Does there exist a sequence of strongly regular graphs with parameters $(n,d,\lambda,\mu)$ (so every pair of adjacent vertices have $\lambda$ common neighbours, and every pair of non-adjacent ones have $\mu$ common neighbours), with $d - \mu = \Omega(n)$ and $\mu - \lambda = \Omega(n)$?



Strongly regular graphs with certain parameters
I think, the answer is negative (I expect that by $\Omega(n)$ you mean something positive of order $n$). Denote $\mu-\lambda=k,d-\mu=\ell$.


The eigenvalues of the adjacency matrix of such a graph $G$ are $d$ and $t\_{1,2}$ which are the roots of a quadratic equation $t^2+kt-\ell=0$. The multiplicities $n\_{1},n\_2$ enjoy the equations $n\_1+n\_2=n-1$, $n\_1t\_1+n\_2t\_2=-d$. Consider two cases.


1. $n\_1=n\_2$. Then $n\_1=n\_2=(n-1)/2$ and $-d=(t\_1+t\_2)\frac{n-1}2=-k\frac{n-1}2$ which is too large (of order $n^2$).
2. $n\_1\ne n\_2$. Then from $t\_1+t\_2=-k$ and $n\_1t\_1+n\_2t\_2=-d$ we get $t\_2(n\_2-n\_1)=kn\_1-d$ and so $t\_2$ is rational, i.e., the discriminant $k^2+4\ell$ of our quadratic equation is a perfect square, let it be equal to $k^2+4\ell=(k+2x)^2$, then $x^2+kx=\ell$. It yields that $x$ is bounded. So, the roots are $t\_2=x$ and $t\_1=-k-x$. And $n\_1t\_1+n\_2t\_2=-d$ reads as $n\_2x+(n-1-n\_2)(-k-x)=-d$, so $n\_2(k+2x)=(n-1)(k+x)-d$ and $n\_2=(n-1)-(n-1)\cdot\frac{x}{k+2x}-\frac{d}{k+2x}=n-O(1)$. In other words, the matrix $A-xI$ has bounded rank. But by Ramsey theorem it contains a large minor with $-x$ along diagonal and either 0's or 1's outside diagonal. This minor does not have full rank only if $x\in \{0,-1\}$ which is impossible.


