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A306567
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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).
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2
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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
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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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.
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PROG
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(PARI) \\ See Links section.
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CROSSREFS
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See A306569 for the multiplicative variant.
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KEYWORD
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AUTHOR
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STATUS
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approved
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