Talk:Palindromic gapful numbers: Difference between revisions
→Nice Recursive Solution/Definition
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==Nice Recursive Solution/Definition==
If I let f<sub>1</sub> = 0..1..9 and f<sub>2</sub> = 0..11..99 then f<sub>3</sub> is xyx where x is 0..1..9 and y is f<sub>1</sub>. In general f<sub>n</sub> is (x<sup>n-1</sup>+x)+10*f<sub>n-2</sub>. For all palindromic numbers x=0 is invalid for outermost pair. Not a problem for this task as you will wish to choose x to be the ending for each set. Each set is then filtered to verify divisibility by 11x. F# can implement this and execute the entire task in less than 35 thousandths of a sec so sufficiently optimal. Obviously this implies that all palindromic gapful numbers are divisible by 11, but can this be used to improve on 35 thousandths of a sec?
--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 15:07, 3 December 2020 (UTC)
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