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1,2
Table of n, a(n) for n=1..10.
a(3) = 57 because 57*3!+1 = 343 = 7^3.
a={}; Do[k = 2; While[ !IntegerQ[(k^n - 1)/n!], k++ ]; AppendTo[a, (k^n-1)/n!], {n, 1, 10}]; a
Sequence in context: A071579 A060497 A092273 * A181437 A156873 A071540
Adjacent sequences: A193742 A193743 A193744 * A193746 A193747 A193748
nonn
Michel Lagneau, Aug 03 2011
approved