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Numbers of the form N = a+b+c such that N^3 = concat(a,b,c); a, b, c > 0.
4

%I #13 Oct 09 2019 07:49:36

%S 8,45,1611,4445,4544,4949,5049,5455,5554,7172,19908,55556,60434,77778,

%T 422577,427868,461539,478115,488214,494208,543752,559846,598807,

%U 664741,757835,791505,807598,4927940,5555555,6183170,25252524,27272728,27282727,28201724,30731977

%N Numbers of the form N = a+b+c such that N^3 = concat(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 Is it a coincidence that a(2)^3 = 91125 also verifies sqrt(91125) = 9*sqrt(1125)?

%C See A328199 for the triples (a,b,c) and A328200 for the cubes / concatenations.

%H Giovanni Resta, <a href="/A328198/b328198.txt">Table of n, a(n) for n = 1..239</a> (terms < 10^12)

%H NĂºmeros 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 5 + 1 + 2 = 512^(1/3) = 8,

%e 9 + 11 + 25 = 91125^(1/3) = 45,

%e 418 + 1062 + 131 = (4181062131)^(1/3) = 1611, ...

%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", ")))

%Y 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 nonn,base

%O 1,1

%A _M. F. Hasler_, Oct 07 2019

%E a(31)-a(35) from _Giovanni Resta_, Oct 09 2019