I think I can prove the bound $a_n> \prod 2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ if $n=\prod p_i^{\alpha_i}$. It is certainly very far of the real $a_n$, but it is more than enough to prove $\lambda_A=0$.

First a lemma:

$\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ is never a square

Proof: If $p=2$ it is obviously true. Wlog $p$ is odd. We actually want to analyze the polynomial $f(x)=\frac{x^p-1}{x-1}$ for $x=2^{p^{d-1}}$. We will prove actually that this polynomial is never a square for "big" $x$. The idea is approximating $\sqrt{f(x)}$ by a polynomal with rational coefficients. Notice that $\sqrt{f(x)}<\sqrt{\frac{x^p}{x-1}}=x^{\frac{p-1}{2}}\sqrt{\frac{x}{x-1}}=x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}$ (by the generalized binomal theorem).

So $\sqrt{f(x)}-x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k} < x^{\frac{p-1}{2}}\sum\limits_{k=\frac{p+1}{2}}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}<\frac{\binom{p+1}{\frac{p+1}{2}}}{2^{p+1}}\frac{1}{x-1}$ (substituting all denominators for the first one, which is greater).

Also, the left side can't be zero, because all the coefficients of $(x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k})^2$ are at most $1$ (the first ones are exactly $1$, while the last ones are strictly less than $1$).

So if $\sqrt{f(x)}$ is an integer, the left side will be at least $\frac{1}{2^{p-1}}$ (because all denominators divide $2^{p-1}$). So, if $f(x)$ is a perfect square, $x-1<\frac{\binom{p+1}{\frac{p+1}{2}}}{2}<2^{p}-1$.

It indeed doesn't happen for $x=2^{p^{d-1}}$ for $d>1$. For $x=2$, $f(2)$ is clearly not a square because $f(2) \equiv 3$ mod $4$.

Now back to the bound.

Let $p^d$ a prime power factor of $n$. Notice that $gcd (\frac{2^{p^d}-1}{2^{p^{d-1}}-1},2^{p^{d-1}}-1)$ must divide $p$, (in general $gcd(\frac{x^p-1}{x-1},x-1)|p$, because $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+1 \equiv p \,(mod \,(x-1))$) therefore this gcd is $1$ (because clearly $p \nmid 2^{p^{d-1}}-1$).

Now we take a prime $q$ with odd exponent in $\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ (such prime exists because this expression is not a square). Since $q$ doesn't divide $2^{p^{d-1}}-1$ (by the earlier $gcd$ condition), $q$ satisfies $ord_q(2)=p^d \Rightarrow 2p^d|q-1 \Rightarrow q>2p^d$. Notice that this prime $q$ is a factor of $a_{p^d}$.

Repeating for $d-1$, $d-2$, ..., $1$ we get that $2^{p^d}-1$ has at least $d$ distinct prime factors that divide it with odd exponent, whose product is at least $2^d p^{\frac{d(d+1)}{2}}$. Since $2^{p_i^{\alpha_i}}-1|2^n-1$ and all $2^{p_i^{\alpha_i}}-1$ are pairwise coprime, we proved the bound previously stated. (notice that the factor $2^{\alpha_1}$ actually doesn't appear for $i=1$ ($p_1=2$), but it won't make a big difference).

Now, for the proof that $\delta_A=0$, we will use that the bound we proved gives $a_n>n 2^{\omega(n)-1}$. This is because $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}\ge 2 p_i^{\alpha_i}$ (for $i=1$, since we don't have the factor $p_i^{\alpha_i}$ we use only $p_1^{\frac{(\alpha_1)(\alpha_1+1)}{2}}\ge p_1^{\alpha_1}$, what explains the $-1$ after $\omega(n)$)

But by http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem, $\omega(n)$ is almost always about $log(log(n))$, so $a_n$ is almost always at least $C n log (n)$, which give a zero density for $A$.

(notice that the factor $2^{\alpha_i}$ in $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ is very important, otherwise we wouldn't be able to handle squarefree $n$)

I think I can prove the bound $a_n> \prod 2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ if $n=\prod p_i^{\alpha_i}$. It is certainly very far of the real $a_n$, but it is more than enough to prove $\delta_A=0$.

First a lemma:

$\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ is never a square

Proof: If $p=2$ it is obviously true. Wlog $p$ is odd. We actually want to analyze the polynomial $f(x)=\frac{x^p-1}{x-1}$ for $x=2^{p^{d-1}}$. We will prove actually that this polynomial is never a square for "big" $x$. The idea is approximating $\sqrt{f(x)}$ by a polynomal with rational coefficients. Notice that $\sqrt{f(x)}<\sqrt{\frac{x^p}{x-1}}=x^{\frac{p-1}{2}}\sqrt{\frac{x}{x-1}}=x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}$ (by the generalized binomal theorem).

So $\sqrt{f(x)}-x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k} < x^{\frac{p-1}{2}}\sum\limits_{k=\frac{p+1}{2}}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}<\frac{\binom{p+1}{\frac{p+1}{2}}}{2^{p+1}}\frac{1}{x-1}$ (substituting all denominators for the first one, which is greater).

Also, the left side can't be zero, because all the coefficients of $(x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k})^2$ are at most $1$ (the first ones are exactly $1$, while the last ones are strictly less than $1$).

So if $\sqrt{f(x)}$ is an integer, the left side will be at least $\frac{1}{2^{p-1}}$ (because all denominators divide $2^{p-1}$). So, if $f(x)$ is a perfect square, $x-1<\frac{\binom{p+1}{\frac{p+1}{2}}}{2}<2^{p}-1$.

It indeed doesn't happen for $x=2^{p^{d-1}}$ for $d>1$. For $x=2$, $f(2)$ is clearly not a square because $f(2) \equiv 3$ mod $4$.

Now back to the bound.

Let $p^d$ a prime power factor of $n$. Notice that $gcd (\frac{2^{p^d}-1}{2^{p^{d-1}}-1},2^{p^{d-1}}-1)$ must divide $p$, (in general $gcd(\frac{x^p-1}{x-1},x-1)|p$, because $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+1 \equiv p \,(mod \,(x-1))$) therefore this gcd is $1$ (because clearly $p \nmid 2^{p^{d-1}}-1$).

Now we take a prime $q$ with odd exponent in $\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ (such prime exists because this expression is not a square). Since $q$ doesn't divide $2^{p^{d-1}}-1$ (by the earlier $gcd$ condition), $q$ satisfies $ord_q(2)=p^d \Rightarrow 2p^d|q-1 \Rightarrow q>2p^d$. Notice that this prime $q$ is a factor of $a_{p^d}$.

Repeating for $d-1$, $d-2$, ..., $1$ we get that $2^{p^d}-1$ has at least $d$ distinct prime factors that divide it with odd exponent, whose product is at least $2^d p^{\frac{d(d+1)}{2}}$. Since $2^{p_i^{\alpha_i}}-1|2^n-1$ and all $2^{p_i^{\alpha_i}}-1$ are pairwise coprime, we proved the bound previously stated. (notice that the factor $2^{\alpha_1}$ actually doesn't appear for $i=1$ ($p_1=2$), but it won't make a big difference).

Now, for the proof that $\delta_A=0$, we will use that the bound we proved gives $a_n>n 2^{\omega(n)-1}$. This is because $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}\ge 2 p_i^{\alpha_i}$ (for $i=1$, since we don't have the factor $p_i^{\alpha_i}$ we use only $p_1^{\frac{(\alpha_1)(\alpha_1+1)}{2}}\ge p_1^{\alpha_1}$, what explains the $-1$ after $\omega(n)$)

Now, set some fixed $t$:

\delta_A=$\lim\limits_{n \to \infty} \frac{\#\{ a_k \in A | a_k<n\}}{n} = \lim\limits_{n \to \infty} \frac{\#\{a_k \in A | a_k<n, \omega(k)<t\}}{n} + \frac{\#\{a_k \in A | a_k<n, \omega(k)\ge t\}}{n}$

By http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem, $\frac{\#\{a_k \in A | a_k<n, \omega(k)<t\}}{n}$ goes to zero, because by our bound, if $a_k<n$, then $k<n$, and one of the consequences of the theorem presented is that the set of numbers with $\omega(k)<t$ has natural density zero.

By our bound $\#\{a \in A | a<n, \omega(a)\ge t\}$ is at most $\frac{n}{2^{t-1}}$, because it has no elements $a_k$ with $k>\frac{n}{2^{t-1}}$ otherwise $a_k>k2^{\omega(k)-1}>\frac{n}{2^{t-1}}2^{t-1}=n $.

Therefore for each fixed $t$, $\delta_A\le\frac{1}{2^{t-1}}$, so we prove that indeed $\delta_A=0$.

(notice that the factor $2^{\alpha_i}$ in $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ is very important, otherwise we wouldn't be able to handle squarefree $n$)

I think I can prove the bound $a_n> \prod 2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ if $n=\prod p_i^{\alpha_i}$. It is certainly very far of the real $a_n$, but it is more than enough to prove $\lambda_A=0$.

First a lemma:

$\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ is never a square

Proof: If $p=2$ it is obviously true. Wlog $p$ is odd. We actually want to analyze the polynomial $f(x)=\frac{x^p-1}{x-1}$ for $x=2^{p^{d-1}}$. We will prove actually that this polynomial is never a square for "big" $x$. The idea is approximating $\sqrt{f(x)}$ by a polynomal with rational coefficients. Notice that $\sqrt{f(x)}<\sqrt{\frac{x^p}{x-1}}=x^{\frac{p-1}{2}}\sqrt{\frac{x}{x-1}}=x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}$ (by the generalized binomal theorem).

So $\sqrt{f(x)}-x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k} < x^{\frac{p-1}{2}}\sum\limits_{k=\frac{p+1}{2}}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}<\frac{\binom{p+1}{\frac{p+1}{2}}}{2^{p+1}}\frac{1}{x-1}$ (substituting all denominators for the first one, which is greater).

Also, the left side can't be zero, because all the coefficients of $(x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k})^2$ are at most $1$ (the first ones are exactly $1$, while the last ones are strictly less than $1$).

So if $\sqrt{f(x)}$ is an integer, the left side will be at least $\frac{1}{2^{p-1}}$ (because all denominators divide $2^{p-1}$). So, if $f(x)$ is a perfect square, $x-1<\frac{\binom{p+1}{\frac{p+1}{2}}}{2}<2^{p}-1$.

It indeed doesn't happen for $x=2^{p^{d-1}}$ for $d>1$. For $x=2$, $f(2)$ is clearly not a square because $f(2) \equiv 3$ mod $4$.

Now back to the bound.

Let $p^d$ a prime power factor of $n$. Notice that $gcd (\frac{2^{p^d}-1}{2^{p^{d-1}}-1},2^{p^{d-1}}-1)$ must $p$, so this gcd is $1$ (clearly $p \nmid 2^{p^{d-1}}-1$). Now we take a prime $q$ that divides exactly $\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ (I mean, $q^2$ doesn't divide it)(it exists since it is not a square). Since it doesn't divide $2^{p^{d-1}}-1$, it satisfies $ord_q(2)=p^d \Rightarrow q>2p^d$. Repeating for $d-1$, $d-2$, ..., $1$ we get that $2^{p^d}-1$ has at least $d$ distinct prime factors that divide it exactly, whose product is at least $2^d p^{\frac{d(d+1)}{2}}$. Since $2^{p_i^{\alpha_i}}-1|2^n-1$ and all $2^{p_i^{\alpha_i}}-1$ are pairwise coprime, we proved the bound previously stated. (notice that the factor $2^{\alpha_1}$ actually doesn't appear for $p_1=2$, but it won't make a big difference).

Now, for the proof that $\delta_A=0$, we will use that the bound we proved gives $a_n>n 2^{\omega(n)-1}$  (notice that the factor $2^{\alpha_i}$ in $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ is very important, otherwise we wouldn't know how to handle squarefree $n$).

But by http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem, $\omega(n)$ is almost always about $log(log(n))$, so $a_n$ is almost always at least $C n log (n)$, which give a zero density for $A$.

I think I can prove the bound $a_n> \prod 2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ if $n=\prod p_i^{\alpha_i}$. It is certainly very far of the real $a_n$, but it is more than enough to prove $\lambda_A=0$.

First a lemma:

$\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ is never a square

Proof: If $p=2$ it is obviously true. Wlog $p$ is odd. We actually want to analyze the polynomial $f(x)=\frac{x^p-1}{x-1}$ for $x=2^{p^{d-1}}$. We will prove actually that this polynomial is never a square for "big" $x$. The idea is approximating $\sqrt{f(x)}$ by a polynomal with rational coefficients. Notice that $\sqrt{f(x)}<\sqrt{\frac{x^p}{x-1}}=x^{\frac{p-1}{2}}\sqrt{\frac{x}{x-1}}=x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}$ (by the generalized binomal theorem).

So $\sqrt{f(x)}-x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k} < x^{\frac{p-1}{2}}\sum\limits_{k=\frac{p+1}{2}}^{\infty} x^{-k}\frac{\binom{2k}{k}}{4^k}<\frac{\binom{p+1}{\frac{p+1}{2}}}{2^{p+1}}\frac{1}{x-1}$ (substituting all denominators for the first one, which is greater).

Also, the left side can't be zero, because all the coefficients of $(x^{\frac{p-1}{2}}\sum\limits_{k=1}^{\frac{p-1}{2}} x^{-k}\frac{\binom{2k}{k}}{4^k})^2$ are at most $1$ (the first ones are exactly $1$, while the last ones are strictly less than $1$).

So if $\sqrt{f(x)}$ is an integer, the left side will be at least $\frac{1}{2^{p-1}}$ (because all denominators divide $2^{p-1}$). So, if $f(x)$ is a perfect square, $x-1<\frac{\binom{p+1}{\frac{p+1}{2}}}{2}<2^{p}-1$.

It indeed doesn't happen for $x=2^{p^{d-1}}$ for $d>1$. For $x=2$, $f(2)$ is clearly not a square because $f(2) \equiv 3$ mod $4$.

Now back to the bound.

Let $p^d$ a prime power factor of $n$. Notice that $gcd (\frac{2^{p^d}-1}{2^{p^{d-1}}-1},2^{p^{d-1}}-1)$ must divide $p$, (in general $gcd(\frac{x^p-1}{x-1},x-1)|p$, because $\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+...+1 \equiv p \,(mod \,(x-1))$) therefore this gcd is $1$ (because clearly $p \nmid 2^{p^{d-1}}-1$).

Now we take a prime $q$ with odd exponent in $\frac{2^{p^d}-1}{2^{p^{d-1}}-1}$ (such prime exists because this expression is not a square). Since $q$ doesn't divide $2^{p^{d-1}}-1$ (by the earlier $gcd$ condition), $q$ satisfies $ord_q(2)=p^d \Rightarrow 2p^d|q-1 \Rightarrow q>2p^d$. Notice that this prime $q$ is a factor of $a_{p^d}$.

Repeating for $d-1$, $d-2$, ..., $1$ we get that $2^{p^d}-1$ has at least $d$ distinct prime factors that divide it with odd exponent, whose product is at least $2^d p^{\frac{d(d+1)}{2}}$. Since $2^{p_i^{\alpha_i}}-1|2^n-1$ and all $2^{p_i^{\alpha_i}}-1$ are pairwise coprime, we proved the bound previously stated. (notice that the factor $2^{\alpha_1}$ actually doesn't appear for $i=1$ ($p_1=2$), but it won't make a big difference).

Now, for the proof that $\delta_A=0$, we will use that the bound we proved gives $a_n>n 2^{\omega(n)-1}$. This is because $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}\ge 2 p_i^{\alpha_i}$ (for $i=1$, since we don't have the factor $p_i^{\alpha_i}$ we use only $p_1^{\frac{(\alpha_1)(\alpha_1+1)}{2}}\ge p_1^{\alpha_1}$, what explains the $-1$ after $\omega(n)$)

But by http://en.wikipedia.org/wiki/Hardy%E2%80%93Ramanujan_theorem, $\omega(n)$ is almost always about $log(log(n))$, so $a_n$ is almost always at least $C n log (n)$, which give a zero density for $A$.

(notice that the factor $2^{\alpha_i}$ in $2^{\alpha_i}p_i^{\frac{(\alpha_i)(\alpha_i+1)}{2}}$ is very important, otherwise we wouldn't be able to handle squarefree $n$)