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Search: id:A062927
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| A062927 |
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Numbers n such that n divides the sum of digits of 9^n. |
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+0
2
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OFFSET
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1,2
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COMMENT
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Next term after 243, if it exists, is greater than 30000. - Ryan Propper (rpropper(AT)stanford.edu), Jun 18 2005
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EXAMPLE
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a(1)=3, so 3 divides sum of digits of 9^3 (i.e.7+2+9=18).
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MATHEMATICA
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Do[If[Mod[Plus @@ IntegerDigits[9^n, 10], n] == 0, Print[n]], {n, 1, 30000}] (Propper)
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CROSSREFS
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Sequence in context: A061929 A019461 A067862 this_sequence A045638 A038224 A133195
Adjacent sequences: A062924 A062925 A062926 this_sequence A062928 A062929 A062930
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KEYWORD
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hard,more,nonn,base
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AUTHOR
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Amarnath Murthy (amarnath_murthy(AT)yahoo.com) and Shyam Sunder Gupta (guptass(AT)rediffmail.com), Feb 16 2002
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