A290977 First n-digit number to appear twice in a row in the decimal expansion of Pi.

3, 59, 209, 9314, 64015, 886287, 7348278, 85105027

Offset: 1

Author

Bobby Jacobs, Aug 16 2017

Keywords

• base
• more
• nonn

Comments

209209 and 305305 appear in Pi before any 2-digit number appears twice in a row.

a(n) (n >= 1) begins at the following decimal places: 24, 413, 326, 8239, 107472, 1632152, 9719518. - Robert G. Wilson v, Aug 23 2017

Examples

			a(1) = 3 because 3 is the first 1-digit number to appear twice in a row in the decimal expansion of Pi = 3.14159265358979323846264(33)...
		

Links

• David G. Andersen, The Pi-Search Page.

Crossrefs

Cf. A000796, A287994, A290984.

Cf. A050279, A096755, A096756, A096757, A096758, A096759, A096760, A096761, A096762, A096763.

Programs

Mathematica

With[{s = Rest@ First@ RealDigits[N[Pi, 10^4]]}, Keys@ Merge[#, Identity] &@ Table[If[Length@ # > 0, TakeSmallest[#, 1], 0 -> 0] &@ Sort[Map[#[[1, 1]] &, DeleteCases[#, {}]]] &@ Map[SequenceCases[#, {a_, b_} /; b == a + n] &, KeyMap[FromDigits, PositionIndex@ Partition[s, n, 1]]], {n, 4}]] (* Michael De Vlieger, Aug 16 2017 *)
pi = StringDrop[ ToString[ N[Pi, 1632200]], 2]; f[n_] := Block[{k = 1}, While[ StringTake[pi, {k, k +n -1}] != StringTake[pi, {k +n, k +2n -1}], k++]; k]; Array[f, 6] (* Robert G. Wilson v, Aug 17 2017 *)

PARI

eva(n) = subst(Pol(n), x, 10)
pistring(n) = default(realprecision, n+10); my(x=Pi); floor(x*10^n)
pidigit(n) = pistring(n)-10*pistring(n-1)
consecpidigits(pos, len) = my(v=vector(len)); for(k=1, len, v[k]=pidigit(pos+k)); v
a(n) = my(v=[], w=[], x=1); while(1, v=consecpidigits(x, n); w=consecpidigits(x+n, n); if(v==w, return(eva(v))); x++) \\ Felix Fröhlich, Aug 16 2017

Python

from sympy import S
# download https://stuff.mit.edu/afs/sipb/contrib/pi/pi-billion.txt, then
# with open('pi-billion.txt', 'r') as f: pi_digits = f.readline()
pi_digits = str(S.Pi.n(3*10**5+2))[:-2] # alternative to above
pi_digits = pi_digits.replace(".", "")
def a(n):
    for k in range(1, len(pi_digits)-n):
        s = pi_digits[k:k+2*n]
        if s[0] != 0 and s[:len(s)//2] == s[len(s)//2:]:
            return int(s[:len(s)//2])
print([a(n) for n in range(1, 6)]) # Michael S. Branicky, Jan 10 2022

Extensions

a(6) from Robert G. Wilson v, Aug 19 2017

a(7) from Bobby Jacobs, Aug 22 2017

a(8) from Michael S. Branicky, Jan 10 2022