

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).


2



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
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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)/91] 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(j1) + 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^k1)/9,
 a(2 * 10^k) = 99 * 10^k + 2 * (10^k1)/9,
 a(3 * 10^k) = 27 * 10^k + 3 * (10^k1)/9,
 a(4 * 10^k) = 99 * 10^k + 4 * (10^k1)/9,
 a(5 * 10^k) = 96 * 10^k + 5 * (10^k1)/9,
 a(6 * 10^k) = 99 * 10^k + 6 * (10^k1)/9,
 a(7 * 10^k) = 63 * 10^k + 7 * (10^k1)/9,
 a(8 * 10^k) = 99 * 10^k + 8 * (10^k1)/9,
 a(9 * 10^k) = 81 * 10^k + 9 * (10^k1)/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



