A328199 -id:A328199 - cp's OEIS frontend

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

8, 45, 1611, 4445, 4544, 4949, 5049, 5455, 5554, 7172, 19908, 55556, 60434, 77778, 422577, 427868, 461539, 478115, 488214, 494208, 543752, 559846, 598807, 664741, 757835, 791505, 807598, 4927940, 5555555, 6183170, 25252524, 27272728, 27282727, 28201724, 30731977

Offset: 1

M. F. Hasler, Oct 07 2019

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.
Is it a coincidence that a(2)^3 = 91125 also verifies sqrt(91125) = 9*sqrt(1125)?
See A328199 for the triples (a,b,c) and A328200 for the cubes / concatenations.

			5 + 1 + 2 = 512^(1/3) = 8,
9 + 11 + 25 = 91125^(1/3) = 45,
418 + 1062 + 131 = (4181062131)^(1/3) = 1611, ...

Giovanni Resta, Table of n, a(n) for n = 1..239 (terms < 10^12)
Números y algo mas, 9 + 11 + 25 = 91125^(1/3) etc, post on facebook.com, Sep 30 2019.

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).

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", ")))

a(31)-a(35) from Giovanni Resta, Oct 09 2019

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

Original entry on oeis.org

512, 91125, 4181062131, 87824421125, 93824221184, 121213882349, 128711132649, 162324571375, 171323771464, 368910352448, 7890107061312, 171471879319616, 220721185826504, 470511577514952, 75460133084214033, 78330233506116032, 98316229404133819, 109294197946170875

Offset: 1

M. F. Hasler, Oct 07 2019

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).

			512^(1/3) = 8 = 5 + 1 + 2,
91125^(1/3) = 45 = 9 + 11 + 25,
4181062131^(1/3) = 1611 = 418 + 1062 + 131, ...

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.

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).

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", ")))

cp's OEIS Frontend

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

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