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A323288
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Largest number that can be obtained from the "Choix de Bruxelles", version 2 (A323460) operation applied to n.
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4
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2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 110, 112, 114, 116, 118, 40, 42, 44, 46, 48, 210, 212, 214, 216, 218, 60, 62, 64, 66, 68, 310, 312, 314, 316, 318, 80, 82, 84, 86, 88, 410, 412, 414, 416, 418, 100, 102, 104, 106, 108, 510, 512, 514, 516, 518
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OFFSET
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1,1
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COMMENTS
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Equally, this is the largest number that can be obtained from the "Choix de Bruxelles", version 1 (A323286) operation applied to n.
Maximal element in row n of irregular triangle in A323460 (or, equally, A323286).
Conjecture: If n contains no digit >= 5, then a(n) = 2*n; otherwise, a(n) is obtained from n by doubling the substring from the last digit >= 5 to the last digit. - Charlie Neder, Jan 19 2019. (This is true. - N. J. A. Sloane, Jan 22 2019)
Corollary: a(n)/n < 10 for all n, and a(n) = 10 - 1/k + O(1/k^2) for n = 10*k+5. - N. J. A. Sloane, Jan 23 2019
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LINKS
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FORMULA
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PROG
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(PARI) a(n, base=10) = { my (d=digits(n, base), v=2*n); for (w=1, #d, for (l=0, #d-w, if (d[l+1], my (h=d[1..l], m=fromdigits(d[l+1..l+w], base), t=d[l+w+1..#d]); v = max(v, fromdigits(concat([h, digits(m*2, base), t]), base))))); v } \\ Rémy Sigrist, Jan 15 2019
(Python)
def a(n):
s, out = str(n), {n}
for l in range(1, len(s)+1):
for i in range(len(s)+1-l):
if s[i] == '0': continue
t = int(s[i:i+l])
out.add(int(s[:i] + str(2*t) + s[i+l:]))
if t&1 == 0: out.add(int(s[:i] + str(t//2) + s[i+l:]))
return max(out)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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