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A328198 Numbers of the form N = a+b+c such that N^3 = concat(a,b,c); a, b, c > 0.

%I A328198 #13 Oct 09 2019 07:49:36
%S A328198 8,45,1611,4445,4544,4949,5049,5455,5554,7172,19908,55556,60434,77778,
%T A328198 422577,427868,461539,478115,488214,494208,543752,559846,598807,
%U A328198 664741,757835,791505,807598,4927940,5555555,6183170,25252524,27272728,27282727,28201724,30731977
%N A328198 Numbers of the form N = a+b+c such that N^3 = concat(a,b,c); a, b, c > 0.
%C A328198 A variant of Kaprekar and pseudo-Kaprekar triples, cf. A006887 and A060768.
%C A328198 Leading zeros as in A006887(4), 26198073 = (26+198+073)^3, are not allowed here.
%C A328198 Is it a coincidence that a(2)^3 = 91125 also verifies sqrt(91125) = 9*sqrt(1125)?
%C A328198 See A328199 for the triples (a,b,c) and A328200 for the cubes / concatenations.
%H A328198 Giovanni Resta, <a href="/A328198/b328198.txt">Table of n, a(n) for n = 1..239</a> (terms < 10^12)
%H A328198 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 A328198 5 + 1 + 2 = 512^(1/3) = 8,
%e A328198 9 + 11 + 25 = 91125^(1/3) = 45,
%e A328198 418 + 1062 + 131 = (4181062131)^(1/3) = 1611, ...
%o A328198 (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 A328198 for( Ln=1,oo, for( n=10^(Ln-1),10^Ln-1, is(n,Ln)&& print1(n", ")))
%Y A328198 Cf. A328199 (corresponding a,b,c), A328200 (cubes / concatenations), A006887 & A291461 (Kaprekar numbers), A060768 (pseudo Kaprekar numbers); A000578 (the cubes), A055642 (number of digits of n).
%K A328198 nonn,base
%O A328198 1,1
%A A328198 _M. F. Hasler_, Oct 07 2019
%E A328198 a(31)-a(35) from _Giovanni Resta_, Oct 09 2019