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%I #22 Apr 26 2021 21:29:09
%S 512,91125,4181062131,87824421125,93824221184,121213882349,
%T 128711132649,162324571375,171323771464,368910352448,7890107061312,
%U 171471879319616,220721185826504,470511577514952,75460133084214033,78330233506116032,98316229404133819,109294197946170875
%N Cubes of the form N^3 = concat(a,b,c) with N = a+b+c; a, b, c > 0.
%C A variant of Kaprekar and pseudo-Kaprekar triples, cf. A006887 and A060768.
%C Leading zeros as in A006887(4), 26198073 = (26 + 198 + 073)^3, are not allowed here.
%C Even though this may be the most relevant sequence concerning this problem, we consider A328198 (sequence of the values N) as the main entry where all other information can be found. See also A328199 for the triples (a,b,c).
%H Giovanni Resta, <a href="/A328200/b328200.txt">Table of n, a(n) for n = 1..239</a>
%H Números y algo mas, <a href="https://www.facebook.com/permalink.php?story_fbid=2467527383315339&id=126559577412143">9 + 11 + 25 = 91125^(1/3) etc</a>, post on facebook.com, Sep 30 2019.
%e 512^(1/3) = 8 = 5 + 1 + 2,
%e 91125^(1/3) = 45 = 9 + 11 + 25,
%e 4181062131^(1/3) = 1611 = 418 + 1062 + 131, ...
%o (PARI) is(n,Ln=A055642(n),n3=n^3,Ln3=A055642(n3))={my(ab,c); for(Lc=Ln3-2*Ln,Ln, [ab,c]=divrem(n3, 10^Lc); n-c<10^(Ln-1) || c < 10^(Lc-1) || for( Lb=Ln3-Ln-Lc,Ln, vecsum(divrem(ab,10^Lb)) == n-c && ab%10^Lb>=10^(Lb-1)&& return(1)))} \\ A055642(n)=logint(n,10)+1 = #digits(n)
%o for( Ln=1,oo, for( n=10^(Ln-1),10^Ln-1, is(n,Ln)&& print1(n^3", ")))
%Y Cf. A328198 (values of N), A328199 (triples a,b,c), A006887 & A291461 (Kaprekar numbers), A060768 (pseudo Kaprekar numbers); A000578 (the cubes), A055642 (number of digits of n).
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Oct 07 2019
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