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

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

Giovanni Resta, Table of n, a(n) for n = 1..239

Números 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

Adjacent sequences: A328197 A328198 A328199 * A328201 A328202 A328203

Keyword

nonn,base

Author

M. F. Hasler, Oct 07 2019

Status

approved