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A249452
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Numbers n such that A249441(n) = 3.
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1
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15, 31, 47, 63, 95, 127, 191, 255, 383, 511, 767, 1023, 1535, 2047, 3071, 4095, 6143, 8191, 12287, 16383, 24575, 32767, 49151, 65535, 98303, 131071, 196607, 262143, 393215, 524287, 786431, 1048575, 1572863, 2097151, 3145727, 4194303, 6291455, 8388607, 12582911
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OFFSET
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1,1
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COMMENTS
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Or n for which none of entries in the n-th row of Pascal's triangle (A007318) is divisible by 4 (cf. comment in A249441).
Using the Kummer carries theorem, one can prove that, for n>=2, a(n) has the form of either 1...1 or 101...1 in base 2.
The sequence is a subset of so-called binomial coefficient predictors (BCP) in base 2 (see Shevelev link, Th. 6 and Cor. 8), which were found also using Kummer theorem and have a very close binary structure.
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LINKS
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FORMULA
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a(n) has either form 2^k - 1 or 3*2^m-1, k, m >= 4 (cf. A000225, A055010). Since, for
k>=5, 2^k-1<3*2^(k-1)-1<2^(k+1)-1, we have
that, for n>=1, a(2*n) = 2^(n+4)-1; a(2*n+1)
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MATHEMATICA
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CoefficientList[Series[(15 + 16 x - 14 x^2 - 16 x^3)/(1 - x -2 x^2 + 2 x^3), {x, 0, 70}], x] (* Vincenzo Librandi, Oct 30 2014 *)
LinearRecurrence[{1, 2, -2}, {15, 31, 47, 63}, 40] (* Harvey P. Dale, Apr 01 2019 *)
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PROG
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(PARI) is(n)=(n+1)>>valuation(n+1, 2)<5 && !setsearch([1, 2, 3, 5, 7, 11, 23], n) \\ Charles R Greathouse IV, Nov 06 2014
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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