A259379 Numbers k of the form a - b + c, such that k^3 equals the decimal concatenation a//b//c and numbers k, b, and c have the same number of digits.

155, 209, 274, 286, 287, 351, 364, 428, 573, 637, 715, 727, 846, 923, 1095, 1096, 2191, 8905, 18182, 18183, 81818, 81819, 326734, 336634, 663367, 673267, 2727273, 2727274, 4545454, 5454547, 7272727, 23529411, 23529412, 76470589

Offset: 1

Author

Pieter Post, Jul 22 2015

Keywords

• nonn
• base

Comments

This sequence is infinite because it has several infinite subsequences. For example:

274, 326734, 332667334, 3..326..673..34 etc.;

364, 336634, 333666334, 3..36..63..34 etc.;

637, 663367, 666333667, 6..63..36..67 etc.;

727, 673267, 667332667, 6..673..326..67 etc.

Note that: 274 + 727 = 364 + 637 = 1001 and 326734 + 673267 = 336634 + 663367 = 1000001.

Many numbers come in pairs, like: (286, 287), (1095, 1096), (18182, 18183) but also bigger number (140017877, 140017878) and (859982123, 859982124).

140017877 + 859982124 = 140017878 + 859982123 = 1000000001.

Examples

			155^3 = 3723875 and 155 = 3 - 723 + 875.
715^3 = 365525875 and 715 = 365 - 525 + 875.
		

Links

• Pieter Post, Table of n, a(n) for n = 1..189

Crossrefs

Cf. A056733, A101311, A118937, A118938, A228381, A257796, A258482.

Programs

PARI

isok(n)=nb = #digits(n, 10); if (a = n^3 \ 10^(2*nb), c = n^3 % 10^nb; b = (n^3 - a*10^(2*nb))\10^nb; n^3 == (a-b+c)^3;); \\ Michel Marcus, Aug 05 2015

Python

def modb(n,m):
    kk = 0
    l = 1
    while n > 0:
        na = n % m
        l += 1
        kk += ((-1)**l) * na
        n //= m
    return kk
for n in range (100, 10**9):
    ll = len(str(n))
    if modb(n**3, 10**ll) == n:
        print(n, end=', ') # corrected by David Radcliffe, May 09 2025