A177900 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.

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

Offset: 1

Author

John W. Layman, Dec 15 2010

Keywords

• nonn

Comments

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."

Examples

			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.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A037020, A005114.

Programs

Maple

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:
seq(A[i],i=1..N); # Robert Israel, Jul 09 2014

Mathematica

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 *)

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

Formula

a(n) = A070015(p_n) where p runs through the primes. - Robert G. Wilson v, Jul 09 2014

Extensions

Offset corrected to 1 by Michel Marcus, Jul 09 2014