A006887 -id:A006887 - cp's OEIS frontend

A060809 Erroneous version of A006887.

1, 8, 10, 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, 495208, 499500, 500500, 517076, 533170, 543752, 559846, 565137, 598807

Offset: 1

dead

Included in accordance with OEIS policy of including published but incorrect sequences, to serve as pointers to the correct versions.

Shyam Sunder Gupta, On Some Marvellous Numbers of Kaprekar, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 9, 275-315.

A291461 Kaprekar triples: q^3 = x10^2n + y10^n + z, with q = x + y + z and 10^n > q > 10^(n-1) (q = 1 allowed for n = 1).

Original entry on oeis.org

1, 512, 91125, 26198073, 12519490248, 20301732352, 87824421125, 93824221184, 121213882349, 128711132649, 162324571375, 171323771464, 368910352448, 19902511000000, 87782935806307, 171471879319616, 220721185826504, 470511577514952, 2977097087043793, 9063181647017784

Offset: 1

M. F. Hasler, Aug 24 2017

Hans Havermann, Table of n, a(n) for n = 1..1000

Cf. A006887.

m=10; for(q=1,1e4,if(qA006887 for slightly more efficient code.

a(n) = A006887(n)^3.

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.

Original entry on oeis.org

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

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", ")))

A171493 "Kaprekar quadruples": digits of X^4 taken D at a time sum to X (where D is number of digits in X.)

Original entry on oeis.org

1, 7, 45, 55, 67, 100, 433, 4950, 5050, 38212, 65068, 190576, 295075, 299035, 310024, 336700, 343333, 394615, 414558, 433566, 448228, 450550, 467236, 475497, 476191, 486486, 499500, 500500, 523513, 534898, 549550, 599743, 622414, 628408, 647362

Offset: 1

Robert Munafo, Dec 10 2009

Referred to as "natural" Kaprekar numbers on Munafo webpage because a(n) and the 4 pieces of a(n)^4 must all have the same number of digits (some of which can be leading zeros). Analogous to A053816 for squares, as opposed to A006886 and A045913 which allow irregular divisions.

			7^4 = 2401 ; 2+4+0+1 = 7. 67^4 = 20151121 ; 20+15+11+21 = 67. 4950^4 = 600372506250000 ; 0600+3725+0625+0000 = 4950.

Robert Gerbicz, Table of n, a(n) for n = 1..10852
Shyam Sunder Gupta, On Some Marvellous Numbers of Kaprekar, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 9, 275-315.
R. Munafo, Kaprekar Sequences

Cf. A006886, A006887, A045913, A053816

Added term a(1)=1, Robert Gerbicz, Jul 28 2011

A171500 "Kaprekar quintuples": digits of X^5 taken D at a time sum to X (where D is number of digits in X.)

Original entry on oeis.org

1, 10, 1000, 7776, 27100, 73440, 95120, 500499, 505791, 540539, 598697, 665335, 697598, 732347, 7607610, 37944478, 46945205, 54995500, 55216205, 56607166, 58106906, 63136413, 66595563, 68167738, 68807564, 69188525, 70667477, 72197730, 73197730, 74145807

Offset: 1

Robert Munafo, Dec 10 2009

Referred to as "natural" Kaprekar numbers on Munafo webpage because a(n) and the 5 pieces of a(n)^5 must all have the same number of digits (some of which can be leading zeros). Analogous to A053816 for squares, as opposed to A006886 and A045913 which allow irregular divisions.

			7776^5 = 28430288029929701376 ; 2843+0288+0299+2970+1376 = 7776.
27100^5 = 14616603103510000000000 ; 146+16603+10351+00000+00000 = 27100.

Robert Gerbicz, Table of n, a(n) for n = 1..26397
Robert Munafo, Kaprekar Sequences

Cf. A006886, A006887, A045913, A053816, A171493.

More terms from Robert Gerbicz, Jul 28 2011

cp's OEIS Frontend

A060809 Erroneous version of A006887.

Views

Author

Keywords

Comments

Links

A291461 Kaprekar triples: q^3 = x*10^2n + y*10^n + z, with q = x + y + z and 10^n > q > 10^(n-1) (q = 1 allowed for n = 1).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

Formula

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

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

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

A171493 "Kaprekar quadruples": digits of X^4 taken D at a time sum to X (where D is number of digits in X.)

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A171500 "Kaprekar quintuples": digits of X^5 taken D at a time sum to X (where D is number of digits in X.)

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A291461 Kaprekar triples: q^3 = x10^2n + y10^n + z, with q = x + y + z and 10^n > q > 10^(n-1) (q = 1 allowed for n = 1).

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.