Displaying 11-20 of 36 results found.
Decimal expansion of the number whose continued fraction expansion is A292106.
+10
5
6, 1, 4, 2, 2, 0, 1, 7, 6, 3, 0, 9, 3, 0, 2, 6, 4, 1, 2, 5, 5, 9, 4, 6, 5, 1, 0, 6, 4, 5, 7, 8, 7, 0, 8, 5, 8, 8, 5, 1, 9, 2, 8, 3, 3, 0, 7, 1, 0, 7, 3, 8, 7, 3, 2, 3, 2, 8, 0, 4, 7, 5, 4, 8, 5, 0, 2, 9, 6, 0, 2, 4, 9, 4, 0, 9, 6, 1, 8, 1, 3, 8, 7, 2, 3, 6, 5, 0, 9, 8, 0, 1, 0, 9, 0, 3, 3, 0, 4, 6, 2, 7, 4, 6
COMMENTS
Another version of the "e*Pi" constant (cf. A292101).
EXAMPLE
6.142201763093026412559465106457...
MATHEMATICA
With[{nmax=100}, First[RealDigits[FromContinuedFraction[ContinuedFraction[E, nmax]ContinuedFraction[Pi, nmax]], 10, nmax]]] (* Paolo Xausa, Oct 23 2023 *)
Continued fraction for e^3.
+10
4
20, 11, 1, 2, 4, 3, 1, 5, 1, 2, 16, 1, 1, 16, 2, 13, 14, 4, 6, 2, 1, 1, 2, 2, 2, 3, 5, 1, 3, 1, 1, 68, 7, 5, 1, 4, 2, 1, 1, 1, 1, 1, 1, 7, 3, 1, 6, 1, 2, 5, 4, 7, 2, 1, 3, 2, 2, 1, 2, 1, 4, 1, 1, 13, 1, 1, 2, 1, 1, 1, 1, 3, 7, 11, 18, 54, 1, 2, 2, 2, 1, 1, 6, 2, 2, 46, 2, 189, 1, 24, 1, 8, 13, 4, 1, 1
LINKS
K. Matthews, Finding the continued fraction of e^(l/m) ["... there is no known formula for the partial quotients of the continued fraction expansion of e^3, or more generally e^(l/m) with l distinct from 1,2 and gcd(l,m)=1..."]
EXAMPLE
20.085536923187667740928529... = 20 + 1/(11 + 1/(1 + 1/(2 + 1/(4 + ...)))). - Harry J. Smith, Apr 30 2009
MAPLE
with(numtheory); Digits:=200: cf:=convert(evalf( exp(3)), confrac); # N. J. A. Sloane, Sep 05 2012
MATHEMATICA
ContinuedFraction[ E^3, 100]
PROG
(PARI) contfrac(exp(1)^3)
(PARI) { allocatemem(932245000); default(realprecision, 21000); x=contfrac(exp(3)); for (n=1, 20001, write("b058282.txt", n-1, " ", x[n])); } \\ Harry J. Smith, Apr 30 2009
a(n) = ceiling(n/3)^2 - floor(n/3)^2.
+10
4
0, 1, 1, 0, 3, 3, 0, 5, 5, 0, 7, 7, 0, 9, 9, 0, 11, 11, 0, 13, 13, 0, 15, 15, 0, 17, 17, 0, 19, 19, 0, 21, 21, 0, 23, 23, 0, 25, 25, 0, 27, 27, 0, 29, 29, 0, 31, 31, 0, 33, 33, 0, 35, 35, 0, 37, 37, 0, 39, 39, 0, 41, 41, 0, 43, 43, 0, 45, 45, 0, 47, 47, 0, 49, 49, 0, 51, 51, 0, 53, 53, 0
COMMENTS
If n is a multiple of 3, then a(n) = 0, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 2*k + 1. - Antti Karttunen, Apr 14 2022
REFERENCES
Maria Paola Bonacina and Nachum Dershowitz, Canonical Inference for Implicational Systems, in Automated Reasoning, Lecture Notes in Computer Science, Volume 5195/2008, Springer-Verlag.
FORMULA
G.f.: x*(1+x+x^3+x^4)/(1-2*x^3+x^6).
a(n) = (2/3)*floor((2*n+1)/3)*(1-cos(2*Pi*n/3)).
a(n) = ([n/3]*2 + 1)*dist(n,3Z). (End)
a(n) = 2*sin(n*Pi/3)*(4*n*sin(n*Pi/3)-sqrt(3)*cos(n*Pi))/9. - Wesley Ivan Hurt, Sep 24 2017 a(n) = 2*a(n-3) - a(n-6), for n > 5. - Chai Wah Wu, Jul 27 2022
MATHEMATICA
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 1, 1, 0, 3, 3}, 90] (* G. C. Greubel, Dec 09 2022 *)
PROG
(Magma) I:=[0, 1, 1, 0, 3, 3]; [n le 6 select I[n] else 2*Self(n-3) - Self(n-6): n in [1..91]]; // G. C. Greubel, Dec 09 2022 (SageMath)
def A102899(n): return (1+2*(n//3))*((n%3)>0)
Second terms of continued fractions for power towers e, e^e, e^e^e, ...
+10
4
COMMENTS
It was conjectured (but remains unproved) that none of the power towers e, e^e, e^e^e, ... are integers. If so, the corresponding continued fractions contain at least 2 terms. If the conjecture fails, let the corresponding a(n) = 0.
LINKS
Eric Weisstein's World of Mathematics, e.
EXAMPLE
a(3) = 9 because floor(1/frac(e^e^e)) = 9, since e^e^e ~ 3814279.10476.
MATHEMATICA
$MaxExtraPrecision = Infinity; terms = 4; Map[Function[x, ContinuedFraction[x, 2][[2]]], NestList[Exp, E, terms - 1]]
CROSSREFS
A056072 yields the first term of the continued fraction.
Left(0)/right(1) turning sequence needed to traverse the Stern-Brocot tree ( A007305, A047679) from the root down to e ( A001113).
+10
4
1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
COMMENTS
The sequence has the following regular pattern: 1 0{0} 1 0 1{2} 0 1 0{4} 1 0 1{6} 0 1 0{8} ... where {r} indicates that the preceding term is repeated r times.
Run lengths of this sequence ( A003417) are the coefficients of the continued fraction for e.
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd Edition, Addison-Wesley, 1989, p. 115-123.
LINKS
Michael De Vlieger, Binary tree of a(n), n = 1..2^14-1, where dark blue represents 0 and red represents 1.
FORMULA
The sequence begins with
floor(e / 1) = A003417(1) ones, followed by floor(1 / (e mod 1)) = A003417(2) zeros, followed by floor((e mod 1) / (1 mod (e mod 1))) = A003417(3) ones, followed by floor((1 mod (e mod 1)) / ((e mod 1) mod (1 mod (e mod 1)))) = A003417(4) zeros ...
Limit_{n->oo} (1/n)*Sum_{i=1..n} a(i) = 1/2. (End)
EXAMPLE
In the initial portion of the Stern-Brocot tree shown below, the arrows indicate the traversing route.
1/1
|
.----------------->-----. Right (1)
| |
1/2 2/1
| |
.-----------. .-------->--. Right (1)
| | | |
1/3 2/3 3/2 3/1
| | | |
.-----. .-----. .-----. .-<---. Left (0)
| | | | | | | |
1/4 2/5 3/5 3/4 4/3 5/3 5/2 4/1
...
The first terms of the sequence are therefore 1, 1, 0.
MATHEMATICA
(* Generate up to 2^22 terms of this sequence from the 2^11 X 2^11 bitmap *)
With[{rows = 12}, ImageData[Import["https://oeis.org/ A342991/a342991.png"]][[1 ;; rows]] /. {0. -> 1, 1. -> 0} // Flatten] (* Michael De Vlieger, Nov 04 2022 *) A342991[i_]:=Flatten[Array[{1, PadRight[{}, 4#], 1, 0, PadRight[{}, 2+4#, 1], 0}&, i, 0]]; (* Each iteration adds six runs of values *)
PROG
(Python)
from itertools import count, islice
def A342991_gen(): # generator of terms a = 0
yield from (1, 1)
for n in count(2):
q, r = divmod(n, 3)
yield from (a, )*(1 if r else q<<1)
a = 1-a
(PARI)
A342991(iter) = concat(vector(iter, i, concat([1, vector((i-1)<<2), 1, 0, vector(2+(i-1)<<2, x, 1), 0]))); \\ Each iteration adds six runs of values
(PARI) a(n) = my(r, s=sqrtint(n-1, &r)); bitand(s + (r<s-1 || r==s), 1); \\ Kevin Ryde, Nov 24 2022
Let z(n) be e = exp(1.0) = 2.7182.... truncated to n decimal digits after the decimal point; sequence gives maximum element in the continued fraction for z(n).
+10
3
2, 3, 4, 12, 9, 10, 12, 11, 9, 10, 8, 22, 13, 13, 15, 12, 35, 30, 48, 18, 166, 166, 68, 40, 73, 137, 57, 1288, 62, 28, 416, 552, 138, 47, 24, 156, 110, 31, 463, 85, 108, 106, 295, 295, 54, 98, 40, 388, 216, 32, 49, 199, 488, 47, 64, 822, 51, 152, 854, 38, 701, 88, 94, 149
EXAMPLE
... Here is Maple's computation of the first four terms of the sequence a:
....C2 := 2
....cf := [2]
....a := [2]
..........27
....C2 := --
..........10
....cf := [2, 1, 2, 3]
....a := [2, 3]
..........271
....C2 := ---
..........100
....cf := [2, 1, 2, 2, 4, 3]
....a := [2, 3, 4]
..........1359
....C2 := ----
..........500
....cf := [2, 1, 2, 1, 1, 4, 1, 12]
....a := [2, 3, 4, 12]
MAPLE
with(numtheory); Digits:=200:
C1 := exp(1.0);
for n from 1 to 100 do
C2:= floor(C1*10^(n-1))/10^(n-1);
cf := convert(evalf(C2), confrac):
a := [op(a), max(cf)];
od:
MATHEMATICA
A081837[n_] := Max[ContinuedFraction[Floor[E*10^n]/10^n]];
First differences of coefficients in the continued fraction for e.
+10
3
2, -1, 1, -1, 0, 3, -3, 0, 5, -5, 0, 7, -7, 0, 9, -9, 0, 11, -11, 0, 13, -13, 0, 15, -15, 0, 17, -17, 0, 19, -19, 0, 21, -21, 0, 23, -23, 0, 25, -25, 0, 27, -27, 0, 29, -29, 0, 31, -31, 0, 33, -33, 0, 35, -35, 0, 37, -37, 0, 39, -39
FORMULA
G.f.: (1-x)*(2+x+2*x^2-3*x^3-x^4+x^6)/(1-2*x^3+x^6).
a(n) = 2*C(0,n) -C(1,n) +2*sin(2*Pi*(n-1)/3)*floor((2*n-1)/3)/sqrt(3). [Sign corrected by M. F. Hasler, May 01 2013] a(0)=2, a(1)=-1, for n>0: a(3*n-1) = 2*n-1, a(3*n) = 1-2*n, a(3*n+1) = 0. - M. F. Hasler, May 01 2013 a(n) = - a(n-1) - a(n-2) + a(n-3) + a(n-4) + a(n-5) for n > 6. - Chai Wah Wu, Jul 27 2022 a(n) = 0 if n mod 3 = 1, a(n) = (2*n-1)/3 if n mod 3 = 2, a(n) = (3-2*n)/3 otherwise, with a(0) = 2, and a(1) = -1. - G. C. Greubel, Dec 28 2022
MATHEMATICA
Join[{2}, Differences[ContinuedFraction[E, 120]]] (* or *) LinearRecurrence[{-1, -1, 1, 1, 1}, {2, -1, 1, -1, 0, 3, -3}, 120] (* Harvey P. Dale, Jun 08 2016 *)
PROG
(PARI) A120691(n)={n<2 && return(2-3*n); n=divrem(n-1, 3); if(n[2], -(1+n[1]*2)*(-1)^n[2])} \\ - M. F. Hasler, May 01 2013 (Magma) R<x>:=PowerSeriesRing(Integers(), 70); Coefficients(R!( (1-x)*(2+x+2*x^2-3*x^3-x^4+x^6)/(1-x^3)^2 )); // G. C. Greubel, Dec 28 2022 (SageMath)
def b(n):
if (n%3==1): return 0
elif (n%3==2): return (2*n-1)/3
else: return (3-2*n)/3
def A120691(n): return b(n) + (-1)^n*int(n<2)
Continued fraction for Pi + e.
+10
2
5, 1, 6, 7, 3, 21, 2, 1, 2, 2, 1, 1, 2, 3, 3, 2, 5, 2, 1, 1, 1, 1, 3, 1, 8, 4, 4, 1, 1, 1, 1, 8, 1, 4, 1, 5, 1, 1, 1, 2, 4, 3, 2, 1, 1, 2, 1, 10, 1, 4, 1, 2, 1, 12, 1, 8, 2, 7, 39, 365, 2, 15, 2, 25, 1, 2, 5, 3, 3, 9, 3, 1, 1, 9, 1, 1, 47, 1, 1, 18, 1, 1, 2, 6, 1, 1, 1, 4, 1, 3, 1, 1, 1, 1, 4, 1, 6, 37
COMMENTS
The question of the transcendence of the number Pi + e is still open.
EXAMPLE
a(1) = 5 because Pi + e = 5.859874482048838473822930854632165381954416493075065395941912220031...
5.859874482048838473822930854... = 5 + 1/(1 + 1/(6 + 1/(7 + 1/(3 + ...)))). - Harry J. Smith, May 31 2009
PROG
(PARI) \p 500; contfrac(Pi+exp(1))
(PARI) { allocatemem(932245000); default(realprecision, 21000); x=contfrac(Pi+exp(1)); for (n=1, 20000, write("b058651.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 31 2009
AUTHOR
Avi Peretz (njk(AT)netvision.net.il), Dec 26 2000
Continued fraction for e/5.
+10
2
0, 1, 1, 5, 4, 2, 2, 2, 2, 2, 1, 1, 9, 1, 1, 3, 3, 2, 3, 3, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 3, 9, 1, 3, 3, 3, 4, 3, 3, 4, 1, 2, 2, 1, 4, 1, 2, 2, 1, 5, 9, 1, 5, 3, 3, 6, 3, 3, 6, 1, 2, 2, 1, 6, 1, 2, 2, 1, 7, 9, 1, 7, 3, 3, 8, 3, 3, 8, 1, 2, 2, 1, 8, 1, 2, 2, 1, 9, 9, 1, 9, 3, 3, 10, 3, 3, 10, 1, 2, 2, 1, 10, 1, 2
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,-1).
FORMULA
First 18 terms: 0, 1, 1, 5, 4, 2, 2, 2, 2, 2, 1, 1, 9, 1, 1, 3, 3, 2.
For k >= 1, a(19k)=a(19k+1)=a(19k+16)=a(19k+17)=3; a(19k+2)=a(19k+7)=2k; a(19k+3)=a(19k+6)=a(19k+8)=a(19k+11)=a(19k+14)=1; a(19k+4)=a(19k+5)=a(19k+9)= a(19k+10)=2; a(19k+12)=a(19k+15)=2k+1; a(19k+18)=2k+2.
MATHEMATICA
ContinuedFraction[E/5, 100] (* Paolo Xausa, Sep 21 2024 *)
Coefficients of series giving the best rational approximations to e.
+10
2
7, 497, 71071, 18107089, 7216769351, 4145592145057, 3243346361740927, 3315690551047089761, 4291382388990897826759, 6858633609184481948847121, 13266034908146716343647359647, 30540929340877940990799507474097
COMMENTS
The series giving the best rational approximations to e is e = 3 - 2/a(1) + 2/a(2) - 2/a(3) + ... The continued fraction for e is [2;1,2,1,1,4,1,1,6, 1,1,8...] and the above best approximations give every third convergent, the convergents deriving from [2;1], [2;1,2,1,1], [2;1,2,1,1,4,1, 1] and so forth.
FORMULA
a(n+3) = (16*n^2 +96*n +141)*a(n+2) + (2*n+7)*(16*n^2 +64*n +61)/(2*n+3) * a(n+1) - (2*n+7)/(2*n+3) * a(n). This recurrence relationship is identical to A122533, for the best approximations to 1/e.
MATHEMATICA
RecurrenceTable[{a[n]== ((2*n-3)*(16*n^2 -3)*a[n-1] +(2*n+1)*(16*(n-1)^2 - 3)*a[n-2] -(2*n+1)*a[n-3])/(2*n-3), a[1]==7, a[2]==497, a[3]==71071}, a, {n, 30}] (* G. C. Greubel, Oct 27 2024 *)
PROG
(Magma) I:=[7, 497, 71071]; [n le 3 select I[n] else ((2*n-3)*(16*n^2 -3) *Self(n-1) +(2*n+1)*(16*(n-1)^2 -3)*Self(n-2) -(2*n+1)*Self(n-3))/(2*n-3): n in [1..30]]; // G. C. Greubel, Oct 27 2024 (SageMath)
@CachedFunction
if n<4: return (0, 7, 497, 71071)[n]
else: return ((2*n-3)*(16*n^2 -3)*a(n-1) +(2*n+1)*(16*(n-1)^2 -3)*a(n-2) -(2*n+1)*a(n-3))/(2*n-3)
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