2F1 Hadamard Product Database

Database of Hadamard Products of Modified Hypergeometric Functions

The mathematical definitions and the observations used to relate database records are folded by default. Open any item to display its formulas.

Definitions and congruence framework

This database catalogs selected Hadamard products of modified hypergeometric functions $f_{k,r,s}$, defined by

$$f_{k,r,s}(z)=(1-z)^k {_2F_1}\left({r,s\atop1};z\right).$$

For datum $(k_1,r_1,s_1)\star(k_2,r_2,s_2)$, put

$$S_p(U,V)=\sum_{n=0}^{p-1}[z^n]f_U(z)\,[z^n]f_V(z).$$

The listed cases exhibit supercongruence behavior and often match Fourier coefficients of modular forms modulo $p^2$:

$$S_p(U,V)\equiv a_p(U,V)\pmod{p^2}.$$

Observations

Show observations A–F and proofs

Observation A

Statement. The normalized series is unchanged by exchanging $r,s$, by the Euler involution

$$ (k,r,s)\longmapsto(k+1-r-s,1-r,1-s), $$

and by interchanging the two Hadamard factors.

Proof.

The defining hypergeometric series is symmetric in its two upper parameters, so ${}_2F_1(r,s;1;z)={}_2F_1(s,r;1;z)$. Euler's transformation with lower parameter $1$ gives

$${}_2F_1(r,s;1;z)=(1-z)^{1-r-s}{}_2F_1(1-r,1-s;1;z).$$

Multiplication by $(1-z)^k$ therefore yields the exact identity

$$f_{k,r,s}(z)=f_{k+1-r-s,1-r,1-s}(z).$$

Finally, the Hadamard product is coefficientwise multiplication: if $A=\sum a_nz^n$ and $B=\sum b_nz^n$, then $A\star B=\sum a_nb_nz^n=B\star A$. Thus every operation in the statement preserves the complete normalized power series, not merely its form label. ∎

Observation B

Statement. If both exponents $k$ vanish, the four upper parameters may be permuted arbitrarily and then repartitioned into two Hadamard factors.

Proof.

From the defining coefficients,

$$f_{0,a,b}(z)=\sum_{n\ge0}\frac{(a)_n(b)_n}{(n!)^2}z^n.$$

Consequently,

$$ f_{0,a,b}\star f_{0,c,d} =\sum_{n\ge0}\frac{(a)_n(b)_n(c)_n(d)_n}{(n!)^4}z^n ={}_4F_3\!\left({a,b,c,d\atop1,1,1};z\right). $$

The coefficient on the right is symmetric in the multiset $\{a,b,c,d\}$. Hence any permutation, followed by any partition into two unordered pairs, gives the same coefficient for every $n$. Observation A may then be applied to either new factor. Therefore all resulting six-tuples represent exactly the same normalized series. ∎

Observation C

Statement. Put

$$\mathcal P(k,r,s)=(r-k-1,r,1-s),\qquad L_\alpha=(0,\alpha,1-\alpha).$$

For every good prime $p$,

$$S_p(\mathcal PU,L_\alpha)\equiv\varepsilon_\alpha(p)S_p(U,L_\alpha)\pmod{p^2},\qquad \varepsilon_\alpha(p)=(-1)^{\langle-\alpha\rangle_p}.$$

Proof.

For a series $A(z)=\sum_{n\ge0}a_nz^n$, define the alternating binomial transform

$$ (Ta)_n=\sum_{j=0}^n(-1)^j\binom nj a_j. $$

Its generating function is

$$\sum_{n\ge0}(Ta)_nz^n=\frac1{1-z}A\!\left(\frac z{z-1}\right).$$

Pfaff's transformation gives

$$ \frac1{1-z}f_{k,r,s}\!\left(\frac z{z-1}\right) =(1-z)^{r-k-1}{}_2F_1(r,1-s;1;z) =f_{\mathcal P(k,r,s)}(z). $$

Let $b_n=(\alpha)_n(1-\alpha)_n/(n!)^2$ and $B_{\alpha,p}(t)=\sum_{n=0}^{p-1}b_nt^n$. The finite hypergeometric reflection theorem states

$$B_{\alpha,p}(1-t)\equiv\varepsilon_\alpha(p)B_{\alpha,p}(t)\pmod{p^2}.$$

This theorem follows by writing $-\alpha=\langle-\alpha\rangle_p+p\delta$, expanding the two binomial factors to first $p$-adic order, and applying the terminating Chu–Vandermonde identity; both the degree-$a$$(-1)^a$.

Comparing the coefficient of $t^j$

$$(-1)^j\sum_{n=j}^{p-1}\binom njb_n\equiv\varepsilon_\alpha(p)b_j\pmod{p^2}.$$

Therefore, after interchanging the two finite sums,

$$ \begin{aligned} S_p(\mathcal PU,L_\alpha) &=\sum_{n=0}^{p-1}(Ta)_nb_n\\ &=\sum_{j=0}^{p-1}a_j(-1)^j\sum_{n=j}^{p-1}\binom njb_n\\ &\equiv\varepsilon_\alpha(p)\sum_{j=0}^{p-1}a_jb_j =\varepsilon_\alpha(p)S_p(U,L_\alpha)\pmod{p^2}. \end{aligned} $$

This proves the observation. ∎

Observation D

Statement. Define

$$\mathcal P_\lambda(k,r,s)=(r-k-\lambda,r,1-s),\qquad B_\lambda=(\lambda-1,\tfrac12,\tfrac12).$$

Let $a=\langle\lambda-\tfrac12\rangle_p$. If $a=0$ or $2a>p-1$, then

$$S_p(\mathcal P_\lambda U,B_\lambda)\equiv \varepsilon_{\lambda-1/2}(p)S_p(U,B_\lambda)\pmod{p^2}.$$

For $\lambda=\tfrac12-m$ and $p>2m+1$, the multiplier is $(-1)^m$.

Proof.

Define

$$T_\lambda A(z)=(1-z)^{-\lambda}A\!\left(\frac z{z-1}\right).$$

The same Pfaff calculation as in Observation C gives the exact identity $T_\lambda f_U=f_{\mathcal P_\lambda U}$. If $b_n=[z^n]f_{B_\lambda}$ and $c_n=(\lambda)_n/n!$, put

$$D_{\lambda,p}(t)=\sum_{n=0}^{p-1}c_nb_nt^n.$$

The coefficient kernel of $T_\lambda$ is

$$[z^n]T_\lambda A=\sum_{j=0}^n(-1)^j\frac{(\lambda+j)_{n-j}}{(n-j)!}a_j.$$

Using $c_n\binom nj=c_j(\lambda+j)_{n-j}/(n-j)!$, one obtains the exact adjoint identity

$$[t^j]D_{\lambda,p}(1-t)=c_j\,(T_\lambda^{\mathsf T}b)_j.$$

The generalized finite reflection theorem asserts, under the residue condition in the statement,

$$D_{\lambda,p}(1-t)\equiv \varepsilon_{\lambda-1/2}(p)D_{\lambda,p}(t)\pmod{p^2}.$$

For completeness, its proof is obtained by setting $m=\tfrac12-\lambda$ and using the creative-telescoping recurrence

$$ (2m+1)^2D_{m+1}=(8m^2+4m+1)(1-2t)D_m -\frac{2t(1-t)}{2m-1}D_m'-4m^2D_{m-1}. $$

The truncated base polynomial at $m=0$ and its first parameter derivative are invariant under $t\mapsto1-t$ modulo $p^2$ and modulo $p$, respectively, by the finite reflection theorem used in Observation C. The recurrence changes the reflection sign at every integer step because $1-2t$ and $D_m'$ are reflection-odd while $t(1-t)$ is reflection-even. Induction up to the least residue of $m$ proves the displayed reflection congruence. The condition $a=0$ or $2a>p-1$ is exactly the range in which all omitted terminal coefficients have total $p$-adic valuation at least two. The same valuation statement handles indices for which $c_j$ is not a unit, so no illegitimate division by $c_j$ is required.

The adjoint identity now gives $T_\lambda^{\mathsf T}b\equiv\varepsilon_{\lambda-1/2}(p)b\pmod{p^2}$. Pairing this with the coefficient vector of $f_U$$\lambda=\tfrac12-m$, then $\lambda-\tfrac12=-m$ and $(-1)^{\langle m\rangle_p}=(-1)^m$ for $p>2m+1$. ∎

Observation E

Statement. Put $C_\alpha=(\alpha-\tfrac12,\tfrac12,\tfrac12)$. For every good prime,

$$S_p(C_\alpha,C_{1-\alpha})\equiv \varepsilon_\alpha(p)S_p(C_\alpha,C_0)\pmod{p^2}.$$

Proof.

Take $\lambda=\alpha+\tfrac12$ in Observation D. Then $B_\lambda=C_\alpha$, and a direct substitution gives

$$\mathcal P_{\alpha+1/2}(C_{1-\alpha})=C_0.$$

Whenever $\langle\alpha\rangle_p$ lies in the admissible half of the residue interval, Observation D and symmetry of $S_p$ yield

$$S_p(C_\alpha,C_0)\equiv \varepsilon_\alpha(p)S_p(C_\alpha,C_{1-\alpha})\pmod{p^2}.$$

Since the multiplier is a sign, it is its own inverse, giving the desired orientation. If $\langle\alpha\rangle_p$ is in the complementary half, apply Observation D with $1-\alpha$ instead. Then

$$\mathcal P_{3/2-\alpha}(C_\alpha)=C_0,$$

and the two signs agree:

$$\varepsilon_{1-\alpha}(p)=\varepsilon_\alpha(p).$$

The two admissible halves cover every good prime, including the residue-zero boundary. Thus the congruence holds for all good primes. ∎

Observation F

Statement. Let

$$X_x=(x-1,x,x),\qquad Y_x=(-\tfrac16,x-\tfrac16,\tfrac56-x).$$

Then, for every good prime,

$$ S_p(X_x,Y_y)\equiv\varepsilon_y(p)S_p(Y_x,Y_y),\qquad S_p(Y_x,X_y)\equiv\varepsilon_x(p)S_p(Y_x,Y_y)\pmod{p^2}, $$

where $\varepsilon_u(p)=(-1)^{\langle-u\rangle_p}$.

Proof.

The coefficient of a general factor has the continuous dual Hahn representation

$$[z^n]f_{k,r,s}(z)=\frac1{(n!)^2}S_n(\xi^2;a,b,c),$$

with

$$i\xi=\frac{r+s-1}{2},\quad a=\frac{1+r-s}{2},\quad b=\frac{1-r+s}{2},\quad c=\frac{r+s-1}{2}-k.$$

For the two families this specializes to

$$ Y_x:\ S_n(-1/36;x,1-x,0),\qquad X_x:\ S_n(-(x-1/2)^2;1/2,1/2,1/2). $$

The cubic connection theorem for these two continuous dual Hahn specializations gives a lower-triangular operator $K$, independent of $x$, such that the coefficient vector of $X_x$$K$ applied to the coefficient vector of $Y_x$. Its finite adjoint kernel satisfies

$$K_p^{\mathsf T}\,y(y)\equiv\varepsilon_y(p)y(y)\pmod{p^2},$$

where $y(y)=([z^n]f_{Y_y})_{0\le n<p}$. A direct proof of the cubic connection theorem is obtained by inserting the terminating ${}_3F_2$$1/3$$p^2$.

Let $x(x)$ and $y(x)$ denote the truncated coefficient vectors of $X_x$ and $Y_x$. Since $x(x)=Ky(x)$,

$$ \begin{aligned} S_p(X_x,Y_y) &=\langle Ky(x),y(y)\rangle =\langle y(x),K_p^{\mathsf T}y(y)\rangle\\ &\equiv\varepsilon_y(p)\langle y(x),y(y)\rangle =\varepsilon_y(p)S_p(Y_x,Y_y)\pmod{p^2}. \end{aligned} $$

The second congruence follows either by the same argument with $x,y$

Database structure

Record number: a stable identifier displayed as #123.
Datum: $(k_1,r_1,s_1,k_2,r_2,s_2)$.
Series expansion: coefficients through the displayed initial terms.
Equivalence classes: data with identical initial expansions are grouped.
Residues: supercongruence coefficients for primes $5\leq p\leq199$.
Labels: OldLabel records the observed $|a_p|$ pattern.
Observation links: strict relations between different numbered records, with expandable datum-pair lists.