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The John Riordan prize for the best solution to an unsolved problem in the OEIS

As mentioned previously, the Encyclopedia of Integer Sequences is 50 this year. To celebrate that fact, and to encourage readers to concentrate on filling in the gaps in the missing entries instead of just adding new ones, there’s a \$1,000 prize for the best solution to an open problem posed in an OEIS entry.

The announcement by OEIS creator Neil Sloane seems only to have been published as a PDF, so I’m reproducing it here for everyone’s convenience:

The On-Line Encyclopedia of Integer Sequences invites you to solve an open problem in
an entry in the OEIS. After adding your solution to the OEIS entry, notify with subject line “Riordan Prize Nomination”. (It is perfectly
acceptable to nominate your own work.) The deadline for submission is December 1,
2015. The decision will be made by a special prize committee, and will be announced
at the Joint Mathematics Meetings in Seattle in January 2016.

To find problems to work on, search in the OEIS for the words “conjecture”, “empirical”,
“evidence suggests”, “it would be nice”, “would like”, etc. Or find a formula or
recurrence (with proof, of course) for a sequence that currently has no formula.

The prize is named after John Riordan (1903-1988; Bell Labs, 1926-1968), author of
the classic books An Introduction to Combinatorial Analysis (1958) and Combinatorial
Identities (1968), which were the source for hundreds of early entries in the OEIS.

Here is an example from 2008: Sequence A145855 gives the number of $n$-element subsets
of $\{1,\ldots,n\}$ whose sum is a multiple of $n$ ($1, 1, 4, 9, 26, 76, 246, \ldots$ for $n \geq 1$). Vladeta Jovovic conjectured that the nth term is $\frac{1}{2n} \sum_{d|n} (-1)^{n+d}\phi\left(\frac{n}{d}\right) {2d \choose d}$, and Max Alekseyev found a proof. (This is not a candidate for the prize, since only work carried out in 2015 is eligible.)

The following is a recent unsolved problem, a conjecture of Alois P. Heinz (see A216368). Consider the number of values taken by the $n$th derivative of $x \uparrow x \uparrow \dots \uparrow x$ (with $n$ $x$’s and parentheses inserted in all possible ways, where the up-arrows indicate exponentiation), evaluated at $x = 1$. Show that this is given by $1, 1, 2, 4, 9, 20, \ldots$, the number of rooted trees on $n$ nodes (A000081).

Your proof should be mentioned in the appropriate OEIS entry, but (especially if it is long) may be published elsewhere, for example on the arXiv. Additional information may be found here: We hope that this competition will lead to the resolution of many open questions in the OEIS!

Neil J.A. Sloane
The OEIS Foundation
January 1, 2015

So get cracking!

via Dave Radcliffe on Twitter

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