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Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n \choose 2}-{ n-k\choose 2} -{k\choose2} $, which suggests to multiply the relation by $ (1-p)^{-{n\choose 2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n\choose 2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k\choose 2 }(1-p)^{-{k\choose2}}{x^{k-1}\over(k-1)!}\cdot (1-p)^{-{n-k\choose 2}} P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x):=\sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}$$ $$f(x):=\sum_{n=0}^\infty (1-p)^{-{n\choose2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}\Big).$$

Incidentally, for $p=1/2$ one has $g(x)=e^x-1$, and we obtain the special values $$P_n(1/2)=2^{-{n\choose 2}}B_n$$ where $B_n$ are the Bell numbers.

Given that in the above formula for $f(x)$ the variable $p$ only enters in the term ${p\over 1-p}$, we may express the polynomials $P_n$ as $$P_n(p )=(1-p)^{n\choose 2}Q_n\big({p\over1-p}\big)$$ where the sequence $Q_n(q)$ is defined correspondingly by the somehow simpler generating series $$\sum_{n=0}^\infty Q_{n}{x^{n}\over n!} = \exp\Big( \sum_{n=1}^\infty q^{n \choose 2 }{x^{n}\over n!}\Big).$$

The first polynomials $Q_n$ are curiously similar to binomial expansions: $$Q_0=1 $$ $$Q_1=1 $$ $$Q_2=q + 1 $$ $$Q_3 = q^3+3q+1 $$ $$Q_4=q^6+4q^3+3q^2+6q+1 $$ $$Q_5=q^{10}+5q^6+10q^4+10q^3+15q^2+10q+1 $$ $$Q_6=q^{15}+6q^{10}+15q^7+25q^6+60q^4+35q^3+45q^2+15q+1 $$ $$Q_7=q^{21}+7q^{15}+21q^{11}+21q^{10}+35q^9+105q^7+105q^6+105q^5+210q^4+140q^3+105q^2+21q+1 $$

Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n \choose 2}-{ n-k\choose 2} -{k\choose2} $, which suggests to multiply the relation by $ (1-p)^{-{n\choose 2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n\choose 2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k\choose 2 }(1-p)^{-{k\choose2}}{x^{k-1}\over(k-1)!}\cdot (1-p)^{-{n-k\choose 2}} P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x)=g(x,p):=\sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}$$ $$f(x)=f(x,p):=\sum_{n=0}^\infty (1-p)^{-{n\choose2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}\Big).$$

Incidentally, for $p=1/2$ one has $g(x)=e^x-1$, and we obtain the special values $$P_n(1/2)=2^{-{n\choose 2}}B_n$$ where $B_n$ are the Bell numbers.

Given that in the above formula for $f(x)$ the variable $p$ only enters in the term ${p\over 1-p}$, we may express the polynomials $P_n$ as $$P_n(p )=(1-p)^{n\choose 2}Q_n\big({p\over1-p}\big)$$ where the sequence $Q_n(q)$ is defined correspondingly by the somehow simpler generating series $$F(x)=F(x,q)=\sum_{n=0}^\infty Q_{n}{x^{n}\over n!} := \exp G(x,q)$$ where $$G(x)=G(x,q)=g\Big(x,{q\over q+1}\Big) := \sum_{n=1}^\infty q^{n \choose 2 }{x^{n}\over n!} $$ is the power series solution to the linear delay differential equation $$G(0)=0$$ $$G'(x)=1+G(qx).$$ The first polynomials $Q_n$ are curiously similar to binomial expansions: $$Q_0=1 $$ $$Q_1=1 $$ $$Q_2=q + 1 $$ $$Q_3 = q^3+3q+1 $$ $$Q_4=q^6+4q^3+3q^2+6q+1 $$ $$Q_5=q^{10}+5q^6+10q^4+10q^3+15q^2+10q+1 $$ $$Q_6=q^{15}+6q^{10}+15q^7+25q^6+60q^4+35q^3+45q^2+15q+1 $$ $$Q_7=q^{21}+7q^{15}+21q^{11}+21q^{10}+35q^9+105q^7+105q^6+105q^5+210q^4+140q^3+105q^2+21q+1 $$

Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n^2\over2}-{ (n-k)^2\over 2} -{k^2\over2} $, which suggests to multiply the relation by $ (1-p)^{-{n^2\over2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n^2\over2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k(k-1)\over 2 }(1-p)^{-{k^2\over 2}}{x^{k-1}\over(k-1)!}\cdot P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x):=\sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}$$ $$f(x):=\sum_{n=0}^\infty (1-p)^{-{n^2\over2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}\Big).$$ This is possibly still far from your needs, but could hopefully be of use. Further steps would require some reference (hopefully somebody here does have it!) e.g. on the function $$G(x,z):=\sum_{n=1}^\infty {z^{n^2}x^{n}\over n!}$$ since $g(x)=G\big(x p^{-{1\over 2}},\big({p\over 1-p}\big)^{1\over2}\ \big)$.

[edit] As a consolation, we obtain the values for $p=1/2$, $$P_n(1/2)=2^{-{n\choose 2}}B_n$$ where $B_n$ are the Bell numbers.

Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n \choose 2}-{ n-k\choose 2} -{k\choose2} $, which suggests to multiply the relation by $ (1-p)^{-{n\choose 2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n\choose 2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k\choose 2 }(1-p)^{-{k\choose2}}{x^{k-1}\over(k-1)!}\cdot (1-p)^{-{n-k\choose 2}} P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x):=\sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}$$ $$f(x):=\sum_{n=0}^\infty (1-p)^{-{n\choose2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty \Big({p\over 1-p}\Big)^{n \choose 2 }{x^{n}\over n!}\Big).$$

Incidentally, for $p=1/2$ one has $g(x)=e^x-1$, and we obtain the special values $$P_n(1/2)=2^{-{n\choose 2}}B_n$$ where $B_n$ are the Bell numbers.

Given that in the above formula for $f(x)$ the variable $p$ only enters in the term ${p\over 1-p}$, we may express the polynomials $P_n$ as $$P_n(p )=(1-p)^{n\choose 2}Q_n\big({p\over1-p}\big)$$ where the sequence $Q_n(q)$ is defined correspondingly by the somehow simpler generating series $$\sum_{n=0}^\infty Q_{n}{x^{n}\over n!} = \exp\Big( \sum_{n=1}^\infty q^{n \choose 2 }{x^{n}\over n!}\Big).$$

The first polynomials $Q_n$ are curiously similar to binomial expansions: $$Q_0=1 $$ $$Q_1=1 $$ $$Q_2=q + 1 $$ $$Q_3 = q^3+3q+1 $$ $$Q_4=q^6+4q^3+3q^2+6q+1 $$ $$Q_5=q^{10}+5q^6+10q^4+10q^3+15q^2+10q+1 $$ $$Q_6=q^{15}+6q^{10}+15q^7+25q^6+60q^4+35q^3+45q^2+15q+1 $$ $$Q_7=q^{21}+7q^{15}+21q^{11}+21q^{10}+35q^9+105q^7+105q^6+105q^5+210q^4+140q^3+105q^2+21q+1 $$

Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n^2\over2}-{ (n-k)^2\over 2} -{k^2\over2} $, which suggests to multiply the relation by $ (1-p)^{-{n^2\over2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n^2\over2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k(k-1)\over 2 }(1-p)^{-{k^2\over 2}}{x^{k-1}\over(k-1)!}\cdot P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x):=\sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}$$ $$f(x):=\sum_{n=0}^\infty (1-p)^{-{n^2\over2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}\Big).$$ This is possibly still far from your needs, but could hopefully be of use. Further steps would require some reference (hopefully somebody here does have it!) e.g. on the function $$G(x,z):=\sum_{n=1}^\infty {z^{n^2}x^{n}\over n!}$$ since $g(x)=G\big(x p^{-{1\over 2}},\big({p\over 1-p}\big)^{1\over2}\ \big)$.

Here is a generating function approach. Note that the RHS is almost a convolution $\sum_{k=0}^n a_{k}b_{n-k}$, which suggests the use of generating functions, after some bricolage on the the recursion aimed to get a convolution form (whence a Cauchy product of power series).

To start with, note that the first term on the RHS may be naturally included in the sum for $k=1$. The binomial has already a suitable form; and then note the exponent of $(1-p)$: it can be written $k(n-k)= {n^2\over2}-{ (n-k)^2\over 2} -{k^2\over2} $, which suggests to multiply the relation by $ (1-p)^{-{n^2\over2}}{x^{n-1}\over(n-1)!}$ : $$(1-p)^{-{n^2\over2}}P_{n}{x^{n-1}\over (n-1)!}= p^{k(k-1)\over 2 }(1-p)^{-{k^2\over 2}}{x^{k-1}\over(k-1)!}\cdot P_{n-k}{x^{n-k}\over (n-k)!} $$

If we introduce the generating functions

$$g(x):=\sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}$$ $$f(x):=\sum_{n=0}^\infty (1-p)^{-{n^2\over2}}P_{n}{x^{n}\over n!} $$ the recurrence relations write $f'=g'f$, whence $$f(x)=\exp\Big( \sum_{n=1}^\infty p^{n(n-1)\over 2 }(1-p)^{-{n^2\over 2}}{x^{n}\over n!}\Big).$$ This is possibly still far from your needs, but could hopefully be of use. Further steps would require some reference (hopefully somebody here does have it!) e.g. on the function $$G(x,z):=\sum_{n=1}^\infty {z^{n^2}x^{n}\over n!}$$ since $g(x)=G\big(x p^{-{1\over 2}},\big({p\over 1-p}\big)^{1\over2}\ \big)$.

[edit] As a consolation, we obtain the values for $p=1/2$, $$P_n(1/2)=2^{-{n\choose 2}}B_n$$ where $B_n$ are the Bell numbers.