A197123 - OEIS

%I #54 Sep 24 2024 09:26:09

%S 1,26,592,582,60943,949129,8530614,52637962,201890888,4392366484,

%T 89879780761,756130190263,3186120489507,18220874234996,

%U 276854551127715,8230687217052243,93415455347042966,13724950651727463,1350168131352524443,84756845106452435773,585270898631522188621,2761994111668451704865,64722721994615606186022

%N a(n) is the first n-digit substring to repeat in the decimal expansion of Pi.

%C a(4) is written in the sequence as a 3-digit number 582 because the repeating substring is the 4-digit number 0582.

%C a(18) should also have a leading zero: 013724950651727463. This value starts at digit 378,355,223 and at digit 1,982,424,643. This computation was performed by Richard Tobin. - _Clive Tooth_, Mar 06 2012

%H Dave Andersen, <a href="http://www.angio.net/pi/piquery">The Pi-Search Page.</a>

%H David H. Bailey, <a href="http://crd.lbl.gov/~dhbailey/dhbpapers/pi.pdf">The computation of pi to 29,360,000 decimal digits using Borweins' quartically convergent algorithm</a>, Mathematics of Computation 50 (1988), pp. 283-296.

%H MIT Student Information Processing Board, <a href="https://stuff.mit.edu/afs/sipb/contrib/pi/">One billion digits of Pi</a>.

%H Jeff Sponaugle, <a href="https://sponaugle.com/wp/math_pi_repeat/">Calculating a(19)-a(22) in A197123</a>.

%e For n=2 the a(2)=26 solution is because if we look at all the 2-digit substrings 14,41,15,59,92,26,... of the decimal expansion of Pi=3.1415926535897932384626 we find that the first 2-digit substring to appear twice is 26.

%e From _Bobby Jacobs_, Dec 24 2016: (Start)

%e 1 appears at positions 1 and 3.

%e 26 appears at positions 6 and 21.

%e 592 appears at positions 4 and 61.

%e 0582 appears at positions 50 and 132.

%e 60943 appears at positions 397 and 551.

%e 949129 appears at positions 496 and 1296.

%e 8530614 appears at positions 4167 and 4601.

%e ... (End)

%o (Python)

%o # download https://stuff.mit.edu/afs/sipb/contrib/pi/pi-billion.txt, then

%o # with open('pi-billion.txt', 'r') as f: digits_of_pi = f.readline()

%o from sympy import S; digits_of_pi = str(S.Pi.n(3*10**5)) # alternatively

%o def a(n):

%o   global digits_of_pi

%o   seen = set()

%o   for i in range(2, len(digits_of_pi)-n):

%o     ss = digits_of_pi[i:i+n]

%o     if ss in seen: return int(ss)

%o     seen.add(ss)

%o for n in range(1, 11):

%o   print(a(n), end=", ") # _Michael S. Branicky_, Jan 26 2021

%Y Cf. A000796 (Pi), A159345 (the number of digits of Pi required to include the repeated string), A279860.

%K base,nonn,changed

%O 1,2

%A _Peter de Rivaz_, Oct 10 2011

%E a(16)-a(18) from _Clive Tooth_, Mar 06 2012

%E a(19)-a(22) from _Jeff Sponaugle_, Aug 22 2024

%E a(23) from _Jeff Sponaugle_, Sep 23 2024