A323288 Largest number that can be obtained from the "Choix de Bruxelles", version 2 (A323460) operation applied to n.

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

Offset: 1

Author

N. J. A. Sloane, Jan 15 2019

Keywords

• nonn
• base

Comments

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

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

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane, "Choix de Bruxelles": A New Operation on Positive Integers, arXiv:1902.01444 [math.NT], Feb 2019; Fib. Quart. 57:3 (2019), 195-200.
• Eric Angelini, Lars Blomberg, Charlie Neder, Remy Sigrist, and N. J. A. Sloane,, "Choix de Bruxelles": A New Operation on Positive Integers, Local copy.

Crossrefs

Cf. A323286, A323460, A017329.

Programs

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)
print([a(n) for n in range(1, 60)]) # Michael S. Branicky, Jul 24 2022

Formula

a(n) >= 2*n. - Rémy Sigrist, Jan 15 2019

Extensions

More terms from Rémy Sigrist, Jan 15 2019