A112373 -id:A112373 - cp's OEIS frontend

A114550 Decimal expansion of the constant Sum_{n>=0} 1/A112373(n), where the partial quotients of the continued fraction A114551 satisfy A114551(2n) = A112373(n) and A114551(2n+1) = A112373(n+1)/A112373(n).

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

Offset: 1

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.
Thus the sum of unit fractions 1/A112373(n) converges rapidly.

			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).

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, A114551 (continued fraction), A114552.

dm = 5; digits = 105;
b[n_] := b[n] = If[n < 2, 1, (b[n - 1]^3 + b[n - 1]^2)/b[n - 2]];
s[m_] := s[m] = N[Sum[1/b[n], {n, 0, m}], digits + 5];
s[m = dm];
s[m += dm];
While[RealDigits[s[m]] != RealDigits[s[m - dm]], m += dm];
RealDigits[s[m], 10, digits][[1]] (* Jean-François Alcover, Sep 30 2019 *)
c[0]=2; c[1] = c[2] = 1; c[n_] := c[n] = c[n-1] c[n-2] + Mod[n, 2] c[n-2];
RealDigits[FromContinuedFraction[c /@ Range[0, 14]], 10, 105][[1]] (* Jean-François Alcover, Oct 01 2019 *)

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.

Original entry on oeis.org

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

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

A259644 Numerators of sum(1/A112373(k): k=0..n), denominators = A112373.

Original entry on oeis.org

1, 2, 5, 31, 2419, 176795035, 883922739668546300971, 1511516294872733607299090320742127160367108420362968907

Offset: 0

Reinhard Zumkeller, Jul 02 2015

			Sum(1/A112373(k)) = 1, 2, 5/2, 31/12, 2419/936, 176795035/68408496, ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..10
Andrew N. W. Hone, Update on A112373, SeqFan list, July 02 2015.
Andrew N. W. Hone, Curious continued fractions, nonlinear recurrences and transcendental numbers, arXiv:1507.00063 [math.NT], 2015.

Cf. A112373.

import Data.Ratio (numerator)
a259644 n = a259644_list !! n
a259644_list = map numerator $
               scanl1 (+) $ map (recip . fromIntegral) a112373_list

(* b = A112373 *)
b[n_] := b[n] = If[n < 2, 1, (b[n-1]^3 + b[n-1]^2)/b[n-2]];
a[n_] := Sum[1/b[k], {k, 0, n}] // Numerator;
Table[a[n], {n, 0, 7}] (* Jean-François Alcover, Dec 15 2018 *)

A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 10, 4, 1, 10, 30, 28, 8, 1, 15, 70, 112, 72, 16, 1, 21, 140, 336, 360, 176, 32, 1, 28, 252, 840, 1320, 1056, 416, 64, 1, 36, 420, 1848, 3960, 4576, 2912, 960, 128, 1, 45, 660, 3696, 10296, 16016, 14560, 7680, 2176, 256

Offset: 0

Peter Bala, Apr 30 2012

Let b(n,x) = Sum_{k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := (1/2)*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159 (x = -4).

The polynomials R(n,x) factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u * T(2*n+1,u), where u = sqrt((x+2)/2). Hence R(n,x) = 1/u * T(n,u) * T(n+1,u).

			Triangle begins
.n\k.|..0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = =
..0..|..1
..1..|..1....1
..2..|..1....3....2
..3..|..1....6...10....4
..4..|..1...10...30...28....8
..5..|..1...15...70..112...72...16
..6..|..1...21..140..336..360..176...32

Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.

T(n,0) = 1; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1).
O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
Removing the first column from the triangle produces the Riordan array (x/(1-x)^3, 2*x/(1-x)^2).
The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*Sum_{k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x.
Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
With P(n,x) as defined in the Comments section we have (x+2)/x - {Sum_{k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {Sum_{k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently Sum_{k >= 0} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
R(n,2) = A011900(n); R(n,-2) = (-1)^n * A109613(n); R(n,3) = A182432;
R(n,-3) = (-1)^n * A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n * A084159(n).

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

A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 2, 10, 505, 12878813, 4229958765311886322, 5876687051603582015287706866081267480733704277890

Offset: 0

Andrew Hone, Dec 12 2005

nonn

A second-order recurrence with the Laurent property. This property is satisfied by any second-order recurrence of the form a(n+2) = f(a(n+1))/a(n) with f being a polynomial of the form f(x) = x*p(x) where p is a polynomial of degree d with integer coefficients such that p(0)=1 and p has the reciprocal property x^d*p(1/x) = p(x). Hence if a(0) = a(1) = 1 then a(n) is an integer for all n.

As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) or about 0.962 (A202543).

Seiichi Manyama, Table of n, a(n) for n = 0..10
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.

Cf. A101879, A112373, A202543.

a[0]:=1; a[1]:=1; f(x):=x^3+x;
for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od;
s[3]:=ln(10); s[4]:=ln(505);
for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-2*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...
f:=proc(n) option remember; local i,j,k,t1,t2,t3; if n <= 1 then RETURN(1); fi; (f(n-1)^3+f(n-1))/f(n-2); end;
# N. J. A. Sloane

nxt[{a_,b_}]:={b,(b^3+b)/a}; NestList[nxt,{1,1},10][[All,1]] (* Harvey P. Dale, Jun 26 2017 *)

def A(l, m, n)
  a = Array.new(2 * m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(:*) + a[m] ** l
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A112449(n)
  A(3, 1, n)
end # Seiichi Manyama, Nov 20 2016

a(1-n) = a(n). - Seiichi Manyama, Nov 20 2016

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

A114550 Decimal expansion of the constant Sum_{n>=0} 1/A112373(n), where the partial quotients of the continued fraction A114551 satisfy A114551(2n) = A112373(n) and A114551(2n+1) = A112373(n+1)/A112373(n).

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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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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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A259644 Numerators of sum(1/A112373(k): k=0..n), denominators = A112373.

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A211955 Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.

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A270121 Denominators in a perturbed Engel series.

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A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

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

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