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.

A046000 a(n) is the largest number m equal to the sum of digits of m^n.

Original entry on oeis.org

1, 9, 9, 27, 36, 46, 64, 68, 63, 81, 117, 108, 108, 146, 154, 199, 187, 216, 181, 207, 207, 225, 256, 271, 288, 337, 324, 307, 328, 341, 396, 443, 388, 423, 463, 477, 424, 495, 469, 523, 502, 432, 531, 572, 603, 523, 592, 666, 667, 695, 685, 685, 739, 746, 739, 683, 684, 802, 754, 845, 793, 833, 865
Offset: 0

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Author

David W. Wilson and Patrick De Geest

Keywords

Comments

Cases a(n) = 1 begin: 0, 105, 164, 186, 194, 206, 216, 231, 254, 282, 285, ... Cf. A133509. - Jean-François Alcover, Jan 09 2018

Examples

			a(3) = 27 because 27 is the largest number with 27^3 = 19683 and 1+9+6+8+3 = 27.
a(5) = 46 because 46 is the largest number with 46^5 = 205962976 and 2+0+5+9+6+2+9+7+6 = 46.
		

References

  • Amarnath Murthy, The largest and the smallest m-th power whose digits sum /product is its m-th root. To appear in Smarandache Notions Journal, 2003.
  • Amarnath Murthy, e-book, "Ideas on Smarandache Notions" MS.LIT
  • Joe Roberts, "Lure of the Integers", The Mathematical Association of America, 1992, p. 172.

Crossrefs

Programs

  • Mathematica
    meanDigit = 9/2; translate = 900; upperm[1] = translate;
    upperm[n_] := Exp[-ProductLog[-1, -Log[10]/(meanDigit*n)]] + translate;
    (* assuming that upper bound of m fits the implicit curve m = Log[10, m^n]*9/2 *)
    a[0] = 1; a[n_] := (For[max = m = 0, m <= upperm[n], m++, If[m == Total[IntegerDigits[m^n]], max = m]]; max);
    Table[a[n], {n, 0, 1000}] (* Jean-François Alcover, Jan 09 2018, updated Jul 07 2022 *)
  • Python
    def ok(k, n): return sum(map(int, str(k**n))) == k
    def a(n):
        d, lim = 1, 1
        while lim < n*9*d: d, lim = d+1, lim*10
        return next(k for k in range(lim, 0, -1) if ok(k, n))
    print([a(n) for n in range(63)]) # Michael S. Branicky, Jul 06 2022

Formula

a(n) = A061211(n)^(1/n), for n > 0.

Extensions

More terms from Asher Auel, Jun 01 2000
More terms from Franklin T. Adams-Watters, Sep 01 2006
Edited by N. J. A. Sloane at the suggestion of David Wasserman, Dec 12 2007