Take $ f_0 (n) $ as a polynomial function. Set $ S_0 = (f_0 (1), f_0 (2), f_0 (3), ...) $. Let $ f_1 (n) $ be the sum of the first $ n $ of the sequence $ S_0 $ and let $ S_1 = (f_1 (1), f_1 (2), f_1 (3), ...) $. Let $ f_k (n) $ be the sum of the first $ n $ terms of the sequence $ S_ {k-1} $ and let $ S_k = (f_k (1), f_k (2), f_k (3), ...) $ for every positive integer $ k $. Explicit $ f_k (n) $ as a function of $ k, n $ and the coefficients of $ f_0 (n) $
I believe there is a formula for calculating calculate $\sum_n^k$ because j ^ k, k is constant and $ j $ ranges from $ 1 $ to $ n $. Is there a formula that has connection with bernoulli numbers
Here: If $ f_0=\sum_{i=0}^t a_{t-i}X^i, $ then: $$ f_k(n)=\sum_{j_k=1}^n\sum_{j_{k-1}=1}^{j_k} \sum_{j_{k-2}=1}^{j_{k-1}}\cdots\sum_{j_2=1}^{j_3}\sum_{j_1=1}^{j_2} \sum_{i=0}^{t} a_{t-i} j_1^i= $$$$ =\sum_{i=0}^{t} \left( a_{t-i} \sum_{j_k=1}^n\sum_{j_{k-1}=1}^{j_k} \sum_{j_{k-2}=1}^{j_{k-1}}\cdots\sum_{j_2=1}^{j_3}\sum_{j_1=1}^{j_2} j_1^i \right) $$Now, using the method of diagonalizing consecutive sums two by two (this tool is used in large cardinal theory also, by the way) , we obtain $$ f_k(n)=\frac{1}{n!}\sum_{i=0}^{t} \left( a_{t-i} \sum_{j=1}^n \left( j^i\prod_{q=1}^{k-1}\left( n-j+q \right) \right) \right) $$
Does it serve as a prof?