A114550 -id:A114550 - cp's OEIS frontend

A114551 Continued fraction expansion of the constant (A114550) equal to Sum_{n>=0} 1/A112373(n) such that the partial quotients satisfy a(2n) = A112373(n) for n > 0 and a(2n+1) = A112373(n+1)/A112373(n) for n >= 0.

2, 1, 1, 2, 2, 6, 12, 78, 936, 73086, 68408496, 4999703411742, 342022190843338960032, 1710009514450915230711940280907486, 584861200495456320274313200204390612579749188443599552

Offset: 0

Paul D. Hanna, Dec 08 2005

A112373 is defined by the recurrence: let b(n) = A112373(n), then

b(n) = (b(n-1)^3 + b(n-1)^2)/b(n-2) for n >= 2 with b(0)=b(1)=1.

			2.584401724019776724812076147153331342112382090467969...
= Sum_{n>=0} 1/A112373(n) = 1/1 + 1/1 + 1/2 + 1/12 + 1/936 + 1/68408496 + ...
= [2;1,1,2,2,6,12,78,936,73086,68408496,...] (continued fraction).
The recurrence of partial quotients is demonstrated by:
(odd-index) a(7) = 78 = a(6)*a(5) + a(5) = 12*6 + 6;
(even-index) a(8) = 936 = a(7)*a(6) = 78*12.

Michael De Vlieger, Table of n, a(n) for n = 0..20
Taras Goy and Mark Shattuck, Determinant Formulas of Some Hessenberg Matrices with Jacobsthal Entries Jacobsthal Entries, Applications and Appl. Math. (2021) Vol. 16, Issue 1, Art. 10.
A. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015 and J. Int. Seq. 18 (2015) # 15.8.4.

Cf. A112373, A114550 (constant), A114552 (bisection).

a[0] = 2; a[1] = a[2] = 1;
a[n_] := a[n] = a[n-1] a[n-2] + Mod[n, 2] a[n-2];
a /@ Range[0, 14] (* Jean-François Alcover, Oct 01 2019 *)

a(n)=if(n<0,0,if(n<3,[2,1,1][n+1],a(n-1)*a(n-2)+(n%2)*a(n-2)))

a(2n) = a(2n-1)*a(2n-2) for n>=2, a(2n+1) = a(2n)*a(2n-1) + a(2n-1) for n>=1, with a(0)=2, a(1)=a(2)=1. - Jeffrey Shallit

A112373 a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 2, 12, 936, 68408496, 342022190843338960032, 584861200495456320274313200204390612579749188443599552

Offset: 0

Andrew Hone, Dec 02 2005

nonn

As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) = A104457.
The Laurent property is satisfied by any second-order recurrence of the form a(n+2)=f(a(n+1))/a(n) where f is a polynomial of the form f(x)=x^m*p(x) with m a positive integer >= 2 and p arbitrary. In that case a(0)=a(1)=1 generates a sequence of integers and the ratios a(n+1)/a(n) and a(n+1)*a(n-1)/a(n)^2 are integers for all n. - Andrew Hone, Dec 12 2005
Also denominators of Sum_{k=0..n} 1/a(k) with numerators = A259644. - Reinhard Zumkeller, Jul 02 2015
The next term (a(8)) has 141 digits. - Harvey P. Dale, Apr 05 2019

Reinhard Zumkeller, Table of n, a(n) for n = 0..10
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics, Vol. 43, No. 3 (2015), pp. 589-633.
Sergey Fomin and Andrei Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, Vol. 28, No. 2 (2002) pp. 119-144.
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbersJournal of Integer Sequences, Vol. 18 (2015), Article #15.8.4; arXiv preprint, arXiv:1507.00063 [math.NT], 2015.
Andrew N. W. Hone, Continued fractions for some transcendental numbers, Monatshefte für Mathematik, Vol. 182 (2017), pp. 33-38; arXiv preprint, arXiv:1509.05019 [math.NT], 2015-2016.

Cf. A114550, A114551, A114552.

Cf. A259644.

a112373 n = a112373_list !! n
a112373_list = 1 : 1 : zipWith (\u v -> (u^3 + u^2) `div` v)
                               (tail a112373_list) a112373_list
-- Reinhard Zumkeller, Jul 02 2015

I:=[1,1]; [n le 2 select I[n] else (Self(n-1)^3+Self(n-1)^2)/Self(n-2): n in [1..10]]; // Vincenzo Librandi, Jul 02 2015

a[0]:=1; a[1]:=1; f(x):=x^3+x^2; for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od; s[3]:=ln(2^2*3); s[4]:=ln(2^3*3^2*13); for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));
# s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...

RecurrenceTable[{a[n] == (a[n - 1]^3 + a[n - 1]^2)/a[n - 2], a[0] == a[1] == 1}, a, {n, 0, 7}] (* Michael De Vlieger, Jul 02 2015 *)
nxt[{a_,b_}]:={b,(b^3+b^2)/a}; NestList[nxt,{1,1},8][[All,1]] (* Harvey P. Dale, Apr 05 2019 *)

{a(n) = my(a=self()); if(n<0, a(1-n), n<2, 1, a(n-1)^2 * (1 + a(n-1)) / a(n-2))}; /* Michael Somos, Apr 19 2017 */

a(n+1) / a(n) = A114552(n), a(n) = a(1-n) for all n in Z. - Michael Somos, Apr 19 2017

Sum_{n>=0} 1/a(n) = A114550. - Amiram Eldar, Nov 13 2020

A114552 Odd-indexed bisection of A114551, which is the continued fraction expansion of Sum_{n>=0} 1/A112373(n); also, a(n) = A112373(n+1)/A112373(n) for n>=0.

Original entry on oeis.org

1, 2, 6, 78, 73086, 4999703411742, 1710009514450915230711940280907486, 1000118217480414644596377710149364954622541926875237885435586055466747491994623223953758

Offset: 0

Paul D. Hanna, Dec 08 2005

nonn

A112373 is defined by the recurrence: let b(n) = A112373(n), then
b(n) =(b(n-1)^3 + b(n-1)^2)/b(n-2) for n>=2 with b(0)=b(1)=1.
The next term 58499...45086 has 228 digits and is too large to include.

			a(5) = 4999703411742 = 73086*(1 + 1*2*6*78*73086).
a(5) = a(4)*(1 - a(4) + a(4)^2/a(3)) = 73086*(1 - 73086 + 73086^2/78).

Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015 and J. Int. Seq. 18 (2015) # 15.8.4.

Cf. A112373, A114550 (constant), A114551 (continued fraction).

a(n)=if(n==0,1,if(n==1,2,a(n-1)*(1-a(n-1)+a(n-1)^2/a(n-2))))

a(n) = a(n-1)*(1 + a(0)*a(1)*a(2)*...*a(n-1)) for n>=1, with a(0)=1.

a(n) = a(n-1)*(1 - a(n-1) + a(n-1)^2/a(n-2)) for n>=2, with a(0)=1, a(1)=2.

a(7) corrected by Georg Fischer, Aug 26 2022

A270121 Denominators in a perturbed Engel series.

Original entry on oeis.org

7, 112, 403200, 1755760043520000, 53695136666462381094317154204367872000000

Offset: 1

Andrew Hone, Mar 11 2016

nonn

The sum of the series 6/a(1)+1/a(2)+1/a(3)+... is a transcendental number, and has a continued fraction expansion whose coefficients are given explicitly in terms of the sequence a(n) and the ratios a(n+1)/a(n).

Amiram Eldar, Table of n, a(n) for n = 1..8
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
Andrew N. W. Hone, Continued fractions for some transcendental numbers, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.

Cf. A112373, A114552, A114550, A270137.

a[1] = 7; a[2] = 112;
a[n_] := a[n] = (a[n-1]^2 (1+(n-1)a[n-1]))/a[n-2];
Array[a, 5] (* Jean-François Alcover, Dec 16 2018 *)

The sequence is generated by taking a(n+1)=b(n-1)*a(n)*(1+n*a(n)), b(n)=a(n+1)/a(n) for n>=1 with initial values a(1)=7,b(0)=2. Alternatively, if a(1)=7,a(2)=112 are given then a(n+1)*a(n-1)=a(n)^2*(1+n*a(n)) for n>=2.

Sum_{n>=1} 1/a(n) = -5/7 + A270137. - Amiram Eldar, Nov 20 2020

A270137 Decimal expansion of the constant 6/A270121(1) + Sum_{n>=2} 1/A270121(n).

Original entry on oeis.org

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

Offset: 1

Andrew Hone, Mar 11 2016

A270121 is defined by the following recurrence: if A270121(n)=x(n) then x(n+1)*x(n-1)=x(n)^2*(1+n*x(n)) for n>=1, with x(1)=7, x(2)=112; and for A270124, if A270124(n)=y(n) then y(0)=2 and y(n)=x(n+1)/x(n) for n>=1. Both of these sequences appear in the continued fraction expansion of this number, which is transcendental.

			0.86607390873015929971... = 6/A270121(1) + Sum_{n>=2} 1/A270121(n) = 6/7 + 1/112 + 1/403200 + 1/1755760043520000 + ... = [0; 1, 6, 2, 7, 32, 112, 10800, 403200, 17418254400, ...] = [0; 1, 6, A270124(0), A270121(1), 2*A270124(1), A270121(2), 3*A270124(2), A270121(3), 4*A270124(3), ...] (continued fraction).

A. N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
A. N. W. Hone, Continued fractions for some transcendental numbers, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.
Index entries for transcendental numbers

Cf. A112373, A114550, A114551, A114552.

The continued fraction expansion takes the form

[0; 1, 6, A270124(0), A270121(1), ..., n*A270124(n-1), A270121(n), (n+1)*A270124(n), A270121(n+1), ...].

More terms from Jon E. Schoenfield, Nov 12 2016

A270138 Continued fraction expansion of the constant 6/A270121(1)+Sum_{n>=2}1/A270121(n).

Original entry on oeis.org

0, 1, 6, 2, 7, 32, 112, 10800, 403200, 17418254400, 1755760043520000

Offset: 0

Andrew Hone, Mar 11 2016

A270121 is defined by the following recurrence: if A270121(n)=x(n) then x(n+1)*x(n-1)=x(n)^2*(1+n*x(n)) for n>=1, with x(1)=7, x(2)=112; and for A270124, if A270124(n)=y(n) then y(0)=2 and y(n)=x(n+1)/x(n) for n>=1. Both of these sequences appear in this continued fraction expansion, which defines a transcendental number.

			6/A270121(1)+Sum_{n>=2}1/A270121(n)=6/7+1/112+1/403200+1/1755760043520000+...
=[0;1,6,2,7,32,112,10800,403200,17418254400,...]
=[0;1,6,A270124(0),A270121(1),2*A270124(1),A270121(2),3*A270124(2),A270121(3),4*A270124(3),...] (continued fraction).

A. N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.8.4.
A. N. W. Hone, Continued fractions for some transcendental numbers, arXiv:1509.05019 [math.NT], 2015-2016, Monatsh. Math. DOI: 10.1007/s00605-015-0844-2.

Cf. A112373, A114550, A114551, A114552.

a(2*n+1) = n*A270124(n-1), a(2*n+2) = A270121(n) for n>=1.

cp's OEIS Frontend

A114551 Continued fraction expansion of the constant (A114550) equal to Sum_{n>=0} 1/A112373(n) such that the partial quotients satisfy a(2n) = A112373(n) for n > 0 and a(2n+1) = A112373(n+1)/A112373(n) for n >= 0.

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A112373 a(n+2) = (a(n+1)^3 + a(n+1)^2)/a(n) with a(0)=1, a(1)=1.

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A114552 Odd-indexed bisection of A114551, which is the continued fraction expansion of Sum_{n>=0} 1/A112373(n); also, a(n) = A112373(n+1)/A112373(n) for n>=0.

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A270121 Denominators in a perturbed Engel series.

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A270137 Decimal expansion of the constant 6/A270121(1) + Sum_{n>=2} 1/A270121(n).

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A270138 Continued fraction expansion of the constant 6/A270121(1)+Sum_{n>=2}1/A270121(n).

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