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Search: id:A141707
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| A141707 |
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Least k>0 such that (2n-1)k is palindromic in base 2. |
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3
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| 1, 1, 1, 1, 1, 3, 5, 1, 1, 27, 1, 89, 13, 1, 49, 1, 1, 13, 69, 5, 25, 3, 1, 103, 29, 1, 63, 3, 9, 103, 7, 1, 1, 19, 37, 147, 1, 13, 3, 19, 11, 45, 1, 37, 23, 3, 1, 27, 61, 1, 233, 47, 13, 1, 21, 23, 59, 525, 5, 1, 93, 23, 41, 1, 1, 49, 27, 13, 187, 87, 269, 15, 111, 13, 29, 7, 1, 13, 3
(list; graph; listen)
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OFFSET
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1,6
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COMMENT
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Even numbers cannot be palindromic in base 2 (unless leading zeros are considered), that's why we search only for odd numbers 2n-1 the k-values such that k(2n-1) is palindromic in base 2. Obviously they are necessarily also odd.
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EXAMPLE
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a(1..5)=1 since 1,3,5,7,9 are already palindromic in base 2.
a(6)=3 since 2*6-1=11 and 2*11=22 are not palindromic in base 2, but 3*11=33 is.
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PROGRAM
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(PARI) A141707(n, L=10^9)={ n=2*n-1; forstep(k=1, L, 2, binary(k*n)-vecextract(binary(k*n), "-1..1")|return(k))}
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CROSSREFS
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Cf. A050782, A141708.
Sequence in context: A143303 A091084 A016610 this_sequence A010261 A005699 A127250
Adjacent sequences: A141704 A141705 A141706 this_sequence A141708 A141709 A141710
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KEYWORD
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base,easy,nice,nonn
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Jul 17 2008
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