A328198 -id:A328198 - cp's OEIS frontend

A060768 Pseudo-Kaprekar triples: q such that if q=x+y+z, then q^3=x10^i + y10^j + z, where (y*10^j+z < 10^i) and z < 10^j.

1, 8, 10, 45, 100, 134, 297, 783, 972, 1000, 1368, 1611, 2322, 2710, 2728, 3086, 4445, 4544, 4949, 5049, 5455, 5554, 7172, 10000, 19908, 21268, 27100, 44443, 55556, 60434, 76581, 77778, 100000, 103239, 133334, 143857, 199728, 208494, 226071

Offset: 1

Larry Reeves (larryr(AT)acm.org), Apr 24 2001

True Kaprekar triples (A006887) must have j=n and i=2n, where n is the number of digits in q.

			134^3=2406104 and 134=24+06+104. 134 is not a Kaprekar triple since the three terms of the sum would need to be 2, 406 and 104. 134 is not a term of A328198 because one addend (06) begins with '0'.

Giovanni Resta, Table of n, a(n) for n = 1..280

Cf. A006887, A328198.

Offset changed to 1 by Giovanni Resta, Oct 09 2019

A328199 Triples (a,b,c) such that (a+b+c)^3 = concat(a,b,c), a, b, c > 0, ordered by size of this value.

Original entry on oeis.org

5, 1, 2, 9, 11, 25, 418, 1062, 131, 878, 2442, 1125, 938, 2422, 1184, 1212, 1388, 2349, 1287, 1113, 2649, 1623, 2457, 1375, 1713, 2377, 1464, 3689, 1035, 2448, 7890, 10706, 1312, 17147, 18793, 19616, 22072, 11858, 26504, 47051, 15775, 14952

Offset: 1

M. F. Hasler, Oct 07 2019

The sequence can be considered as a table with rows of length 3, row(n) = a(3n-2 .. 3n).
A variant of Kaprekar and pseudo-Kaprekar triples, cf. A006887 and A060768.
See A328198 and A328200 (sequence of the values a+b+c and concatenated triples) for more information.

			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..717
Números y algo mas, 9 + 11+ 25 = 91125^(1/3) etc, post on facebook.com, Sep 30 2019.

Cf. A328198 (row sums), A328200 (rows concatenated), 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(concat(divrem(ab,10^Lb)~,c))))} \\ A055642(n)=logint(n,10)+1 = #digits(n)
for( Ln=1,oo, for( n=10^(Ln-1),10^Ln-1, (t=is(n,Ln))&& print1(t", ")))

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

A060768 Pseudo-Kaprekar triples: q such that if q=x+y+z, then q^3=x10^i + y10^j + z, where (y*10^j+z < 10^i) and z < 10^j.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A328199 Triples (a,b,c) such that (a+b+c)^3 = concat(a,b,c), a, b, c > 0, ordered by size of this value.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

cp's OEIS Frontend

A060768 Pseudo-Kaprekar triples: q such that if q=x+y+z, then q^3=x*10^i + y*10^j + z, where (y*10^j+z < 10^i) and z < 10^j.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A328199 Triples (a,b,c) such that (a+b+c)^3 = concat(a,b,c), a, b, c > 0, ordered by size of this value.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A060768 Pseudo-Kaprekar triples: q such that if q=x+y+z, then q^3=x10^i + y10^j + z, where (y*10^j+z < 10^i) and z < 10^j.