A242208 -id:A242208 - cp's OEIS frontend

A329337 Continued fraction of A328907 = 0.6009668516..., solution to 1 + 3^x = 6^x.

0, 1, 1, 1, 1, 40, 1, 3, 2, 1, 2, 23, 1, 13, 1, 8, 1, 15, 2, 3, 103, 4, 10, 4, 2, 2, 2, 1, 1, 84, 1, 4, 1, 3, 1, 1, 5, 1, 7, 23, 8, 1, 8, 24, 1, 1, 2, 12, 39, 14, 19, 3, 4, 8, 3, 2, 1, 4, 1, 8, 1, 1, 1, 2, 1, 10, 1, 35, 1, 10, 2, 2, 2, 1, 1, 15, 2, 3, 1, 4, 7, 5, 1, 9, 1, 1, 1, 1, 2, 3, 3, 2, 1, 4, 54, 4, 1, 3, 2, 1, 1, 1, 1, 4, 22, 1, 4, 3, 1, 1, 1, 2, 6, 1, 1, 4, 1, 8, 1, 20

Offset: 0

M. F. Hasler, Nov 11 2019

			0.6009668516... = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(40 + 1/(1 + 1/(3 + ...)))))))

Cf. A328912 (cont. frac. of A242208: 1 + 2^x = 4^x), A328913 (cont. frac. of A328900: 2^x + 3^x = 4^x), A329335 (cont. frac. of A328905: 1 + 2^x = 5^x).

ContinuedFraction[x/.FindRoot[1+3^x==6^x,{x,1},WorkingPrecision->150]] (* Harvey P. Dale, Jun 13 2022 *)

contfrac(c=solve(x=0,1, 1+3^x-6^x))[^-1] \\ discarding possibly incorrect last term. Use e.g. \p999 to get more terms. - M. F. Hasler, Oct 31 2019

A242047 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal "sextinomial" triangle mod 2, where coefficients are from (1+x+x^2+x^3+x^4+x^5)^n.

Original entry on oeis.org

8, 1, 9, 4, 6, 9, 4, 6, 2, 1, 6, 5, 5, 4, 0, 1, 4, 6, 5, 9, 5, 9, 3, 7, 6, 8, 4, 3, 7, 7, 2, 8, 5, 5, 9, 8, 6, 1, 5, 1, 2, 4, 6, 2, 3, 5, 4, 3, 1, 4, 1, 2, 0, 9, 3, 4, 7, 1, 1, 4, 6, 7, 7, 5, 7, 8, 5, 6, 7, 0, 3, 2, 5, 0, 0, 8, 1, 1, 7, 9, 4, 1, 6, 6, 7, 6, 7, 6, 7, 8, 7, 9, 3, 5, 8, 1, 0, 9, 6, 6, 7, 4, 7, 4, 6

Offset: 0

Jean-François Alcover, Aug 13 2014

			0.819469462165540146595937684377285598615124623543141209347...

Steven Finch, Pascal Sebah and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle (arXiv:0802.2654) p. 12.

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.

mu = Sort[Table[Root[x^6 - 4*x^5 + x^4 - x^3 + 8*x^2 + 11*x + 8, x, n], {n, 1, 5}], N[Abs[#1]] < N[Abs[#2]] &] // Last; RealDigits[Log[mu]/Log[2] - 1, 10, 104] // First

log(abs(mu))/log(2) - 1, where mu is the root of x^6 - 4*x^5 + x^4 - x^3 + 8*x^2 + 11*x + 8 with maximum modulus.

A242049 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.

Original entry on oeis.org

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, 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, 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

Offset: 0

Jean-François Alcover, Aug 13 2014

			0.429947433342452720114697035519922323324065011589304617040276...
= log(1.53717671718235794959014032895522160250150809343236...)

Steven Finch, Pascal Sebah and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle, arXiv:0802.2654 [math.NT], 2008, p. 14.
Sara Kropf and Stephan Wagner, q-Quasiadditive functions, arXiv:1605.03654 [math.CO], 2016. See section 5 example 8 mean mu for the case s_n is the Jacobsthal sequence.
Kevin Ryde, vpar examples/complete-binary-matchings.gp calculations and code in PARI/GP, see log(C).

Cf. A338294.

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.

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

(1/4)*suminf(k=1, (log((1/3)*(2^(k+2) - (-1)^k))/2^k)) \\ Michel Marcus, May 14 2020

Equals (1/4)*Sum_{k >= 1} (log((1/3)*(2^(k+2) - (-1)^k))/2^k).
From Kevin Ryde, Feb 13 2021: (Start)
Equals log(A338294).
Equals Sum_{k>=1} (1/k)*( 1/(1+(-2)^(k+1)) - 1/(-3)^k ) (an alternating series).
(End)

A329336 Continued fraction of A328906 = 0.4895363211996..., solution to 1 + 2^x = 6^x.

Original entry on oeis.org

0, 2, 23, 2, 1, 1, 4, 1, 1, 27, 4, 12, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 2, 1, 2, 1, 2, 6, 1, 10, 4, 3, 4, 1, 2, 1, 1, 43, 69, 1, 2, 41, 1, 3, 2, 3, 3, 1, 5, 4, 1, 1, 1, 7, 1, 1, 1, 11, 13, 2, 3, 1, 1, 1, 118, 2, 1, 1, 12, 1, 2, 2, 2, 6, 2, 3, 1, 4, 1, 8, 1, 1, 18, 2, 21, 1, 4, 1, 3, 1, 51, 6, 1, 1, 18, 2, 1, 1, 2, 56, 1, 1, 5, 4, 1, 4, 7, 1, 2, 2, 1, 9, 76, 2, 1, 3, 1, 5, 3, 1, 7, 6

Offset: 0

M. F. Hasler, Nov 11 2019

			0.4895363211996... = 0 + 1/(2 + 1/(23 + 1/(2 + 1/(1 + 1/(1 + 1/(4 + 1/...))))))

Cf. A328912 (cont. frac. of A242208: 1 + 2^x = 4^x), A328913 (cont. frac. of A328900: 2^x + 3^x = 4^x), A329334 (cont. frac. of A328904: 1 + 3^x = 5^x).

ContinuedFraction[x/.FindRoot[1+2^x==6^x,{x,.4},WorkingPrecision->1000],150] (* Harvey P. Dale, Oct 15 2022 *)

contfrac(c=solve(x=0,1, 1+2^x-6^x))[^-1] \\ discarding possibly incorrect last term. Use e.g. \p999 to get more terms. - M. F. Hasler, Oct 31 2019

A242048 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal "septinomial" triangle mod 2, where coefficients are from (1+x+...+x^5+x^6)^n.

Original entry on oeis.org

8, 3, 1, 7, 9, 6, 3, 9, 6, 7, 3, 4, 4, 4, 0, 6, 8, 9, 9, 9, 3, 8, 9, 3, 1, 0, 7, 4, 5, 8, 6, 6, 8, 9, 5, 7, 3, 2, 5, 9, 2, 8, 5, 5, 8, 5, 0, 2, 1, 3, 7, 7, 2, 2, 0, 5, 5, 3, 5, 0, 0, 4, 2, 1, 6, 0, 7, 8, 0, 6, 2, 5, 8, 3, 6, 6, 4, 4, 6, 5, 7, 6, 3, 6, 4, 8, 7, 7, 5, 2, 3, 1, 9, 6, 9, 8, 8, 6, 0, 3, 0, 6

Offset: 0

Jean-François Alcover, Aug 13 2014

			0.83179639673444068999389310745866895732592855850213772205535...

Steven Finch, Pascal Sebah and Zai-Qiao Bai, Odd Entries in Pascal's Trinomial Triangle (arXiv:0802.2654) p. 13.

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.

mu = Sort[Table[Root[x^6 - x^5 - 2*x^4 - 28*x^3 + 16*x + 64, x, n], {n, 1, 5}], N[Abs[#1]] < N[Abs[#2]]&] // Last; RealDigits[Log[mu]/Log[2] - 1, 10, 102] // First

log(abs(mu))/log(2) - 1, where mu is the root of x^6 - x^5 - 2*x^4 - 28*x^3 + 16*x + 64 with maximum modulus.

A318057 a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.

Original entry on oeis.org

0, -2, 3, 2, 5, 2, 6, 9, 10, 9, 13, 12, 15, 16, 19, 16, 20, 22, 24, 25, 27, 29, 28, 30, 33, 32, 36, 32, 38, 32, 41, 42, 44, 45, 46, 47, 50, 48, 52, 54, 53, 56, 53, 58, 59, 60, 64, 62, 66, 62, 67, 69, 71, 73, 75, 74, 77, 78, 80, 82, 81, 84, 81, 87, 81, 88, 90

Offset: 1

A.H.M. Smeets, Aug 14 2018

The correct binary value of the golden section is given in A068432; the continued fraction terms of the golden section is given in A000012.
For the number of correct decimal digits of the golden section see A318058.
The denominator of the k-th convergent obtained from a continued fraction tend to k*A001622; the error between the k-th convergent and the constant itself tends to 1/(2*k*A001622), or in binary digits 2*k*log(A001622)/log(2) bits after the binary point.
The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).

			   n   convergent         binary expansion       a(n)
  ==  =============  ==========================  ====
   1    1 / 1          1.0                         0
   2    2 / 1         10.0                        -2
   3    3 / 2          1.1000                      3
   4    5 / 3          1.101                       2
   5    8 / 5          1.100110                    5
   6   13 / 8          1.101                       2
   7   21 / 13         1.1001110                   6
   8   34 / 21         1.1001111001                9
   9   55 / 34         1.10011110000              10
  10   89 / 55         1.1001111001                9
  oo  lim = A068432    1.1001111000110111011110   --

Cf. A000012, A001622, A002390, A068432, A202543, A242208, A318058.

p, q, i, base = 1, 1, 0, 2
while i < 20200:
    p, q, i = p+q, p, i+1
a0, p, q = p//q, q, p
i, p, dd = 0, p*base, [0]
while i < 30000:
    d, p, i = p//q, (p%q)*base, i+1
    dd = dd+[d]
n, pn, qn = 0, 1, 0
while n < 20000:
    n, pn, qn = n+1, pn+qn, pn
    if pn//qn != a0:
        print(n, "- manual!")
    else:
        i, p, q, di = 0, (pn%qn)*base, qn, 0
        while di == dd[i]:
            i, di, p = i+1, p//q, (p%q)*base
        print(n, i-1)

Lim {n -> oo} a(n)/n = 2*log(A001622)/log(2) = 2*A002390/log(2) = A202543/log(2) = 2*A242208.

A329335 Continued fraction of A328905 = 0.5638955242599..., solution to 1 + 2^x = 5^x.

Original entry on oeis.org

0, 1, 1, 3, 2, 2, 2, 1, 3, 3, 1, 5, 3, 1, 1, 3, 1, 1, 6, 1, 36, 20, 18, 3, 1, 3, 2, 1, 2, 9, 3, 2, 1, 1, 1, 2, 7, 1, 1, 5, 1, 112, 2, 1, 6, 2, 1, 1, 1, 1, 2, 44, 1, 2, 3, 70, 1, 1, 1, 12, 3, 1, 5, 6, 1, 1, 10, 4, 4, 2, 3, 1, 7, 1, 4, 1, 1, 1, 5, 2, 1, 5, 1, 4, 3, 1, 1, 1, 1, 2, 1, 1, 4, 6, 7, 2

Offset: 0

M. F. Hasler, Nov 11 2019

			0.5638955242599... = 0 + 1/(1 + 1/(1 + 1/(3 + 1/(2 + 1/(2  + 1/(2 + 1/...))))))

Cf. A328912 (cont. frac. of A242208: 1 + 2^x = 4^x), A328913 (cont. frac. of A328900: 2^x + 3^x = 4^x), A329337 (cont. frac. of A328907: 1 + 3^x = 6^x).

contfrac(c=solve(x=0,1, 1+2^x-5^x))[^-1] \\ discarding possibly incorrect last term. Use e.g. \p999 to get more terms. - M. F. Hasler, Oct 31 2019

cp's OEIS Frontend

A329337 Continued fraction of A328907 = 0.6009668516..., solution to 1 + 3^x = 6^x.

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A242047 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal "sextinomial" triangle mod 2, where coefficients are from (1+x+x^2+x^3+x^4+x^5)^n.

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A242049 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.

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A329336 Continued fraction of A328906 = 0.4895363211996..., solution to 1 + 2^x = 6^x.

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A242048 Decimal expansion of the asymptotic growth rate of the number of odd coefficients in Pascal "septinomial" triangle mod 2, where coefficients are from (1+x+...+x^5+x^6)^n.

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A318057 a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.

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A329335 Continued fraction of A328905 = 0.5638955242599..., solution to 1 + 2^x = 5^x.

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