucas sequences? Lucas himself was searching for this in his research into primality testing as described in Edouard Lucas and Primality Testing by Hugh C. Williams. So we are looking for sequences which satisfy recursion relations and which depend on parameters that reduce to the integers in some limiting case.

One direction would be to use classical special functions to define the sequence. For example, $$ T_n = T_n(x) := \tan(n\ x)/x \tag{4} $$ where $\ T_n(x) \to n\ $ as $\ x \to 0.\ $ One of many recursions it satisfies is $$ T_{n+1}(T_1 + 2T_{n-1} - T_n) = (T_1 + T_n)T_{n-1}. \tag{5} $$ Other special functions include the Jacobi $\ sn\ $ and $\ \theta\ $ functions, the Weierstrass $\ \sigma\ $ function and a few others. My own efforts in this direction are a part of my collection of special algebraic identities.

A related special case is Elliptic divisibilty sequences which satisfy many recurrence relations and naturally come from multiples of points on elliptic curves. My own efforts in this direction are partly desribed in my WXYZ project files.

  • $\begingroup$ Thank you for the answer. $\endgroup$ – Leox Apr 14 at 8:06

I've not seen one, but let's cook something up. If you don't mind, I'll add a comma betwen the two suffixes in the second deformation, so $pq$ doesn't look like a $1$-index product.

Since $[n]_{p,\,q}=\left(\frac{p}{q}\right)^{(n-1)/2}[n/2]_{p/q}$ expresses a double deformation in terms of a single deformation, one option is to define$$[n]_{p,\,q,\,r}:=\left(\frac{p}{q}\right)^{(n-1)/2}[n/2]_{p/q,\,r}=\left(\frac{p}{q}\right)^{(n-1)/2}\left(\frac{p}{qr}\right)^{(n/2-1)/2}[n/4]_{p/(qr)}$$etc., i.e. $$[n]_{p_1,\,\cdots,\,p_j}=\left(\prod_{k=2}^j\left(\frac{p_1}{\prod_{l=2}^k P_l}\right)^{(n/2^{k-2}-1)/2}\right)[n/2^{j-1}]_{p_1/\prod_{k\ge 2}p_k}.$$I'll leave you to invent an alternative in which $[n]_{p,\,q,\,r}:=\left(\frac{q}{r}\right)^{(n-1)/2}[n/2]_{p,\,q/r}$ etc. instead.

  • $\begingroup$ Thank you...... $\endgroup$ – Leox Apr 13 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.