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%I #44 Aug 28 2015 15:56:06
%S 155,209,274,286,287,351,364,428,573,637,715,727,846,923,1095,1096,
%T 2191,8905,18182,18183,81818,81819,326734,336634,663367,673267,
%U 2727273,2727274,4545454,5454547,7272727,23529411,23529412,76470589
%N Numbers n of the form a - b + c, such that n^3 equals the decimal concatenation a//b//c and numbers n, b, and c have the same number of digits.
%C This sequence is infinite because it has several infinite subsequences. For example:
%C 274, 326734, 332667334, 3..326..673..34 etc.;
%C 364, 336634, 333666334, 3..36..63..34 etc.;
%C 637, 663367, 666333667, 6..63..36..67 etc.;
%C 727, 673267, 667332667, 6..673..326..67 etc.
%C Note that: 274 + 727 = 364 + 637 = 1001 and 326734 + 673267 = 336634 + 663367 = 1000001.
%C Many numbers come in pairs, like: (286, 287), (1095, 1096), (18182, 18183) but also bigger number (140017877, 140017878) and (859982123, 859982124).
%C 140017877 + 859982124 = 140017878 + 859982123 = 1000000001.
%H Pieter Post, <a href="/A259379/b259379.txt">Table of n, a(n) for n = 1..189</a>
%e 155^3 = 3723875 and 155 = 3 - 723 + 875.
%e 715^3 = 365525875 and 715 = 365 - 525 + 875.
%o (Python)
%o def modb(n,m):
%o ...kk = 0
%o ....l=1
%o ....while n > 0:
%o ........na=int(n%m)
%o ........l=l+1
%o ........kk= kk+((-1)**l)*na
%o ........n =int(n//m)
%o ....return kk
%o for n in range (100,10**9):
%o ....ll= len(str(n))
%o ....if modb(n**3,10**ll)==n:
%o .........print (n)
%o (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
%Y Cf. A056733, A101311, A118937, A118938, A228381, A257796, A258482.
%K nonn,base
%O 1,1
%A _Pieter Post_, Jul 22 2015
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