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A328200 Cubes of the form N^3 = concat(a,b,c) with N = a+b+c; a, b, c > 0.

%I A328200 #22 Apr 26 2021 21:29:09
%S A328200 512,91125,4181062131,87824421125,93824221184,121213882349,
%T A328200 128711132649,162324571375,171323771464,368910352448,7890107061312,
%U A328200 171471879319616,220721185826504,470511577514952,75460133084214033,78330233506116032,98316229404133819,109294197946170875
%N A328200 Cubes of the form N^3 = concat(a,b,c) with N = a+b+c;  a, b, c > 0.
%C A328200 A variant of Kaprekar and pseudo-Kaprekar triples, cf. A006887 and A060768.
%C A328200 Leading zeros as in A006887(4), 26198073 = (26 + 198 + 073)^3, are not allowed here.
%C A328200 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 A328200 Giovanni Resta, <a href="/A328200/b328200.txt">Table of n, a(n) for n = 1..239</a>
%H A328200 Números y algo mas, <a href="https://www.facebook.com/permalink.php?story_fbid=2467527383315339&amp;id=126559577412143">9 + 11 + 25 = 91125^(1/3) etc</a>, post on facebook.com, Sep 30 2019.
%e A328200 512^(1/3) = 8 = 5 + 1 + 2,
%e A328200 91125^(1/3) = 45 = 9 + 11 + 25,
%e A328200 4181062131^(1/3) = 1611 = 418 + 1062 + 131, ...
%o A328200 (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 A328200 for( Ln=1,oo, for( n=10^(Ln-1),10^Ln-1, is(n,Ln)&& print1(n^3", ")))
%Y A328200 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 A328200 nonn,base
%O A328200 1,1
%A A328200 _M. F. Hasler_, Oct 07 2019