let n>1 be a +ve integer then the largest integer m such that (nm {n raised to power m} +1) divides ( 1 +n+n2 ...................+n127)is

Dear Anju,

The first seven consecutive perfect numbers seen in this form are:
6 = (2)(3) = (2^(2-1))*(2^2 - 1)
28 = (4)(7) = (2^(3-1))*(2^3 - 1)
496 = (16)(31) = (2^(5-1))*(2^5 - 1)
8,128 = (64)(127) = (2^(7-1))*(2^7 - 1)
33,550,336 = (4096)(8191) = (2^(13-1))*(2^13 - 1)
8,589,869,056 = (65536)(131071) = (2^(17-1))*(2^17 - 1)
137,438,691,328 = (262144)(524287) = (2^(19-1))*(2^19 - 1)

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