A291461 -id:A291461 - cp's OEIS frontend

A006887 Kaprekar triples: q such that q = x + y + z and q^3 = x10^2n + y10^n + z, where z < 10^n and n is the number of digits in q. q is not a power of 10 (except q=1).

1, 8, 45, 297, 2322, 2728, 4445, 4544, 4949, 5049, 5455, 5554, 7172, 27100, 44443, 55556, 60434, 77778, 143857, 208494, 226071, 279720, 313390, 324675, 329967, 346060, 368928, 395604, 422577, 427868, 461539, 472823, 478115, 488214, 494208

Offset: 1

Robert Munafo

The initial term a(1) = 1 is somewhat conventional: it is the only term with x = y = 0 and q = z = 10^k, which is explicitly allowed only for k = 0 and forbidden for k > 0. In all other cases, 0 < x, y, z < q, and q^3 has the same number of digits as x*10^2n. - M. F. Hasler, Aug 24 2017

			1 = 0 + 0 + 1 and 1^3 = (00)1 (cf. comment),
8 = 5 + 1 + 2 and 8^3 = 512,
45 = 9 + 11 + 25, and 45^3 = 91125. - _M. F. Hasler_, Aug 24 2017

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, NY, 1986, p. 151.

Jack Brennen and Hans Havermann, Table of n, a(n) for n = 1..1000 (first 200 terms from Giovanni Resta)
Futility Closet's "Math Notes", Shows the cubes of a(9) to a(13)
Shyam Sunder Gupta, On Some Marvellous Numbers of Kaprekar, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 9, 275-315.
Hans Havermann, Cube wonders
Douglas E. Iannucci and Bertrum Foster, Kaprekar Triples, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.8.
Robert Munafo, Kaprekar Sequences

Cf. A291461.

ok[n_] := n==1 || Block[{k = 10^IntegerLength[n], m = n^3}, n == Mod[m, k] + Floor[ m/k^2] + Mod[Floor[m/k], k] && ! IntegerQ@ Log10@ n]; Select[ Range@ 500000, ok] (* Giovanni Resta, Aug 23 2017 *)

m=1;for(n=1,6,for(q=m+(n>1),-1+m*=10,q==sumdigits(q^3,m)&&print1(q","))) \\ M. F. Hasler, Aug 24 2017

Entry revised by Larry Reeves (larryr(AT)acm.org), Apr 25 2001 and Dec 08 2002

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

Original entry on oeis.org

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

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

A006887 Kaprekar triples: q such that q = x + y + z and q^3 = x*10^2n + y*10^n + z, where z < 10^n and n is the number of digits in q. q is not a power of 10 (except q=1).

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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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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.

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

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A006887 Kaprekar triples: q such that q = x + y + z and q^3 = x10^2n + y10^n + z, where z < 10^n and n is the number of digits in q. q is not a power of 10 (except q=1).