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A261009
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Write 2^n in base 3, add up the "digits".
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5
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1, 2, 2, 4, 4, 4, 4, 6, 4, 8, 8, 10, 10, 8, 10, 16, 12, 14, 12, 16, 14, 18, 16, 12, 10, 12, 14, 20, 20, 22, 24, 26, 24, 22, 22, 22, 18, 20, 26, 28, 28, 28, 26, 30, 30, 30, 26, 26, 26, 32, 38, 40, 38, 38, 28, 34, 40, 42, 38, 40, 46, 40, 38, 42, 48, 44, 42, 40, 42, 48, 48, 44
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OFFSET
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0,2
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COMMENTS
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Comment from Jean-Paul Allouche, Oct 25 2015: As mentioned by Holdum et al. (2015) the following problem, cited in "Concrete Mathematics" by Graham, Knuth, and Patashnik (1994), is still open: prove that for all n > 256, binomial(2n,n) is either divisible by 4 or by 9 (cf. A000984). This can be easily reduced to show that, for all k >= 9, 2*a(k) - a(k+1) >= 4. This has been proved up to huge values of k (Holdum et al. mention k = 10^{13}).
For additional information about the divisibility of binomial(2n,n) by squares see the comments and references in A000984, - N. J. A. Sloane, Oct 29 2015
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LINKS
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Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994
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FORMULA
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EXAMPLE
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2^7 = 128_10 = 11202_3, so a(7) = 1+1+2+0+2 = 6.
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MAPLE
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S:=n->add(i, i in convert(2^n, base, 3)); [seq(S(n), n=0..100)];
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MATHEMATICA
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Table[Total@ IntegerDigits[2^n, 3], {n, 0, 100}] (* Giovanni Resta, Aug 14 2015 *)
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PROG
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(PARI) a(n) = vecsum(digits(2^n, 3)); \\ Michel Marcus, Aug 14 2015
(Haskell)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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