%I
%S 3,1,2,14,4,15,5,6,7,9,8,10,26,11,12,50,13,23,16,17,25,18,19,20,80,21,
%T 22,24,29,27,28,30,37,90,31,32,33,34,35,200,36,43,84,60,201,38,61,39,
%U 40,41,42,430,53,48,44,320,45,46,79,47,49,51,52
%N a(n) is the lexically first sequence of distinct nonzero integers such that if S(n) is the string formed from the digits of a(1)a(2)...a(n), then dividing S(n) into substrings with lengths equal to the successive digits of S(n) (treating 0 as 10) results in substrings beginning with the successive digits of Pi (A000796).
%C Erase the punctuation:
%C S(Pi) = 312144155679810261112501323161725181920802122242927283037903132333435...
%C Divide into chunks  the size of each chunk is given by the successive DIGITS of S(Pi):
%C 312.1.44.1.5567.9810.2.61112.50132.316172.5181920.802122242.92728303.7.9031323334.35
%C (the "0" digits produce a 10digit chunk)
%C Replace all dots (.) with carriage returns:
%C 312
%C 1
%C 44
%C 1
%C 5567
%C 9810
%C 2
%C 61112
%C 50132
%C 316172
%C 5181920
%C 802122242
%C 92728303
%C 7
%C 9031323334
%C 35
%C ...
%C The first column shows Pi!
%C a(63) = 52 is the last term, a(64) would have to begin with a 0.  _Charlie Neder_, Jun 24 2018
%H Eric Angelini, <a href="http://www.cetteadressecomportecinquantesignes.com/PiFou.htm">Un nombre de ouf!</a>
%H E. Angelini, <a href="/A171059/a171059.pdf">Un nombre de ouf!</a> [Cached copy, with permission]
%K nonn,base,fini,full
%O 1,1
%A _N. J. A. Sloane_, Sep 04 2010, based on a posting to the Sequence Fans Mailing List by _Eric Angelini_, Aug 24 2010
%E a(40)a(63) from _Charlie Neder_, Jun 24 2018
