This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Do[ If[ n^2 == Ceiling[n*3.1415926535897932346264*Floor[n/3.1415926535897932346264]], Print[n]], {n, 0, 10^8}] (* Robert G. Wilson v, Feb 26 2004 *)
for(x=0,2000000,if(x^2==ceil(Pi*x*floor(x/Pi)),print1(x,",")))
The convergents are 3/1, 6/2, 22/7, 44/14, 66/21, 179/57, 355/113, 710/226, 1065/339, 1775/565, 3550/1130, 6745/2147, 13135/4181, 25915/8249, 102928/32763, ... = A325158/A325159.
is_A332095(n)={tan(n)*n < 1 && n*tan(n) >= 0} for(n=0,oo, n*abs(tan(n))<1 && print1(sign(tan(n))*n", ")) /* Much faster: apply to numerators of convergents of Pi the function check(n) which prints all nonzero k*n in the sequence and returns the largest such k, largest in magnitude, possibly negative. N.B.: stops when (k+1)n is not in the sequence, so e.g., n = 11 (in convergents of Pi/2) does not give 22 and 44! */ print1(0); apply( {check(n)=for(i=1,oo,abs(i*n*tan(i*n))<1||return(sign(tan(n))*(i-1)); print1(", "sign(tan(n*i))*i*n))}, contfracpnqn(c=contfrac(Pi),#c)[1,]) \\ M. F. Hasler, Oct 09 2020
The rational numbers are 3, 22/7, 22/7, 201/64, 333/106, 355/113, 355/113, 75948/24175, ...
Table[Numerator[Rationalize[Pi,10^(-n)]],{n,0,27}]
The first few values of abs((cos n)^n), n >= 1, are: abs(cos(1)^1) = 0.5403023058 abs(cos(2)^2) = 0.1731781895 abs(cos(3)^3) = 0.9702769379 abs(cos(4)^4) = 0.1825425480 abs(cos(5)^5) = 0.0018365688 and the record high points are at n = 1, 3, 22, ...
Module[{x, y, runningMax = 0, positions = {}},
x = Range[10^6]; y = Abs[Cos[x]^x];
Do[If[y[[i]] > runningMax, runningMax = y[[i]]; AppendTo[positions, i]; ], {i, Length[y]}];
positions
]
import numpy as np x = np.arange(1, 1+10**8) y = abs(np.cos(x) ** x) A382564 = sorted([1+int(np.where(y==m)[0][0]) for m in set(np.maximum.accumulate(y))])
from mpmath import mp mp.dps = 1 running_max, A382564 = 0, [] for n in range(1, 1+10**5): while ((y:=abs(mp.cos(n)**n)) == 1): mp.dps += 1 if y > running_max: running_max = y A382564.append(n)
The first 4 rational approximations to Pi are 3/1, 22/7, 333/106, 355/113, 103993/33102 where 3 and 103993 are primes.
Select[Numerator[Convergents[Pi,100]],PrimeQ] (* Harvey P. Dale, Dec 23 2018 *)
\\ Continued fraction rational approximation of numeric functions
cfrac(m,f) = x=f; for(n=0,m,i=floor(x); x=1/(x-i); print1(i,","))
cfracnumprime(m,f) = { cf = vector(100000); x=f; for(n=0,m, i=floor(x); x=1/(x-i); cf[n+1] = i; ); for(m1=0,m, r=cf[m1+1]; forstep(n=m1,1,-1, r = 1/r; r+=cf[n]; ); numer=numerator(r); denom=denominator(r); if(isprime(numer),print1(numer,",")); ) }
default(realprecision,10^5);
cf=contfrac(Pi);
n=0;
{ for(k=1, #cf, \\ generate b-file
pq = contfracpnqn( vector(k,j, cf[j]) );
p = pq[1,1]; q = pq[2,1];
if ( ispseudoprime(p), n+=1; print(n," ",p) ); \\ A086785
\\ if ( ispseudoprime(q), n+=1; print(n," ",q) ); \\ A086788
); }
/* Joerg Arndt, Apr 21 2013 */
a(2) = 106 since A002485(4)/A002486(4) = 333/106, A002485(2)/A002486(2) = 3/1, and 333/106 - 3/1 = 15/106 (see table below). Pi = 3/1 + 15/106 + 73/877203 + 1/2195225334 + 2/17599271777 + 3/360950005720 + 7/17348726394920 + .... . n Q(n) = A002485(2n)/A002486(2n) Q(n) - Q(n-1) a(n) - ------------------------------ ------------- ------ 0 0/1 = 0 - - 1 3/1 = 3 3/1 1 2 333/106 = 3.1415094339... 15/106 106 3 103993/33102 = 3.1415926530... 73/877203 877203
3+1/a(0)=22/7 3+1/a(0)+1/a(1)=333/106 3+1/a(0)+1/a(1)+1/a(2)=355/113
c0=3; for (k=2,30,m=contfracpnqn(contfrac(Pi,k));c1=m[1,1]/m[2,1];print1(1/(c1-c0),", ");c0=c1;)
Select[Range[3200], Abs[Sin[#]] < 1/# &] (* Amiram Eldar, Aug 25 2020 *)
print1(1);apply( n-> forstep(n=n,oo,n,abs(sin(n))<1/n||return; print1(","n)), contfracpnqn(c=contfrac(Pi),#c)[1,]); \\ M. F. Hasler, Oct 09 2020
import numpy as np
for x in range(1, 10**9):
if np.abs(np.sin(x)) < 1/x:
print(x, end=", ")
The first few values of (cos k)^k, k >= 1, are: cos(1)^1 = 0.540302305868139 cos(2)^2 = 0.173178189568194 cos(3)^3 = -0.97027693792150 cos(4)^4 = 0.182542548055270 cos(5)^5 = 0.001836568887601 cos(6)^6 = 0.783591241730686 cos(7)^7 = 0.138422055397017 cos(8)^8 = 0.000000200865224 cos(9)^9 = -0.43273721139612 and the record high points are at k = 1, 6, 19, ...
Module[{x, y, runningMax = 0, positions = {}},
x = Range[1, 10^6]; y = Cos[x]^x;
Do[If[y[[i]] > runningMax, runningMax = y[[i]]; AppendTo[positions, i]; ], {i, Length[y]}];
positions
]
import numpy as np x = np.arange(1, 1+10**8) y = np.cos(x) ** x A383541 = sorted([1+int(np.where(y==m)[0][0]) for m in set(np.maximum.accumulate(y))])
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