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A182292
Smallest odd number k such that is equal to the sum of its proper divisors greater than k^(1/n), or 0 if none exist.
0
34155, 407715, 8415
OFFSET
2,1
COMMENTS
a(8) = 159030135. There is no n > 4 for which a(n) is smaller unless a(n) = 0. - Charles R Greathouse IV, Apr 25 2012
Other than a(2) to a(4) and a(8), there is no solution < 2*10^10 for a(n) up to a(1000). - Donovan Johnson, Aug 23 2012
From Alexander Violette, Feb 29 2024: (Start)
a(7) <= 7650499534755.
a(14) <= 221753170660847595. (End)
EXAMPLE
The sum proper divisors of 407715 greater than 407715^(1/3) is 77 + 105 + 165 + 231 + 353 + 385 + 1059 + 1155 + 1765 + 2471 + 3883 + 5295 + 7413 + 11649 + 12355 + 19415 + 27181 + 37065 + 58245 + 81543 + 135905 = 407715.
MATHEMATICA
t={}; d[n_]:= Select[Drop[Divisors[n], -1], #1>n^(1/p)&]; Do[s=Select[Range[1, 5*10^5, 2], #==Plus@@d[#]&];
AppendTo[t, s], {p, 2, 4}]; Flatten[t]
PROG
(PARI) a(n)=my(t, k=8413); while(k+=2, if(sigma(k, -1)>2, if(ispower(k, n, &t), , t=k^(1/n)\1); if(sumdiv(k, d, if(d>t, d))==2*k, return(k)))) \\ Charles R Greathouse IV, Apr 25 2012
CROSSREFS
See A182147 for more details for 34155.
Sequence in context: A344666 A227699 A234820 * A206214 A227487 A204411
KEYWORD
nonn,bref
AUTHOR
Manuel Valdivia, Apr 24 2012
STATUS
approved