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A328200
Cubes of the form N^3 = concat(a,b,c) with N = a+b+c; a, b, c > 0.
3
512, 91125, 4181062131, 87824421125, 93824221184, 121213882349, 128711132649, 162324571375, 171323771464, 368910352448, 7890107061312, 171471879319616, 220721185826504, 470511577514952, 75460133084214033, 78330233506116032, 98316229404133819, 109294197946170875
OFFSET
1,1
COMMENTS
A variant of Kaprekar and pseudo-Kaprekar triples, cf. A006887 and A060768.
Leading zeros as in A006887(4), 26198073 = (26 + 198 + 073)^3, are not allowed here.
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).
LINKS
NĂºmeros y algo mas, 9 + 11 + 25 = 91125^(1/3) etc, post on facebook.com, Sep 30 2019.
EXAMPLE
512^(1/3) = 8 = 5 + 1 + 2,
91125^(1/3) = 45 = 9 + 11 + 25,
4181062131^(1/3) = 1611 = 418 + 1062 + 131, ...
PROG
(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)
for( Ln=1, oo, for( n=10^(Ln-1), 10^Ln-1, is(n, Ln)&& print1(n^3", ")))
CROSSREFS
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).
Sequence in context: A013701 A187461 A291461 * A181244 A181252 A016749
KEYWORD
nonn,base
AUTHOR
M. F. Hasler, Oct 07 2019
STATUS
approved