User:Richard Penner

Unpaid sole moderator of http://physforum.com/ and sometime correspondent for the 2010 appeal of the anti-LHC lawsuit in US Federal Court. Very minor contributor to this site and a very weak amateur mathematician. Favorite sequences: A010553, A036450, A036452, A036453, and A178790.

Elementary Proof(s)

• Trace[Inverse[n * HilbertMatrix[n]]]
• = (1/n) * Trace[Inverse[HilbertMatrix[n]]] /* this is A189766(n)/n */
• $={\frac {1}{n}}\sum _{k=1}^{n}(2k-1){n+k-1 \choose n-k}^{2}{2k-2 \choose k-1}^{2}$ /* From explicit formula for inverse in Weisstein, Eric W., Hilbert Matrix, from MathWorld—A Wolfram Web Resource. */
• $={\frac {1}{n}}\sum _{k=0}^{n-1}(2k+1){n+k \choose n-k-1}^{2}{2k \choose k}^{2}$ /* substitution of k+1 for k */
• $=\sum _{k=0}^{n-1}{\frac {(n-1)!}{n!}}{\frac {(2k+1)!}{2k!}}{\frac {(n+k)!}{(n-k-1)!(2k+1)!}}{\frac {(n+k)!}{(n-k-1)!(2k+1)!}}{\frac {(2k)!}{k!k!}}{2k \choose k}$ /* expansion in terms of factorial */
• $=\sum _{k=0}^{n-1}{\frac {(2k+1)!(2k)!}{(2k+1)!(2k)!}}{\frac {(n-1)!(n+k)!(n+k)!}{n!(n-k-1)!(n-k-1)!(2k+1)!k!k!}}{2k \choose k}$
• $=\sum _{k=0}^{n-1}{\frac {(n-1)!(n+k)!}{n!(n-k-1)!k!k!}}{n+k \choose 2k+1}{2k \choose k}$
• $=\sum _{k=0}^{n-1}{\frac {(n-1)!}{(n-k-1)!k!}}{n+k \choose k}{n+k \choose 2k+1}{2k \choose k}$
• $=\sum _{k=0}^{n-1}{n-1 \choose k}{n+k \choose k}{n+k \choose 2k+1}{2k \choose k}$ /* Thus A189766(n)/n is an integer sequence */
• $={\frac {1}{n}}\sum _{k=0}^{n-1}(2k+1)A_{k}$ /* Proof of Theorem 1.1 from Sun, Zhi-Wei. “On Apery numbers and generalized central trinomial coefficients”. arΧiv:1006.2776.  */
• = a(n) /* A178790 */

Thus the sum of (2k+1) A005259(k) from 0 to n-1 = A189766(n).