cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Views

Author

Pieter Post, Jul 22 2015

Keywords

Comments

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.

Examples

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

Crossrefs

Programs

  • PARI
    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
  • Python
    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

Views

Author

Pieter Post, Jul 22 2015

Keywords

Comments

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).

Examples

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

Crossrefs

Programs

  • Mathematica
    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 *)
  • Python
    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
Showing 1-2 of 2 results.