A227070 Powers n such that the set s(n) = {k > 0 such that k^n ends with k} does not occur for smaller n.

1, 2, 3, 5, 6, 9, 11, 17, 21, 26, 33, 41, 51, 65, 81, 101, 126, 129, 161, 201, 251, 257, 321, 401, 501, 513, 626, 641, 801, 1001, 1025, 1251, 1281, 1601, 2001, 2049, 2501, 2561, 3126, 3201, 4001, 4097, 5001, 5121, 6251, 6401, 8001, 8193, 10001

Offset: 1

Author

T. D. Noe, Jul 29 2013

Keywords

• nonn
• hard
• more
• base

Comments

These numbers might be called automorphic powers because the sets s(n) are called automorphic numbers. It appears that all numbers of the form 1 + 5^i are here. In fact, these appear to produce the only even numbers here. The set s(4) equals s(2). The set s(7) equals s(3). The set s(9) does not differ from s(5) until k = 10443. The set s(17) does not differ from s(9) until k = 108307. The sequence also has 126, 201, 251, 501, and 626, but there may be missing numbers.

Entries a(17)-a(49) have been tentatively obtained by comparing the terms < 10^30 in the sets s(n), for 2 <= n <= 10001. - Giovanni Resta, Jul 30 2013

Crossrefs

Cf. A003226 (n=2), A033819 (n=3), A068407 (n=5), A068408 (n=6).

Cf. A072496 (n=11), A072495 (n=21), A076650 (n=26).

Cf. A227071.

Programs

Mathematica

ts = {}; t = {}; Do[s = Select[Range[11000000], PowerMod[#, n, 10^IntegerLength[#]] == # &]; If[! MemberQ[ts, s], Print[n]; AppendTo[ts, s]; AppendTo[t, n]], {n, 2, 101}]; t = Join[{1}, t]

Formula

Conjecture: a(n+1) = A003592(n) + 1. - Eric M. Schmidt, Jul 30 2013

Extensions

a(17)-a(49) from Giovanni Resta, Jul 30 2013