%I #49 Jan 20 2022 00:34:20
%S 1,2,3,4,5,10,11,12,13,14,15,20,21,22,23,24,25,30,31,32,33,34,35,40,
%T 41,42,43,44,45,100,101,102,103,104,105,110,111,112,113,114,115,120,
%U 121,122,123,124,125,130,131,132,133,134,135,140,141,142,143,144,145
%N a(n) is built digit-by-digit (see comments for details).
%C a(n) is built digit-by-digit as a_i ... a_3 a_2 a_1.
%C Note that in this case, the definition of "digit" is a nonnegative integer. If i > 3, the number of digits of a_i may be greater than 1.
%C Successively, we have:
%C a_1 = n mod 6;
%C a_2 = ((n - a_1)/primorial(2)) mod prime(2+1);
%C a_3 = ((n - a_1 - a_2*primorial(2))/primorial(3)) mod prime(3+1);
%C ...
%C a_i = ((n - a_1 - a_2*primorial(2)-...-a_(i-1)*primorial(i-1))/primorial(i)) mod prime(i+1).
%C So that finally, n = a_1 + a_2*primorial(2) + ... + a_i*primorial(i).
%H Lear Young, <a href="/A241213/b241213.txt">Table of n, a(n) for n = 1..100000</a>
%e a(2287) = 10611.
%e 10611 is built digit-by-digit as a_4 a_3 a_2 a_1 = 10 6 1 1.
%e And a_1 + a_2*primorial(2) + a_3*primorial(3) + a_4*primorial(4) = 1 + 1*6 + 6*30 + 10*210 = 2287.
%e (The definition of "digit" is a nonnegative integer. See comments for how to get a_1, a_2, a_3, a_4.)
%o (Sage)
%o Pr = Primes()
%o c = oeis(2110)[:10]
%o def bjz(a):
%o d = len(str(a)) + 1
%o b = [0] * (d)
%o b[0] = a % 6
%o s = 0
%o for x in range(1, d):
%o if x > 1:
%o s += c[x] * b[x-1]
%o b[x] = ((a - b[0] - s) / c[x+1] ) % Pr.unrank(x+1)
%o return int(''.join(map(str, b[::-1])))
%o [ bjz(x) for x in range(1, 101)] # _Lear Young_, Apr 17 2014
%K nonn,base
%O 1,2
%A _Lear Young_, Apr 17 2014