Consider the sequence of [Apéry numbers](http://oeis.org/A005259)  

$$
A\_n = \sum\_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum\_{j=0}^k \binom{k}{j}^3
= \sum\_{k=0}^n \binom{n}{k}^2\binom{n+k}{k}^2 .
$$
In an email, physicist [Alan Sokal](http://www.physics.nyu.edu/faculty/sokal/) conjectures that it is a [Stieltjes moment sequence](http://en.wikipedia.org/wiki/Stieltjes_moment_problem). That is, that there exists a probability measure $\mu$ on $[0,+\infty)$ so that
$$
A\_n = \int\_{[0,+\infty)} s^n\;d\mu(s)
\tag{1}$$
for $n = 0,1,2,\dots$. [Of course you can equivalently say that $\mu$ is a nondecreasing function with $\mu(0)=0$ and $\lim\_{x\to+\infty} \mu(s) = 1$ and that (1) is a Stieltjes integral, rather than a "measure" integral.] 


**Is that conjecture correct? Is $A\_n$ a Stieltjes moment sequence?** 


[This question is a follow-up to [A conjectured formula for Apéry numbers](https://mathoverflow.net/questions/178790/) , where a formula for $A\_n$ was established.]



Is the sequence of Apéry numbers a Stieltjes moment sequence?
Here are some computations that may be useful. Here below, $\big[ \,f\,\big]\_a^b$ denotes $f(b)-f(a)$, and $n^{(k)}:=n(n-1)\dots(n-k+1)$.


**1. (An integration by parts).** *Let $w$ be a solution of the third order linear ODE on the interval $[a,b]$:
$$(x^4-34x^3+x^2)w''' + 3(2x^3-51x^2+x)w''+(7x^2-112x+1)w'+(x-5)w=0,$$
and put $$M(n):=\int\_a^b x^nw(x)dx$$ for any integer $n\ge0$.
Then, for any $n\ge 2$, 
$$n^3M(n)- (34n^3-51n^2+27n-5)M(n-1) + (n-1)^3M(n-2)= \Big[A(x)(x^n)'' + B(x)(x^n)' + C(x)x^n \Big]\_a^b,$$
where 
$$ A:=pw \qquad B:= qw- (pw)' \qquad C:= (pw )'' - (qw)'+ rw ,$$
and* $$p(x):=x^3-34x^2+x\qquad q(x):=3x^2-51x\qquad r(x):=x+10-x^{-1}.$$


**Proof**. We express the above polynomials of $n$ in the base $n^{(k)}$, then we absorb the latter terms as coefficients of derivatives of $x^n$, and finally we integrate by parts. 


We have:
$$n^3M(n)- (34n^3-51n^2+27n-5)M(n-1) + (n-1)^3M(n-2)=$$
$$\int\_a^b\Big\{n^3x^nw -(34n^3-51n^2+27n-5)x^{n-1}w +(n-1)^3x^{n-2}w\Big\}dx=$$
$$\int\_a^b\Big\{(n^ {(3)}+3n^ {(2)}+ n)x^nw -(34n^{(3)}+51n ^{(2)}+10n-5)x^{n-1}w +(n^{(3)}+n-1)x^{n-2}w\Big\}dx=$$
$$\int\_a^b\Big\{(x^n)'''pw+(x^n)''qw+(x^n)'rw+x^n(5x^{-1} -x^{-2} )w\}dx=$$
$$ \int\_a^bx^n \Big\{-(pw)'''+(qw)''-(rw)'+ (5x^{-1} -x^{-2} )w\}dx
+\Big[ (x^n)''pw- (x^n)'(pw)'+ x^n(pw)'' +(x^n)'qw-x^n(qw)' +rw \Big]\_a^b=$$
$$ -\int\_a^bx^{n-1} \Big\{(x^4-34x^3+x^2)w''' + 3(2x^3-51x^2+x)w''+(7x^2-112x+1)w'+(x-5)w\Big\}dx+$$
$$+\Big[ (x^n)''pw+ (x^n)'\big(qw- (pw)'\big)+ x^n\big( (pw )'' - (qw)'+ rw \big) \Big]\_a^b=$$
$$=\Big[A(x)(x^n)'' + B(x)(x^n)' + C(x)x^n \Big]\_a^b.\qquad\square$$


**2.(Consequence).** *Let $w$ a solution of the above linear equation on $(0,c)\setminus\{c\_0\}$, with $\int\_0^c w(x)dx=1$, and and assume it verifies the following linear boundary conditions, expressed in terms of the above coefficients $A,B,C$:* 


i) $A(x)=o(1),\quad B(x)=o(1),\quad C(x)=O(1)$, as $x\to0$;


ii) $A(x),\quad B(x),\quad C(x)$, are continuous at $x=c\_0$ ,


iii) $A(x)=o(1) ,\quad B(x)=o(1) ,\quad C(x)=o(1)$, as $x\to c$.


*Then the corresponding $M(n)$ are the Apéry sequence.*


Indeed, computing the integral on $[0,c]$ for $M(n)$ as limit of integrals on $[\epsilon, c\_0-\epsilon]\cup[c\_0+\epsilon, c-\epsilon]$ as $\epsilon\to0$, and applying the above integration by parts formula, one gets that $M(n)$ satisfy the Apéry's recurrence, with $M(0)=1$ (note that $M(1)=5M(0)$ follows from the recurrence as well).


**rmk.** *This also include the case where $w$ vanishes identically on $(0,c\_0)$, and the condition is simply that $A, B, C$ should vanish both at $x=c\_0$ and at $x=c$*. 


**3. (Positive solutions of the third order ODE).** *Assume $u$ solves the second order linear ODE 
$$(x^3-34x^2+x)u''+(2x^2-51x+1)u'+\frac{1}{4}(x-10)u=0.$$
Then $w:=u^2$ solves*
$$(x^4-34x^3+x^2)w''' + 3(2x^3-51x^2+x)w''+(7x^2-112x+1)w'+(x-5)w=0.$$


**Proof**. Put $$P:=x^4-34x^3+x^2\qquad Q:=\frac{x^2}{2}-5x,$$
so the equation for $u$ (multiplied by $2x$) writes:
$$2Pu''+P'u'+Qu=0,$$
and we have
$$0=(2Pu''+P'u'+Qu)'u+ 3(2Pu''+P'u'+Qu)u'=$$
$$=(2Pu'''+2P'u''+P'u''+P''u'+Qu'+Q'u)u+3(2Pu''+P'u'+Qu)u'=$$
$$=P(2u'''u+6u''u')+3P'(u''u+u'^2)+(P''+4Q)u'u+Q'u^2=$$
$$=Pw''' +\frac{3}{2}P'w''+\Big(\frac{P''}{2}+2Q\Big)w'+Q'w,$$
which is the above third order equation for $w$.


