A177813 Numbers k such that k^3 divides 13^(k^2) + 1.

1, 7, 203, 11977, 154553, 353423, 3049963, 4482037, 5192537, 7170569, 9904979, 20851957, 33461911, 35852033, 69262991, 88448927, 160274303, 264440183, 306359683, 381231473, 423063571, 978466699, 1974252749, 2115269947, 4647954787, 5218486693, 6824905927, 7803226417, 9040206917, 10041409007, 11055712717, 12483960629, 17244568439, 47773414171, 57280493557, 57729535241, 58257187051, 62418389453, 67340133077

Offset

1,2

Comments

7 divides a(n) for n > 1.

Prime factors of a(n) in the order of their appearance are {7, 29, 59, 22079, 1741, 435709, 25579, 35323, 827, 164837, 176611, 5783, 435709, 22896329, 54461639, 4820033, ...}. - Alexander Adamchuk, May 16 2010

Links

Table of n, a(n) for n=1..39.

Crossrefs

Cf. A127263 (k^3 divides 2^(k^2) + 1).

Cf. A128678, A128679, A128680, A128681, A128682, A128683, A128684, A128685.

Cf. A177813, A177814, A177816, A177817, A177818, A177819, A177820.

Cf. A128677 (least k > p such that (k*p)^3 divides (p-1)^(k*p)^2+1, where p = prime(n) > 2).

Sequence in context: A361141 A128680 A292332 * A292338 A216456 A243144

Adjacent sequences: A177810 A177811 A177812 * A177814 A177815 A177816

Keyword

nonn

Author

Alexander Adamchuk, May 14 2010

Extensions

Corrected and extended by Max Alekseyev, May 16 2010

a(8) = 4482037 (suggested by Max Alekseyev) missing in original sequence inserted by Alexander Adamchuk, May 16 2010

More terms from Max Alekseyev, Feb 14 2012

Status

approved