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A217971
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a(n) = 2^(2*n+1) * (2*n+1)*n^(2*n).
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2
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24, 2560, 653184, 301989888, 220000000000, 231818611654656, 333360204766740480, 627189298506124754944, 1495163506861268427866112, 4404019200000000000000000000, 15705682358754099640245749284864, 66686788842514206222454073642188800, 332430457331186494783020411573611003904
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OFFSET
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1,1
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COMMENTS
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Let S_(2*n+1)(m) denote difference between multiples of 2*n+1 in interval [0,m), m>=1, with even and odd digit sums in base 2*n. As is shown in the Shevelev and Moses link, a recursion for S_(2*n+1)(m) is connected with the periodicity of a special digit function, the smallest period of which is a(n).
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LINKS
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MATHEMATICA
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Table[2^(2*n + 1)*(2*n + 1)*n^(2*n), {n, 15}] (* Wesley Ivan Hurt, Apr 28 2020 *)
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PROG
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(PARI) a(n) = {2^(2*n+1) * (2*n+1)*n^(2*n)} \\ Andrew Howroyd, Apr 28 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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