A249452 Numbers k such that A249441(k) = 3.

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

Author

Vladimir Shevelev, Oct 29 2014

Keywords

• nonn
• easy

Comments

Or k for which none of entries in the k-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
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2).

Crossrefs

Cf. A000225, A029744, A048278, A052955, A055010, A089633, A249441.

Programs

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 *)

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

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 2014, 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

Extensions

More terms from Peter J. C. Moses, Oct 29 2014