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Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum_{j=0}^b\binom bj(1-t)^{-(b+j)} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$ Another observation that might be useful: $$ \sinh(t)\cosh^{a-2}(t)\sinh((a+1)t)=\sum_{b\leqslant a}b!S(a,b)\sinh^{2b}(t) $$

Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum_{j=0}^b\binom bj(1-t)^{-(b+j)} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$ Another observation that might be useful: $$ \cosh^{a-2}(t)\sinh((a+1)t)=\sum_{b=1}^ab!S(a,b)\sinh^{2b-1}(t) $$

Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum_{j=0}^b\binom bj(1-t)^{-(b+j)} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$

Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum_{j=0}^b\binom bj(1-t)^{-(b+j)} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$ Another observation that might be useful: $$ \sinh(t)\cosh^{a-2}(t)\sinh((a+1)t)=\sum_{b\leqslant a}b!S(a,b)\sinh^{2b}(t) $$

Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=\sum_{j=0}^b\binom bj(1-t)^{-j} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$

Probably this is not very helpful, but it is an explicit expression after all.

I get $$ S(a,b)=\frac{(-1)^{a+1}}{b!}{}_2F_1(-a+1,b+1;2b-a+1;2)\binom b{2b-a}2^{2b-a}. $$ This follows from $$ S(a,b)=\sum_{\ell=2b-a}^b(-1)^\ell\binom{a+\ell-1}{2b-1}\binom b\ell2^\ell, $$ which in turn I derived from the generating function $$ \sum_{a,b}S(a,b)t^au^b=e^{\frac{2-t}{(1-t)^2}tu}. $$

Note also that $S(a,b)$ is defined for other values of $a$ and $b$. In particular, for $a\geqslant2b$ one obtains $$ S(a,b)=\frac{(-1)^b}{b!}{}_2F_1(a,-b;a-2b+1;2)\binom{a-1}{2b-1} $$

In fact, from that generating function, $b!S(a,b)$ is the coefficient of $\left(2t+3t^2+4t^3+5t^4+...\right)^b$ at $t^a$. Here is the table of few of the $b!S(a,b)$: $$ \begin{array}{ccccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 3 & 4 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 4 & 12 & 8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 25 & 36 & 16 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 6 & 44 & 102 & 96 & 32 & 0 & 0 & 0 & 0 & 0 \\ 0 & 7 & 70 & 231 & 344 & 240 & 64 & 0 & 0 & 0 & 0 \\ 0 & 8 & 104 & 456 & 952 & 1040 & 576 & 128 & 0 & 0 & 0 \\ 0 & 9 & 147 & 819 & 2241 & 3400 & 2928 & 1344 & 256 & 0 & 0 \\ 0 & 10 & 200 & 1372 & 4712 & 9290 & 11040 & 7840 & 3072 & 512 & 0 \\ 0 & 11 & 264 & 2178 & 9108 & 22363 & 34332 & 33488 & 20224 & 6912 & 1024 \\ \end{array} $$

Here is, in fact, a version that works for any $a$ and $b$, and does not contain any powers of $-1$ or $2$: represent $$ \frac{2-t}{(1-t)^2}=(1-t)^{-2}+(1-t)^{-1}; $$ then from the above we obtain that $b!S(a,b)$ is the coefficient at $t^a$ of the series $t^b((1-t)^{-2}+(1-t)^{-1})^b$, i. e. of $$ t^b\sum_{j=0}^b\binom bj(1-t)^{-2j}(1-t)^{-(b-j)}=t^b\sum_{j=0}^b\binom bj(1-t)^{-(b+j)} $$ It then follows easily that $$ S(a,b)=\frac1{b!}\sum_{j=0}^b\binom bj\binom{a+j-1}{a-b}=\frac1{b!}{}_2\!\!\ F_1(a,-b;b;-1)\binom{a-1}{a-b}. $$