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A297560
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Numbers with nonzero Conway base-13 function value.
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1
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1715, 1728, 1741, 1754, 1767, 1780, 1793, 1806, 1819, 1847, 1848, 1849, 1850, 1851, 1852, 1853, 1854, 1855, 1884, 1897, 1910, 1923, 1936, 1949, 1962, 1975, 1988, 2016, 2017, 2018, 2019, 2020, 2021, 2022, 2023, 2024
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OFFSET
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1,1
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COMMENTS
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All terms below 13^k are of two forms: (a) 10*13^k + Sum_{i=0..k-1} d_i*13^i or 11*13^k + Sum_{i=0..k-1} d_i*13^i, where exactly one of d_0, d_1, ..., d_(k-1) is equal to 12, and the other k - 2 digits are from 0 to 9 (but not all 0); (b) d*13^k + a, where 0 <= d <= 12, and a is any term below 13^(k-1). - Jianing Song, Jun 28 2018
Let b(k) be the number of terms below 13^k, then by the comment above, b(k) = 13*b(k-1) + 2*(k-1)*(10^(k-2)-1), where b(1) = 0. So we have b(k) = (15*13^k - (48*k+160)*10^(k-1) + 12*k + 1)/72, which means that there are (5/24)*N + o(N) terms in this sequence that are no greater than N. - Jianing Song, Jun 04 2019
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LINKS
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FORMULA
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a(i*b(k)+j) = i*13^n + a(j), where b(k) is the number of terms below 13^n (see comment), 0 <= i <= 9, 1 <= j <= b(k).
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EXAMPLE
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1715 in decimal equals A1C in base 13 and has Conway base-13 function value 1 by definition, so 1715 is a term.
2024 in decimal equals BC9 in base 13 and has Conway base-13 function value -0.9, so 2024 is a term.
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PROG
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(PARI) Conway(n) = my(v=digits(n, 13), d=#v, s=0, p=0); forstep(i=d, 1, -1, if(v[i]==10 || v[i]==11, s=i; break())); forstep(i=d, 1, -1, if(v[i]==12, p=i; break())); [s, p]; if(p>s && s && !sum(i=s+1, p-1, v[i]==12), (sum(i=s+1, p-1, v[i]*10^(p-1-i)) + sum(i=p+1, d, v[i]*10^(p-i)))*if(v[s]==10, 1, -1), 0)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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