• Let $a(n)$ be A001515, i.e., the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here $$ a(n) = (2n-1)a(n-1) + a(n-2), \\ a(0) = 1, a(1) = 2 $$ The closed form is $$ a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{2k}\frac{(2k)!}{k!2^k} $$ Also exponential generating function is $$ \frac{\exp(1-\sqrt{1-2x})}{\sqrt{1-2x}} $$ and generating function using continued fraction is $\frac{1}{G(0)}$ where $$ G(j)=1-x-\frac{(j+1)x}{G(j+1)} $$
• Let $$ R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; v1[2]=2; for(i=2, n, v1[i+1]=(2*i-1)*v1[i] + v1[i-1]); v1
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q, (j+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=a_upto(n)==R_upto(n)

UPD:

• Let $T(n,k)$ be A001498, i.e., triangle of coefficients of Bessel polynomials $y_n(x)$. Here $$ T(n,k)=\frac{(n+k)!}{2^k(n-k)!k!} $$
• Let $$ R_1(n,q,z)=(q+1)zR_1(n-1,q+1,z)+\sum\limits_{j=0}^{q}(jz+1)R_1(n-1,j,z), \\ R_1(0,q,z)=1 $$

I conjecture that $$T(n,k)=[z^k]R_1(n,0,z)$$

Here is the PARI/GP prog to check it numerically:

T(n, k)=if(k<0||k>n, 0, binomial(n, k)*(n+k)!/2^k/n!)
y(n)=sum(j=0, n, x^j*T(n,j))
R1_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*x*v1[q+2] + sum(j=0, q, (j*x+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test1(n)=vector(n+1, i, y(i-1))==R1_upto(n)

Is there a way to prove it?

• Let $a(n)$ be A001515 i.e. the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here $$ a(n) = (2n-1)a(n-1) + a(n-2), \\ a(0) = 1, a(1) = 2 $$ The closed form is $$ a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{2k}\frac{(2k)!}{k!2^k} $$ Also exponential generating function is $$ \frac{\exp(1-\sqrt{1-2x})}{\sqrt{1-2x}} $$ and generating function using continued fraction is $\frac{1}{G(0)}$ where $$ G(j)=1-x-\frac{(j+1)x}{G(j+1)} $$
• Let $$ R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; v1[2]=2; for(i=2, n, v1[i+1]=(2*i-1)*v1[i] + v1[i-1]); v1
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q, (j+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=a_upto(n)==R_upto(n)

UPD:

• Let $T(n,k)$ be A001498, i.e., triangle of coefficients of Bessel polynomials $y_n(x)$. Here $$ T(n,k)=\frac{(n+k)!}{2^k(n-k)!k!} $$
• Let $$ R_1(n,q,z)=(q+1)zR_1(n-1,q+1,z)+\sum\limits_{j=0}^{q}(jz+1)R_1(n-1,j,z), \\ R_1(0,q,z)=1 $$

I conjecture that $$T(n,k)=[z^k]R_1(n,0,z)$$

Here is the PARI/GP prog to check it numerically:

T(n, k)=if(k<0||k>n, 0, binomial(n, k)*(n+k)!/2^k/n!)
y(n)=sum(j=0, n, x^j*T(n,j))
R1_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*x*v1[q+2] + sum(j=0, q, (j*x+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test1(n)=vector(n+1, i, y(i-1))==R1_upto(n)

Is there a way to prove it?

• Let $a(n)$ be A001515, i.e., the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here $$ a(n) = (2n-1)a(n-1) + a(n-2), \\ a(0) = 1, a(1) = 2 $$ The closed form is $$ a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{2k}\frac{(2k)!}{k!2^k} $$ Also exponential generating function is $$ \frac{\exp(1-\sqrt{1-2x})}{\sqrt{1-2x}} $$ and generating function using continued fraction is $\frac{1}{G(0)}$ where $$ G(j)=1-x-\frac{(j+1)x}{G(j+1)} $$
• Let $$ R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; v1[2]=2; for(i=2, n, v1[i+1]=(2*i-1)*v1[i] + v1[i-1]); v1
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q, (j+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=a_upto(n)==R_upto(n)

Is there a way to prove it?

• Let $a(n)$ be A001515, i.e., the Bessel polynomial $y_n(x)$ evaluated at $x=1$. Here $$ a(n) = (2n-1)a(n-1) + a(n-2), \\ a(0) = 1, a(1) = 2 $$ The closed form is $$ a(n)=\sum\limits_{k=0}^{n}\binom{n+k}{2k}\frac{(2k)!}{k!2^k} $$ Also exponential generating function is $$ \frac{\exp(1-\sqrt{1-2x})}{\sqrt{1-2x}} $$ and generating function using continued fraction is $\frac{1}{G(0)}$ where $$ G(j)=1-x-\frac{(j+1)x}{G(j+1)} $$
• Let $$ R(n,q)=(q+1)R(n-1,q+1)+\sum\limits_{j=0}^{q}(j+1)R(n-1,j), \\ R(0,q)=1 $$

I conjecture that $$R(n,0)=a(n).$$

Here is the PARI/GP prog to check it numerically:

a_upto(n)=my(v1); v1=vector(n+1, i, 0); v1[1]=1; v1[2]=2; for(i=2, n, v1[i+1]=(2*i-1)*v1[i] + v1[i-1]); v1
R_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*v1[q+2] + sum(j=0, q, (j+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test(n)=a_upto(n)==R_upto(n)

UPD:

• Let $T(n,k)$ be A001498, i.e., triangle of coefficients of Bessel polynomials $y_n(x)$. Here $$ T(n,k)=\frac{(n+k)!}{2^k(n-k)!k!} $$
• Let $$ R_1(n,q,z)=(q+1)zR_1(n-1,q+1,z)+\sum\limits_{j=0}^{q}(jz+1)R_1(n-1,j,z), \\ R_1(0,q,z)=1 $$

I conjecture that $$T(n,k)=[z^k]R_1(n,0,z)$$

Here is the PARI/GP prog to check it numerically:

T(n, k)=if(k<0||k>n, 0, binomial(n, k)*(n+k)!/2^k/n!)
y(n)=sum(j=0, n, x^j*T(n,j))
R1_upto(n)=my(v1, v2, v3); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; for(i=1, n, for(q=0, n-i, v2[q+1]=(q+1)*x*v1[q+2] + sum(j=0, q, (j*x+1)*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test1(n)=vector(n+1, i, y(i-1))==R1_upto(n)

Is there a way to prove it?