%I #24 Dec 07 2021 07:24:07
%S 1,1,10,213,133130,50044104412,1456053604226211530303,
%T 1355606752437672176235012441560305430335663,
%U 211028537470000781652623227715306164580285678106041347266088244412145807188883237767
%N Base conversion sequence: a(1) = 1; a(n) is the concatenation of all the previous terms, evaluated in base n-1, written in base n.
%C This will only work for n <= 10. To get a sequence that is defined for all n, it will be necessary to replace a(n) by a list of its "digits". So the result will be a triangle: 1 / 1 / 1,0 / 2,1,3 / ..., in which row n is a list of the digits written in base n. This should be an additional sequence with a cross-reference to this one. - _N. J. A. Sloane_, Sep 21 2019
%C See A349918 for the corresponding triangle. - _Rémy Sigrist_, Dec 05 2021
%H Rémy Sigrist, <a href="/A309737/b309737.txt">Table of n, a(n) for n = 1..10</a>
%H Rémy Sigrist, <a href="/A309737/a309737.gp.txt">PARI program for A309737</a>
%F a(1) = 1; a(n) is the concatenation of all the previous terms, evaluated in base n-1, written in base n.
%e For a(3) the previous terms are {1,1}. Evaluating the concatenation of those terms in base n-1 = 2 gives 11_2 = 3; converting that to base n = 3 gives 10_3, so a(3) = 10.
%e n=4: 1110_3 = 39_10 = 213_4, so a(4) = 213.
%o (PARI) See Links section.
%o (Python)
%o from sympy.ntheory.digits import digits
%o def fromdigits(d, b):
%o n = 0
%o for di in d: n *= b; n += di
%o return n
%o def afull():
%o alst, diglst = [1], [1]
%o for n in range(2, 11):
%o andigs = digits(fromdigits(diglst, n-1), n)[1:]
%o alst.append(int("".join(map(str, andigs))))
%o diglst.extend(andigs)
%o return alst
%o print(afull()) # _Michael S. Branicky_, Dec 05 2021
%Y Cf. A349918.
%K nonn,base,full,fini
%O 1,3
%A _Moshe Levy_, Aug 14 2019
%E More terms from _Rémy Sigrist_, Dec 05 2021