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Search: id:A147553
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| A147553 |
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Numbers n such that n^2 divides n.n where dot"." means concatenation. |
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+0
2
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| 1, 143, 142857143, 142857142857143, 142857142857142857143, 142857142857142857142857143, 142857142857142857142857142857143, 142857142857142857142857142857142857143
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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I proved that for n>0, a(n)=(10^(6n-3)+1)/7. Namely for n>1, a(n) is of the form 142857.142857. ... .142857.143. Except the first term 11 divides all other terms, so there is no prime p such that p^2 divides p.p. For n>0 a(n).a(n)/(a(n)*a(n))=7.
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EXAMPLE
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143*143|143.143 (143143/(143*143)=7) so 143 is in the sequence.
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MATHEMATICA
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a[0]=1; a[n_]:=(10^(6n-3)+1)/7; Table[a[k], {k, 0, 8}]
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CROSSREFS
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Cf. A147554.
Sequence in context: A135946 A029555 A046179 this_sequence A030122 A057404 A093159
Adjacent sequences: A147550 A147551 A147552 this_sequence A147554 A147555 A147556
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KEYWORD
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base,easy,nice,nonn
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AUTHOR
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Farideh Firoozbakht (mymontain(AT)yahoo.com), Dec 23 2008
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