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A249452
Numbers n such that A249441(n) = 3.
1
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
OFFSET
1,1
COMMENTS
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.
LINKS
E. E. Kummer, Über die Ergänzungssätze zu den allgemeinen Reciprocitätsgesetzen, J. Reine Angew. Math. 44 (1852), 93-146.
V. Shevelev, Binomial Coefficient Predictors, Journal of Integer Sequences, Vol. 14 (2011), Article 11.2.8
FORMULA
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)
= 3*2^(n+3)-1. - Vladimir Shevelev, - Oct 29, Nov 06 2014
a(1) = 15, and for n>1, a(n) = A052955(n+6). [Follows from above] - Antti Karttunen, Nov 03 2014
G.f.: (15+16*x-14*x^2-16*x^3)/(1-x-2*x^2+2*x^3); a(n) = 16*A029744(n)-1. - Peter J. C. Moses, Oct 30 2014
MATHEMATICA
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 *)
PROG
(PARI) a(n)=if(n==1, 15, (n%2+2)<<(n\2+3)-1) \\ Charles R Greathouse IV, Nov 06 2014
(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
KEYWORD
nonn,easy
AUTHOR
Vladimir Shevelev, Oct 29 2014
EXTENSIONS
More terms from Peter J. C. Moses, Oct 29 2014
STATUS
approved