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

1,2

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).

Links

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

Example

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.)

Prog

(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

Crossrefs

Sequence in context: A266117 A037473 A007092 * A362931 A047596 A199502

Adjacent sequences: A241210 A241211 A241212 * A241214 A241215 A241216

Keyword

nonn,base

Author

Lear Young, Apr 17 2014

Status

approved