NovaDenizen says the polynomial sequence i wanted to know about has these two recurrence relations
(1) $p_n(x+1) = \sum_{i=0}^{n} (x+1)^{n-i}p_i(x)$
(2) $p_{n+1}(x) = \sum_{i=1}^{x} ip_n(i)$
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i was trying to calculate the probability of something and i came upon them. i needed to know what this was equal to:
$$p_n(x)=\sum_{k_n=k_{n-1}}^{x}....\sum_{k_3=k_2}^{x} \sum_{k_2=k_1}^{x} \sum_{k_1=1}^{x}k_1k_2...k_n$$
$k \in (1,2,...,x)$.
if you make it continuous and over the reals instead of over the natural numbers then its not too hard to see what that equals.
$$p_n(x) \approx \int_{k_n=0}^{x}...\int_{k_3=k_2}^{x}\int_{k_2=k_1}^{x}k_1k_2k_ndk_1dk_2...dk_n =\dfrac{(x)^{2n}}{2^n n!}$$
i computed some of these and got
$$p_1(x) \approx \frac{x^2}{2}, p_2(x) \approx \frac{x^4}{8}, p_3(x) \approx \frac{x^6}{48}, p_4(x) \approx \frac{x^8}{384}, p_5(x) \approx \frac{x^{10}}{3840},...$$ so im assuming that's the formula.
from the summation formula its easy to see that $$p_1(x)=\sum_{k=1}^x k=\frac{x(x+1)}{2}$$
i spent some time to compute $$p_2(x)=\sum_{k_2=k_1}^x\sum_{k_1=1}^x k_1k_2=x^4/8+(5 x^3)/12+(3 x^2)/8+x/12=x (3 x+1) (x+1) (x+2)/24$$
these agree with the approximations from integrating, which im guessing gives the first terms of $p_n(x)$.
also i think its might be fair to say that $\dfrac{(x)^{2n}}{2^n n!} < p_n(x) < \dfrac{(x+1)^{2n}}{2^n n!}$ if you can use the integral approximation to get lower and upper estimates of $p_n(x)$.
but im wondering what this sequence of polynomials is. i think i can just use the first terms of them to calculate the probabilities i wanted to know well enough, but it wouldn't hurt to know if this sequence of polynomials has a name. thanks.
Let $q_n(a,b) = \sum_{k_1=a}^b \sum_{k_2 = k_1}^b \dots \sum_{k_n=k_{n-1}}^b k_1 k_2 \dots k_n$.
$p_n(x) = q_n(1,x)$
$q_1(a,b) = \dfrac{b(b+1)}{2} - \dfrac{a(a-1)}{2} $
$q_n(a,a) = a^n$
$q_{n+1}(a,b) = \sum_{i=a}^b iq_n(a,i) = \sum_{i=a}^b iq_n(i,b)$
$q_{n_1 + n_2 + 1}(a,b) = \sum_{i=a}^b q_{n_1}(a,i)iq_{n_2}(i,b)$
$q_2(a,b) = \sum_{i=a}^b iq_1(a,i) = \sum_{i=a}^b i\left(\dfrac{i(i+1)}{2} - \dfrac{a(a-1)}{2}\right) = \sum_{i=a}^b \dfrac{i^3}{2} + \dfrac{i^2}2 + i\dfrac{a(a-1)}{2}$
That last expression for $q_2$ is solvable as a power sum.
It seems to me that a solution for general $n$ wouldn't be any simpler than Faulhaber's formula.