

A135244


Largest m such that the sum of the aliquot parts of m (A001065) equals n, or 0 if no such number exists.


4



0, 4, 9, 0, 25, 8, 49, 15, 14, 21, 121, 35, 169, 33, 26, 55, 289, 77, 361, 91, 38, 85, 529, 143, 46, 133, 28, 187, 841, 221, 961, 247, 62, 253, 24, 323, 1369, 217, 81, 391, 1681, 437, 1849, 403, 86, 493, 2209, 551, 94, 589, 0, 667, 2809, 713, 106, 703, 68, 697, 3481
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

2,2


COMMENTS

Previous name: Aliquot predecessors with the largest values.
Find each node's predecessors in aliquot sequences and choose the largest predecessor.
Climb the aliquot trees on shortest paths (see A135245 = Climb the aliquot trees on thickest branches).
The sequence starts at offset 2, since all primes satisfy sigma(n)n = 1.  Michel Marcus, Nov 11 2014


LINKS



EXAMPLE

a(25) = 143 since 25 has 3 predecessors (95,119,143), 143 being the largest.
a(5) = 0 since it has no predecessors (see Untouchables  A005114).


MATHEMATICA

seq[max_] := Module[{s = Table[0, {n, 1, max}], i}, Do[If[(i = DivisorSigma[1, n]  n) <= max, s[[i]] = Max[s[[i]], n]], {n, 2, (max  1)^2}]; Rest @ s]; seq[50]


PROG

(PARI) lista(nn) = {for (n=2, nn, k = (n1)^2; while(k && (sigma(k)k != n), k); print1(k, ", "); ); } \\ Michel Marcus, Nov 11 2014


CROSSREFS

Cf. A001065, A005114, A125601, A135245, A057709, A057710, A063769, A080907, A121507, A037020, A126016.


KEYWORD

nonn


AUTHOR

Ophir Spector (ospectoro(AT)yahoo.com), Nov 25 2007


EXTENSIONS



STATUS

approved



