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A046000
a(n) is the largest number m equal to the sum of digits of m^n.
9
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
0,2
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
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.
LINKS
T. D. Noe, Table of n, a(n) for n = 0..1000 (a(105), a(164), a(186), ... corrected by Mohammed Yaseen)
FORMULA
a(n) = A061211(n)^(1/n), for n > 0.
EXAMPLE
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.
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 *)
PROG
(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
KEYWORD
base,nonn,nice
AUTHOR
David W. Wilson and Patrick De Geest
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
STATUS
approved