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.
a(1)=1 because the initial digit "3" of Pi is prime. a(2)=a(6)=a(38)=1 because the first 2, 6, and 38 digits of Pi (including the initial 3) also form the primes 31, 314159 and 31415926535897932384626433832795028841, cf. A005042 and A060421.
With[{pd=RealDigits[Pi,10,1000][[1]]},Table[Position[Partition[pd,n,1],?(PrimeQ[FromDigits[#]]&&#[[1]]!=0&),{1},1,Heads->False],{n,60}]]// Flatten (* _Harvey P. Dale, Apr 25 2016 *)
A198344(n)=for(c=0,9e9,ispseudoprime(Pi\.1^(n+c-1)%10^n)&Pi\.1^c%10&return(c+1)) /* Replace upper limit 9e9 by "default(realprecision)-n" to avoid an error message and return 0 in case no n-digit prime is found */
# uses code in A104841 print([A104841_A198344(n)[1] for n in range(1, 65)]) # Michael S. Branicky, Dec 28 2022
This sequence starts 2,4,0,0,... since the 1st and 2nd terms of the continued fraction expansion of Pi, A001203 = (3, 7, 15, 1, ...) are the 2nd resp. 4th primes, while the next two terms are not primes.
default(realprecision,1000); t=contfrac(Pi); vector(#t,i,isprime(t[i])*primepi(t[i]))
The first digit is 3, which is prime, so a(1) = 3. The second digit is 1, which is no prime, but 31 is prime, so a(2) = 31. The third digit is 4, which does not end any prime. The fourth digit is 1, not prime, but 41 is prime, so a(3) = 41.
v=[3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 3, 8, 3] for(n=1,#v,x=0;p=1;forstep(k=n,1,-1,x+=p*v[k];p*=10;if(v[k]&&isprime(x),print1(x", "))))
The first a(3)=10 binary digits of Pi are 1101110001_2 which is prime 881_10.
p = First[ RealDigits[ Pi, 10, 10^5]]; p = p[[ Select[ Range[10^5], p[[ # ]] == 0 || p[[ # ]] == 1 & ]]]; Do[ If[ PrimeQ[ FromDigits[ Take[p, n], 2]], Print[n]], {n, 1, Length[p] } ]
a(5) = A011546(1902) = 314159...066118631 is a prime with 1902 digits. a(6) = A011546(3971) = 314159...411010447. a(7) = A011546(5827) = 314159...690496521. a(8) = A011546(16208) = A005042(5) = 314159....943936307. For n<=17511, there are eight primes in sequence A011546(n). In addition, because of round(Pi*10^47576) = floor(Pi*10^47576), and A011546(47577)(=A005042(6)) is a prime. Thus, A011546(47577) will appear in here. A011546(613373)(=A005042(8)) as well. But A011546(78073)(=A005042(7)+1) is not prime.
Do[If[PrimeQ[Round[Pi*10^(n-1)]],Print[Round[Pi*10^(n-1)]]],{n,17511}]
Select[Module[{nn=20,pid},pid=RealDigits[Pi,10,nn+2][[1]];Table[Floor[(FromDigits[ Take[ pid,n+1]])/10+1/2],{n,nn}]],PrimeQ] (* Harvey P. Dale, Jan 01 2023 *)
Do[If[PrimeQ[Round[Pi*10^(n-1)]],Print[n]],{n,17511}]
default(realprecision, 10^5); x=Pi; is(k) = ispseudoprime(round(x*10^k--)); \\ Jinyuan Wang, Mar 27 2020
a(4) = 1592653, because starting at the 4th digit in the expansion, the smallest substring of the digits of Pi forming a prime number is 3.14|1592653|589...
N:= 1000: # to use up to N+1 digits of pi.
nmax:= 30: # to get up to a(nmax), if possible.
S:= floor(10^N*Pi):
L:= ListTools:-Reverse(convert(S,base,10)):
for n from 1 to nmax do
p:= L[n];
for k1 from n+1 to N+1 do
p:= 10*p + L[k1];
if isprime(p) then break fi
od:
if k1 > N+1 then
A[n]:= "Ran out of digits";
break
else
A[n]:= p
end
od:
seq(A[i],i=1..n-1); # Robert Israel, Aug 27 2014
from sympy.mpmath import * from sympy import isprime def A245571(n): mp.dps = 1000+n s = nstr(pi,mp.dps)[:-1].replace('.','')[n-1:] for i in range(len(s)-1): p = int(s[:i+2]) if p > 10 and isprime(p): return p else: return 'Ran out of digits' # Chai Wah Wu, Sep 16 2014, corrected Chai Wah Wu, Sep 24 2014
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