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Largest number that can be obtained from the "Choix de Bruxelles", version 2 (A323460) operation applied to n.
4

%I #35 Jan 09 2025 13:02:10

%S 2,4,6,8,10,12,14,16,18,20,22,24,26,28,110,112,114,116,118,40,42,44,

%T 46,48,210,212,214,216,218,60,62,64,66,68,310,312,314,316,318,80,82,

%U 84,86,88,410,412,414,416,418,100,102,104,106,108,510,512,514,516,518

%N Largest number that can be obtained from the "Choix de Bruxelles", version 2 (A323460) operation applied to n.

%C Equally, this is the largest number that can be obtained from the "Choix de Bruxelles", version 1 (A323286) operation applied to n.

%C Maximal element in row n of irregular triangle in A323460 (or, equally, A323286).

%C 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)

%C 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

%C The high-water marks for a(n)/n occur at n = 1,15,25,35,45,..., cf. A017329. - _N. J. A. Sloane_, Jan 23 2019

%H Rémy Sigrist, <a href="/A323288/b323288.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane, <a href="http://arxiv.org/abs/1902.01444">"Choix de Bruxelles": A New Operation on Positive Integers</a>, arXiv:1902.01444 [math.NT], Feb 2019; Fib. Quart. 57:3 (2019), 195-200.

%H Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane,, <a href="/A307635/a307635.pdf">"Choix de Bruxelles": A New Operation on Positive Integers</a>, Local copy.

%F a(n) >= 2*n. - _Rémy Sigrist_, Jan 15 2019

%o (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

%o (Python)

%o def a(n):

%o s, out = str(n), {n}

%o for l in range(1, len(s)+1):

%o for i in range(len(s)+1-l):

%o if s[i] == '0': continue

%o t = int(s[i:i+l])

%o out.add(int(s[:i] + str(2*t) + s[i+l:]))

%o if t&1 == 0: out.add(int(s[:i] + str(t//2) + s[i+l:]))

%o return max(out)

%o print([a(n) for n in range(1, 60)]) # _Michael S. Branicky_, Jul 24 2022

%Y Cf. A323286, A323460, A017329.

%K nonn,base

%O 1,1

%A _N. J. A. Sloane_, Jan 15 2019

%E More terms from _Rémy Sigrist_, Jan 15 2019