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Jun
25
comment How to prove this polynomial always has integer values at all integers?
That must have been what I meant.
Jun
25
comment How to prove this polynomial always has integer values at all integers?
$\sum_{i=0}^m\sum_{j=k}^m\frac{3 (-1)^{k+j}{2k \choose k }{j \choose i}{ m \choose i }{i \choose m-j }}{2(2i-1)(2j+1)(2m-2i-1)}$ equals $\sum_{i=0}^m\sum_{j=0}^{k-1}\frac{3 (-1)^{k+j}{2k \choose k }{j \choose i}{ m \choose i }{i \choose m-j }}{2(2i-1)(2j+1)(2m-2i-1)}$ for $k>1$.
Jun
25
comment How to prove this polynomial always has integer values at all integers?
Experimentally ${x+j\choose j}{x-1\choose j}$ equals $\sum_{k=0}^j (-1)^{k+j} {2k\choose k} B_k(x)/2$. So $P_m(x)=\sum_{k=0}^m \sum_{i=0}^m\sum_{j=k}^m\frac{ 3(-1)^{k+j}{2k \choose k }{j \choose i}{ m \choose i }{i \choose m-j }}{2(2i-1)(2j+1)(2m-2i-1)} B_k(x)$ and one must show that $\sum_{i=0}^m\sum_{j=k}^m\frac{ 3(-1)^{k+j}{2k \choose k }{j \choose i}{ m \choose i }{i \choose m-j }}{2(2i-1)(2j+1)(2m-2i-1)}$ is an integer for $1\leq k\leq m$. Is this any easier?
Jun
24
comment How to prove this polynomial always has integer values at all integers?
@Wadim Zudilin. This $B_k$ basis is great. It shows for instance that it suffices to prove that $P_m(x)$ is an integer for integer $x$ between 0 and $m$.
Jun
21
comment How to prove this polynomial always has integer values at all integers?
@Allen Knutson. My $q_{2n}$ span the space of even polynomials with integer values and with even value at zero. One sees this by showing that the determinant of the $(q_{2n}(i))$ matrix, $i=1,\dots,m$, $n=1,\dots,m$ equals one. Then your suspicion follows "easily".
Jun
19
comment How to prove this polynomial always has integer values at all integers?
Put $q_n(x)={x\choose n}+{-x\choose n}$. Notice that $P_4(x)=-936q_2(x)+1522q_4(x)-704q_6(x)+118q_8(x)$. So it looks like the $P_n$ are integer combinations of the $q_{2n}$.
Jun
1
revised finite generation of a certain type of subring
rewritten entirely
Jun
1
comment finite generation of a certain type of subring
But notice the silly example where $R=k[x,y]/(xy)$, $I=xR$. Here $I$ is not of finite colength and yet $A$ is finitely generated as a $k$-algebra.
May
31
comment finite generation of a certain type of subring
@Neil Epstein. It is not true that $R$ is module-generated over $A$ by any lifting of any $k$-basis of $R/I$. But if 1 is in the lifting then $R$ is indeed module generated by the lifting and the finite generation of $A$ follows from the wikipedia Artin-Tate lemma.
May
29
revised finite generation of a certain type of subring
more detail
May
29
revised finite generation of a certain type of subring
clarification
May
28
revised finite generation of a certain type of subring
corrected mistakes
May
28
revised finite generation of a certain type of subring
added 692 characters in body
May
28
revised finite generation of a certain type of subring
added 11 characters in body
May
28
answered finite generation of a certain type of subring
Apr
15
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21
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18
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Feb
18
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Sep
29
comment Multiprecision numerical evaluation of integral: Sage vs. PARI/GP vs. mpmath
Indeed the WorkingPrecision->50 should simply be deleted. One should leave that to Mathematica. And tau should not have been made numerical.