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A177900
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Smallest positive integer k such that the sum of the proper divisors of k gives the n-th prime, or 0 if prime(n) belongs to A005114.
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2
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0, 4, 0, 8, 21, 27, 39, 65, 57, 115, 32, 155, 63, 50, 129, 235, 265, 371, 305, 201, 98, 365, 237, 171, 245, 291, 485, 309, 325, 327, 128, 189, 279, 917, 1507, 1529, 242, 785, 489, 835, 865, 1211, 385, 605, 579, 965, 338, 2321, 669, 1115, 687, 1165, 399, 2629, 549, 1285, 595, 1841, 1025, 603, 1385, 715, 3809, 921, 1175, 939, 4121, 1655, 805, 1325
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OFFSET
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1,2
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COMMENTS
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Suggested by the following comment of T. D. Noe (see A037020): "Assuming the Goldbach conjecture, it is easy to show that all primes, except 2 and 5, are the sum of the proper divisors of some number."
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LINKS
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FORMULA
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EXAMPLE
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27 and 35 have the proper divisors {1,3,9} and {1,5,7}, respectively, both of which sum to prime(6)=13, and no positive integer smaller than 27 has this property. Thus a(6)=27.
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MAPLE
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N:= 1000: # to get the first N terms
A:= 'A': A[1]:= 0: A[3]:= 0:
Primes:= [seq(ithprime(i), i=1..N)]:
count:= 0:
for k from 1 while count < N-2 do
m:= numtheory:-sigma(k)-k;
if member(m, Primes, 'n') and not assigned(A[n]) then
count:= count+1;
A[n]:= k;
fi
od:
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MATHEMATICA
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f[n_] := DivisorSigma[1, n] - n; k = 1; t = Table[0, {1000}]; While[ k < 24001, a = f@ k; If[a < 1001 && t[[a]] == 0, t[[a]] = k]; k++]; tt = Table[t[[n]], {n, Prime@ Range@ 168}] (* Robert G. Wilson v, Jul 09 2014 *)
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PROG
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(PARI) a(n) = {p = prime(n); if ((p != 2) && (p != 5), k = 2; while ((sigma(k) - k) != p, k++); k; ); } \\ Michel Marcus, Jul 09 2014
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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