TITLE: Possible pattern on sum of first $n$ $l$-th powers.
QUESTION [6 upvotes]: While discussing on some question with fleablood, we questioned ourselves on a possible pattern for the following sequence:
$$F_n(l):=\sum_{k=1}^n k^l$$
Using Wolframalpha (and well-knows results) we have
\begin{align*}
F_n(1)&=\frac{n(n+1)}{2}\\
F_n(2)&=\frac{n(n+1)(2n+1)}{6}\\
F_n(3)&=F_n(1)^2\\
F_n(4)&=\frac{3n^2+3n-1}{5}F_n(2)\\
F_n(5)&=\frac{2n^2+2n-1}{3}F_n(3)\\
F_n(6)& = \frac{3n^4+6n^3-3n+1}{7}F_n(2)
\end{align*}
Moreover using this question we (can) have
$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k).$$
Is there a sequence to be spotted?
Note: This is sort of a challenge question, I see that I haven't put too much effort into it yet, but I found the question interesting nonetheless and thus wanted to share it with you.

REPLY [4 votes]: As Lonidard has pointed out, expressions for $F_n(l)$ in terms of $F_n(l'), l'<l$ can be obtained by means of the binomial theorem and summing the terms of a telescoping series.
Here I give an alternative form, in terms of the triangular numbers
$$T(i)=F_i(1)=\dfrac{i(i+1)}2$$
and, for even $l$, the square-based pyramidal numbers
$$P(i)=F_i(2)=\dfrac{i(i+1)(2i+1)}6.$$
First, for odd $l=2j-1$: Let
\begin{align*}
A(i)&=[2T(i)]^j\\
&=[i(i+1)]^j\\
&=i^j(i+1)^j\\
\Rightarrow A(i-1)&=i^j(i-1)^j\\
\Rightarrow A(i)-A(i-1)&=i^j[(i+1)^j-(i-1)^j]\\
&=i^j\left[\sum_{r=0}^j \binom j r i^{j-r}-\sum_{r=0}^j(-1)^r\binom{j}r i^{j-r}\right]\\
&=2 \sum_{\substack{r=1\\r \textrm{ odd}}}^j \binom j r i^{2j-r}\\
\Rightarrow [2T(n)]^j&=2 \sum_{\substack{r=1\\r \textrm{ odd}}}^j \binom j r F(2j-r)\tag{telescoping}\\
\Rightarrow 2^{j-1}T^j&=\sum_{\substack{r=1\\r \textrm{ odd}}}^j \binom j r F(2j-r)\\
&=jF(2j-1)+\sum_{\substack{r=3\\r \textrm{ odd}}}^j \binom j r F(2j-r)\\
\Rightarrow F(2j-1)&=\left[2^{j-1}T^j-\sum_{\substack{r=3\\r \textrm{ odd}}}^j \binom j r F(2j-r)\right]/j\\
\Rightarrow F(l)&=\left[2^{j-1}T^j-\sum_{\substack{r=3\\r \textrm{ odd}}}^j \binom j r F(l-r+1)\right]/j
\end{align*}
where $l=2j-1$. This gives expressions for the $F(l)$ for $l$ odd.
A similar technique works for $l$ even, though it is more complicated. First, a result which will help with the somewhat strange way that binomial coefficients need to be paired:
Lemma. $$j\left[\binom{j}r+\binom{j-1}r\right]=(2j-r)\binom{j}r.$$
Proof.
\begin{align*}
j\left[\binom{j}r+\binom{j-1}r\right]&=\dfrac{jj!}{r!(j-r)!}+\dfrac{j(j-1)!}{r!(j-r-1)!}\\
&=[j+(j-r)]\dfrac{j(j-1)!}{r!(j-r)!}\\
&=(2j-r)\binom{j}r. 
\end{align*}
Let $P(i)=F_i(2)=i(i+1)(2i+1)/6$ and
\begin{align*}
A(i)&=2^{j-2}3jP(i)T(i)^{j-2}\\
&=3j[2T(i)]^{j-2}P(i)\\
&=j[i(i+1)]^{j-2}i(i+1)(2i+1)/2\\
&=ji^{j-2}(i+1)^{j-2}i(i+1)(2i+1)/2\\
&=ji^{j-1}(i+1)^{j-1}(2i+1)/2\\
\Rightarrow A(i-1)&=j(i-1)^{j-1}i^{j-1}(2i-1)/2\\
\Rightarrow A(i)-A(i-1)&=ji^{j-1}\left[(2i+1)(i+1)^{j-1}-(2i-1)(i-1)^{j-1}\right]/2\\
&=ji^{j-1}\left[i(i+1)^{j-1}+(i+1)(i+1)^{j-1}-i(i-1)^{j-1}-(i-1)(i-1)^{j-1}\right]/2\\
&=ji^{j-1}\left[i(i+1)^{j-1}+(i+1)^j-i(i-1)^{j-1}-(i-1)^j\right]/2\\
&=ji^{j-1}\left[i(i+1)^{j-1}-i(i-1)^{j-1}+(i+1)^j-(i-1)^j\right]/2\\
&=ji^j\left[(i+1)^{j-1}-(i-1)^{j-1}\right]/2+ji^{j-1}\left[(i+1)^j-(i-1)^j\right]/2\\
&=ji^j\left[\sum_{r=0}^{j-1}\binom{j-1}r i^{j-r-1}-\sum_{r=0}^{j-1}(-1)^r\binom{j-1}r i^{j-r-1}\right]/2\,+\\
&\qquad+ji^{j-1}\left[\sum_{r=0}^j\binom{j}r i^{j-r}-\sum_{r=0}^j (-1)^r\binom{j}r i^{j-r}\right]/2\\
&=ji^j\sum_{\substack{r=1\\r \textrm{ odd}}}^{j-1}\binom{j-1}r i^{j-r-1}+ji^{j-1}\sum_{\substack{r=1\\r \textrm{ odd}}}^j\binom{j}r i^{j-r}\\
&=\sum_{\substack{r=1\\r \textrm{ odd}}}^j j\left[\binom{j-1}r+\binom{j}r\right]i^{2j-r-1}\\
&=\sum_{\substack{r=1\\r \textrm{ odd}}}^j (2j-r) \binom j{r} i^{2j-r-1}\tag{by the lemma}\\
\Rightarrow A(n)&=\sum_{\substack{r=1\\r \textrm{ odd}}}^j (2j-r) \binom j{r} F(2j-r-1)\tag{telescoping}\\
\Rightarrow 2^{j-2}3jPT^{j-2}&=j(2j-1)F(2j-2)+ \sum_{\substack{r=3\\r \textrm{ odd}}}^{j} (2j-r) \binom j{r} F(2j-r-1)\\
\Rightarrow j(2j-1)F(2j-2)&=2^{j-2}3jPT^{j-2} - \sum_{\substack{r=3\\r \textrm{ odd}}}^{j} (2j-r) \binom j{r} F(2j-r-1)\\
\Rightarrow F(2j-2)&=\left[2^{j-2}3jPT^{j-2} - \sum_{\substack{r=3\\r \textrm{ odd}}}^{j} (2j-r) \binom j{r} F(2j-r-1)\right]/j(2j-1)\\
\Rightarrow F(l)&=\left[2^{j-2}3jPT^{j-2} - \sum_{\substack{r=3\\r \textrm{ odd}}}^{j} (l-r+2) \binom j{r} F(l-r+1)\right]/{j(l+1)},
\end{align*}
where $l=2j-2$.
Here are some explicit examples, reduced to their lowest terms and with some further simplification where possible. You might also find that [Knuth] has some useful information.
\begin{align*}
F(3) &= T^2\\
F(4) &= (6PT-F(2))/5=P(6T-1)/5\\
F(5) &= (4T^3-F(3))/3 = T^2(4T-1)/3\\
F(6) &= (12PT^2-5F(4))/7\\
F(7) &= 2T^4-F(5)\\
F(8) &= (24PT^3-14F(6)+F(4))/9\\
F(9) &= (16T^5-10F(7)-F(5))/5\\
&= (16T^5-F(5))/5-2F(7)\\
F(10) &= (48PT^4-30F(8)-7F(6))/11\\
F(11) &= (16T^6-10F(9)-3F(7))/3\\
&= (16T^6-10F(9))/3-F(7)\\
F(12) &= (96PT^5-55F(10)-27F(8)-F(6))/13\\
F(13) &= (64T^7-35F(11)-21F(9)-F(7))/7\\
&= (64T^7-F(7))/7 - 5 F(11) - 3 F(9)\\
F(14) &= (192PT^6-91F(12)-77F(10)-9F(8))/15\\
F(15) &= 16T^8-7F(13)-7F(11)-F(9) \\
&= 16T^8-7(F(13)+F(11))-F(9)\\
F(16) &= (384PT^7-140F(14)-182F(12)-44F(10)-F(8))/17\\
F(17) &= (256T^9-84F(15)-126F(13)-36F(11)-F(9))/9\\
 &= (256T^9-84F(15)-F(9))/9-14F(13)-4F(11)\\
F(18) &= (768PT^8-204F(16)-378F(14)-156F(12)-11F(10))/19\\
F(19) &= (256T^{10}-60F(17)-126F(15)-60F(13)-5F(11))/5\\
 &= (256T^{10}-126F(15))/5-12(F(17)+F(13))-F(11)\\
F(20) &= (1536PT^9-285F(18)-714F(16)-450F(14)-65F(12)-F(10))/21\\
F(21) &= (1024T^{11}-165F(19)-462F(17)-330F(15)-55F(13)-F(11))/11\\
 &= (1024T^{11}-F(11))/11-15F(19)-42F(17)-30F(15)-5F(13)\\
F(22) &= (3072PT^{10}-385F(20)-1254F(18)-1122F(16)-275F(14)-13F(12))/23\\
F(23) &= (512T^{12}-55F(21)-198F(19)-198F(17)-55F(15)-3F(13))/3\\
 &= (512T^{12}-55(F(21)+F(15)))/3-66(F(19)+F(17))-F(13)\\
F(24) &= (6144PT^{11}-506F(22)-2079F(20)-2508F(18)-935F(16)-90F(14)-F(12))/25\\
F(25) &= (4096T^{13}-286F(23)-1287F(21)-1716F(19)-715F(17)-78F(15)-F(13))/13\\
&= (4096T^{13}-F(13))/13-22F(23)-99F(21)-132F(19)-55F(17)-6F(15)\\
F(26) &= (12288PT^{12}-650F(24)-3289F(22)-5148F(20)-2717F(18)-442F(16)-15F(14))/27\\
F(27) &= (4096T^{14}-182F(25)-1001F(23)-1716F(21)-1001F(19)-182F(17)-7F(15))/7\\
&= (4096T^{14}-1716F(21))/7-26(F(25)+F(17))-143(F(23)+F(19))-F(15)\\
F(28) &= (24576PT^{13}-819F(26)-5005F(24)-9867F(22)-7007F(20)-1729F(18)-119F(16)-F(14))/29\\
F(29) &= (16384T^{15}-455F(27)-3003F(25)-6435F(23)-5005F(21)-1365F(19)-105F(17)-F(15))/15\\
&= (16384T^{15}-455F(27)-3003F(25)-5005F(21)-F(15))/15-429F(23)-91F(19)-7F(17)\\
F(30) &= (49152PT^{14}-1015F(28)-7371F(26)-17875F(24)-16445F(22)-5733F(20)-665F(18)-17F(16))/31\\
F(31) &= 2048T^{16}-35F(29)-273F(27)-715F(25)-715F(23)-273F(21)-35F(19)-F(17)\\
&= 2048T^{16}-35(F(29)+F(19))-273(F(27)+F(21))-715(F(25)+F(23))-F(17)
\end{align*}
[Knuth] Donald Knuth. Johann Faulhaber and sums of powers. Mathematics of Computation 61, no.203 (Jul 1993), p.277-294.