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Search: id:A135738
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| A135738 |
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Least positive integer with even digit sum in bases 2..n. |
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+0
1
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| 3, 6, 10, 10, 54, 54, 54, 54, 130, 130, 130, 130, 390, 390, 2000, 2000, 3238, 3238, 4080, 4080, 7326, 7326, 16584, 16584, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 231000, 231000, 231000, 231000, 466352, 466352, 466352, 466352, 3020830
(list; graph; listen)
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OFFSET
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2,1
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COMMENT
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The sequence is obviously increasing. It seems that a(2n+1)=a(2n) for n>1. Is there a simple proof? Is there a simple way to construct a(n)? Notice the pattern in base-N, e.g. 130 = 10000010[2] = 11211[3] = 2002[4] = 1010[5] = 334[6] = 244[7] = 202[8] = 154[9] = 109[11] = {10}{10}[12] = {10}0[13].
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REFERENCES
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http://www.mersenneforum.org/showthread.php?p=120023
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LINKS
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"Davar55" on mersenneforum.org, Puzzles / "Sum of digits".
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EXAMPLE
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a(2)=3 since 1=1[2], 2=10[2], so 3=11[2] is the number > 0 with even digit sum (1+1) in base 2.
a(3)=5 since 4=100[2], 5=12[3], so 6=20[3]=[110][2] is the least N>0 with even digit sum in base 2 and in base 3.
a(4)=a(5)=10=1010[2]=101[3]=22[4]=20[5] is the least N>0 having even digit sum in bases 2 through 4, and has so also in base 5.
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PROGRAM
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(PARI) digitsum(n, b=10, s)={n=[n]; while(n=divrem(n[1], b), s+=n[2]); s} /* n-th element of the sequence: */ A135738(Bmax, n=1)={until(!n++, for(b=2, Bmax, digitsum(n, b)%2&next(2)); return(n))} /* display the list */ t=1; for(b=2, 100, print(b, ":", t=A135738(b, t)))
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CROSSREFS
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Cf. A000120, A053735, A053737, A053824, A053827-A053836, A007953.
Sequence in context: A032570 A130483 A115012 this_sequence A073851 A043321 A123053
Adjacent sequences: A135735 A135736 A135737 this_sequence A135739 A135740 A135741
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KEYWORD
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nonn
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AUTHOR
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M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Dec 06 2007
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