|
%I #26 Feb 14 2021 13:22:29
%S 4,2,9,9,4,7,4,3,3,3,4,2,4,5,2,7,2,0,1,1,4,6,9,7,0,3,5,5,1,9,9,2,2,3,
%T 2,3,3,2,4,0,6,5,0,1,1,5,8,9,3,0,4,6,1,7,0,4,0,2,7,6,0,7,2,5,7,4,2,8,
%U 3,3,7,2,8,3,1,3,9,8,1,0,5,6,8,4,5,6,3,4,9,0,0,7,4,8,4,7,4,2,5,3,6,6,5,4,3
%N Decimal expansion of 'lambda', the Lyapunov exponent characterizing the asymptotic growth rate of the number of odd coefficients in Pascal trinomial triangle mod 2, where coefficients are from (1+x+x^2)^n.
%H Steven Finch, Pascal Sebah and Zai-Qiao Bai, <a href="http://arXiv.org/abs/0802.2654">Odd Entries in Pascal's Trinomial Triangle</a>, arXiv:0802.2654 [math.NT], 2008, p. 14.
%H Sara Kropf and Stephan Wagner, <a href="https://arxiv.org/abs/1605.03654">q-Quasiadditive functions</a>, arXiv:1605.03654 [math.CO], 2016. See section 5 example 8 mean mu for the case s_n is the Jacobsthal sequence.
%H Kevin Ryde, <a href="http://user42.tuxfamily.org/pari-vpar/index.html">vpar</a> examples/complete-binary-matchings.gp calculations and code in PARI/GP, see log(C).
%F Equals (1/4)*Sum_{k >= 1} (log((1/3)*(2^(k+2) - (-1)^k))/2^k).
%F From _Kevin Ryde_, Feb 13 2021: (Start)
%F Equals log(A338294).
%F Equals Sum_{k>=1} (1/k)*( 1/(1+(-2)^(k+1)) - 1/(-3)^k ) (an alternating series).
%F (End)
%e 0.429947433342452720114697035519922323324065011589304617040276...
%e = log(1.53717671718235794959014032895522160250150809343236...)
%t digits = 105; lambda = (1/4)*NSum[Log[(1/3)*(2^(k+2) - (-1)^k)]/2^k, {k, 1, Infinity}, WorkingPrecision -> digits + 5, NSumTerms -> 500]; RealDigits[lambda, 10, digits] // First
%o (PARI) (1/4)*suminf(k=1, (log((1/3)*(2^(k+2) - (-1)^k))/2^k)) \\ _Michel Marcus_, May 14 2020
%Y Cf. A338294.
%Y Cf. A242208 (1+x+x^2)^n, A242021 (1+x+x^3)^n, A242022 (1+x+x^2+x^3+x^4)^n, A241002 (1+x+x^4)^n, A242047 (1+x+...+x^4+x^5)^n, A242048 (1+x+...+x^5+x^6)^n.
%K nonn,cons
%O 0,1
%A _Jean-François Alcover_, Aug 13 2014
|
