A228381 -id:A228381 - cp's OEIS frontend

A118938 Sub-Kaprekar numbers (2): n such that n=r-q and n^2=q*10^m+r, for some m>=1, q>=0, 0<=r<10^m, with n not a power of 10.

78, 287, 364, 1096, 18183, 336634, 2727274, 19138757, 23529412, 25974026, 97744361, 120879122, 140017878, 165991904, 237762239, 288553552, 307692308, 333666334, 405436669, 428571430, 440553516, 447710186, 454545455, 473684212

Offset: 1

Giovanni Resta, May 06 2006

			287^2 = 82369 and 369-82 = 287.
A larger example is 1980198021^2 = 3921184202372316441, and 2372316441-392118420 = 1980198021.

Hans Havermann, Table of n, a(n) for n = 1..1000
Hans Havermann, A large number of terms pairing them with the their A118937 partners.

Cf. A006886, A118936, A118937, A228381.

A228103 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (q-r)^2.

Original entry on oeis.org

100, 121, 6084, 10000, 10201, 82369, 132496, 1000000, 1002001, 1162084, 1201216, 1656369, 1860496, 100000000, 100020001, 123121216, 330621489, 10000000000, 10000200001, 13967221489, 113322449956, 1000000000000, 1000002000001, 1786590449956, 7438023471076

Offset: 1

Hans Havermann, Aug 10 2013

q*10^m+r = (q-r)^2 = A228381^2; q,m>0; 0<=r<10^m. - Hans Havermann, Aug 21 2013

			100 = (10-0)^2.
121 = (12-1)^2.
6084 = (6-084)^2.

Hans Havermann, Table of n, a(n) for n = 1..1016
Daniel Joyce, Robert Israel and Lars Blomberg, Can more of these terms be found? [broken link?]
J. B. Wood, Quick Solution to Puzzle?

Cf. A102766, A118936, A228381.

k=3; While[k<10^8, k++; s=k^2; d=IntegerDigits[s]; l=Length[d]; Do[a=FromDigits[Take[d,{1,i}]]; b=FromDigits[Take[d,{i+1,l}]]; If[k==Abs[a-b], w=ToString[s]; Print[StringTake[w,{1,i}], "'", StringTake[w,{i+1,l}]]], {i,l-1}]] (* Hans Havermann, Aug 10 2013, Aug 20 2013 *)

A259379 Numbers k of the form a - b + c, such that k^3 equals the decimal concatenation a//b//c and numbers k, b, and c have the same number of digits.

Original entry on oeis.org

155, 209, 274, 286, 287, 351, 364, 428, 573, 637, 715, 727, 846, 923, 1095, 1096, 2191, 8905, 18182, 18183, 81818, 81819, 326734, 336634, 663367, 673267, 2727273, 2727274, 4545454, 5454547, 7272727, 23529411, 23529412, 76470589

Offset: 1

Pieter Post, Jul 22 2015

This sequence is infinite because it has several infinite subsequences. For example:
  274, 326734, 332667334, 3..326..673..34 etc.;
  364, 336634, 333666334, 3..36..63..34 etc.;
  637, 663367, 666333667, 6..63..36..67 etc.;
  727, 673267, 667332667, 6..673..326..67 etc.
Note that: 274 + 727 = 364 + 637 = 1001 and 326734 + 673267 = 336634 + 663367 = 1000001.
Many numbers come in pairs, like: (286, 287), (1095, 1096), (18182, 18183) but also bigger number (140017877, 140017878) and (859982123, 859982124).
140017877 + 859982124 = 140017878 + 859982123 = 1000000001.

			155^3 = 3723875 and 155 = 3 - 723 + 875.
715^3 = 365525875 and 715 = 365 - 525 + 875.

Pieter Post, Table of n, a(n) for n = 1..189

Cf. A056733, A101311, A118937, A118938, A228381, A257796, A258482.

isok(n)=nb = #digits(n, 10); if (a = n^3 \ 10^(2*nb), c = n^3 % 10^nb; b = (n^3 - a*10^(2*nb))\10^nb; n^3 == (a-b+c)^3;); \\ Michel Marcus, Aug 05 2015

def modb(n,m):
    kk = 0
    l = 1
    while n > 0:
        na = n % m
        l += 1
        kk += ((-1)**l) * na
        n //= m
    return kk
for n in range (100, 10**9):
    ll = len(str(n))
    if modb(n**3, 10**ll) == n:
        print(n, end=', ') # corrected by David Radcliffe, May 09 2025

A260193 Numbers k of the form abs(a - b + c - d) such that k^4 equals the concatenation of a//b//c//d and numbers k,b,c,d have the same number of digits.

Original entry on oeis.org

198, 220, 221, 287, 352, 364, 484, 562, 627, 638, 672, 715, 716, 780, 793, 858, 901, 1095, 1233, 2328, 8905, 18183, 39753, 60248, 85207, 336734, 2727274, 5893504, 8620777, 17769557, 52818678, 70710735, 76470590, 82230444, 101318734, 101636206, 104263158, 105262158, 109891110, 109942690, 117883117, 119722383, 120826541

Offset: 1

Pieter Post, Jul 22 2015

Leading zeros in b, c, and d are allowed.
Many numbers come in pairs, like: (220, 221), (715, 716), (140017877, 140017878).
Some numbers are also member of A259379, for example: 287, 715, 1095 and also the pair (140017877, 140017878).

			198^4 = 1536953616 and 198 = abs (1 - 536 + 953 - 616 ).
8905^4 = 6288335365950625 and 8905 = abs (6288 - 3353 + 6595 - 0625 ).

Pieter Post, Table of n, a(n) for n = 1..239

Cf. A056733, A101311, A118937, A118938, A228381, A257796, A258482, A259379.

test[n_] := Block[{L=IntegerLength@ n, v}, v = IntegerDigits[ n^4, 10^L]; Length@ v == 4 && Abs@ Total[ {1, -1, 1, -1} v] == n]; Select[Range[10^5], test] (* Giovanni Resta, Aug 12 2015 *)

def modb(n, m):
    kk = 0
    l = 1
    while n > 0:
        na = n % m
        l += 1
        kk += ((-1)**l) * na
        n //= m
    return abs(kk)
for n in range (100, 10**9):
    ll = len(str(n))
    if modb(n**4, 10**ll) == n and n**4 >= 10**(ll*3):
         print (n, end=', ') # corrected by David Radcliffe, May 09 2025

cp's OEIS Frontend

A118938 Sub-Kaprekar numbers (2): n such that n=r-q and n^2=q*10^m+r, for some m>=1, q>=0, 0<=r<10^m, with n not a power of 10.

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A228103 Numbers k whose base-10 digits can be split into two parts, q and r, with k = (q-r)^2.

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A259379 Numbers k of the form a - b + c, such that k^3 equals the decimal concatenation a//b//c and numbers k, b, and c have the same number of digits.

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A260193 Numbers k of the form abs(a - b + c - d) such that k^4 equals the concatenation of a//b//c//d and numbers k,b,c,d have the same number of digits.

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Programs

Mathematica

Python