A306567 a(n) is the largest value obtained by iterating x -> noz(x + n) starting from 0 (where noz(k) = A004719(k) omits the zeros from k).

9, 99, 27, 99, 96, 99, 63, 99, 81, 91, 99, 195, 94, 295, 93, 291, 113, 189, 171, 992, 159, 187, 187, 483, 988, 475, 153, 281, 181, 273, 279, 577, 297, 997, 567, 369, 333, 363, 351, 994, 219, 465, 357, 663, 459, 461, 423, 192, 441, 965, 399, 999, 437, 126, 551

Offset

1,1

Comments

For any n > 0, a(n) is well defined:

- the set of zeroless numbers (A052382) contains arbitrarily large gaps,

- for example, for any k > 0, the interval I_k = [10^k..(10^(k+1)-1)/9-1] if free of zeroless numbers,

- let i be such that #I_i > n,

- let b_n be defined by b_n(0) = 0, and for any j > 0, b_n(j) = noz(b_n(j-1) + n),

- as b_n starts below 10^i and cannot cross the gap constituted by I_i,

- b_n is bounded (and eventually periodic), QED.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Rémy Sigrist, PARI program for A306567

Formula

Empirically, for any k >= 0:

- a( 10^k) = 9 * 10^k + (10^k-1)/9,

- a(2 * 10^k) = 99 * 10^k + 2 * (10^k-1)/9,

- a(3 * 10^k) = 27 * 10^k + 3 * (10^k-1)/9,

- a(4 * 10^k) = 99 * 10^k + 4 * (10^k-1)/9,

- a(5 * 10^k) = 96 * 10^k + 5 * (10^k-1)/9,

- a(6 * 10^k) = 99 * 10^k + 6 * (10^k-1)/9,

- a(7 * 10^k) = 63 * 10^k + 7 * (10^k-1)/9,

- a(8 * 10^k) = 99 * 10^k + 8 * (10^k-1)/9,

- a(9 * 10^k) = 81 * 10^k + 9 * (10^k-1)/9.

Example

For n = 1:
- noz(0 + 1) = 1,
- noz(1 + 1) = 2,
- noz(2 + 1) = 3,
  ...
- noz(7 + 1) = 8,
- noz(8 + 1) = 9,
- noz(9 + 1) = noz(10) = 1,
- hence a(1) = 9.

Prog

See Links section.

Crossrefs

See A306569 for the multiplicative variant.

Cf. A004719, A052382.

Sequence in context: A066557 A289214 A121706 * A061817 A083835 A305670

Adjacent sequences: A306564 A306565 A306566 * A306568 A306569 A306570

Keyword

nonn,look,base

Author

Rémy Sigrist, Feb 24 2019

Status

approved