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Feb 6, 2022 at 16:07 answer added T. Amdeberhan timeline score: 0
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Jan 4, 2021 at 11:46 history edited darij grinberg CC BY-SA 4.0
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Jan 3, 2021 at 9:05 history edited Fedor Petrov CC BY-SA 4.0
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Jan 3, 2021 at 0:18 answer added Ira Gessel timeline score: 5
May 31, 2010 at 20:38 comment added Timothy Chow Without the factor of $k^n/n!$ your expression counts $(k+1)$-ary increasing trees. See for example Bergeron, Flajolet, and Salvy, "Varieties of increasing trees," in CAAP '92 (LNCS vol. 581). This doesn't immediately answer your question but it might be a place to start looking for a combinatorial interpretation.
May 31, 2010 at 17:52 answer added Pietro Majer timeline score: 8
May 30, 2010 at 21:48 comment added Qiaochu Yuan I've wondered this as well. Is there a natural equivalence relation on non-intersecting paths which might reduce the exponent of k?
May 30, 2010 at 21:07 history edited Gjergji Zaimi CC BY-SA 2.5
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May 30, 2010 at 12:53 answer added Steve Huntsman timeline score: 1
May 30, 2010 at 12:10 comment added Wadim Zudilin In fact, Timothy responses more to your question rather than mine. The article from the Fibonacci Quart. he mentions can be downloaded from the author's webpage, but it's about an algorithm. I like your example (with meaning +1 :) ), as it has a combinatorial interpretation "up to a power of $k$".
May 30, 2010 at 12:01 comment added Gjergji Zaimi I was secretly hoping this would be simpler than the hard questions you asked in the other thread. I still have to check the links in that response and the comments carefully though.
May 30, 2010 at 11:50 comment added Wadim Zudilin Gjergji, you ask a question somehow mentioned in Timothy's response to my post mathoverflow.net/questions/26336.
May 30, 2010 at 11:44 history asked Gjergji Zaimi CC BY-SA 2.5