A241213 a(n) is built digit-by-digit (see comments for details).

1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15, 20, 21, 22, 23, 24, 25, 30, 31, 32, 33, 34, 35, 40, 41, 42, 43, 44, 45, 100, 101, 102, 103, 104, 105, 110, 111, 112, 113, 114, 115, 120, 121, 122, 123, 124, 125, 130, 131, 132, 133, 134, 135, 140, 141, 142, 143, 144, 145

Offset: 1

Author

Lear Young, Apr 17 2014

Keywords

• nonn
• base

Comments

a(n) is built digit-by-digit as a_i ... a_3 a_2 a_1.

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.

Successively, we have:

a_1 = n mod 6;

a_2 = ((n - a_1)/primorial(2)) mod prime(2+1);

a_3 = ((n - a_1 - a_2*primorial(2))/primorial(3)) mod prime(3+1);

...

a_i = ((n - a_1 - a_2*primorial(2)-...-a_(i-1)*primorial(i-1))/primorial(i)) mod prime(i+1).

So that finally, n = a_1 + a_2*primorial(2) + ... + a_i*primorial(i).

Examples

			a(2287) = 10611.
10611 is built digit-by-digit as a_4 a_3 a_2 a_1 = 10 6 1 1.
And a_1 + a_2*primorial(2) + a_3*primorial(3) + a_4*primorial(4) = 1 + 1*6 + 6*30 + 10*210 = 2287.
(The definition of "digit" is a nonnegative integer. See comments for how to get a_1, a_2, a_3, a_4.)
		

Links

• Lear Young, Table of n, a(n) for n = 1..100000

Programs

Sage

Pr = Primes()
c = oeis(2110)[:10]
def bjz(a):
    d = len(str(a)) + 1
    b  = [0] * (d)
    b[0] = a % 6
    s = 0
    for x in range(1, d):
        if x > 1:
            s += c[x] * b[x-1]
        b[x] = ((a - b[0] - s) / c[x+1] ) % Pr.unrank(x+1)
    return int(''.join(map(str, b[::-1])))
[ bjz(x)  for x in range(1, 101)] # Lear Young, Apr 17 2014