A289138 a(n) = smallest expomorphic number in base n: least integer k such that n^k ends in k, or 0 if no such k exists.

1, 36, 7, 6, 5, 6, 3, 56, 9, 0, 1, 16, 7, 6, 5, 6, 3, 76, 9, 0, 1, 96, 7, 6, 5, 6, 3, 96, 9, 0, 1, 76, 7, 6, 5, 6, 3, 16, 9, 0, 1, 56, 7, 6, 5, 6, 3, 36, 9, 0, 1, 36, 7, 6, 5, 6, 3, 56, 9, 0, 1, 16, 7, 6, 5, 6, 3, 76, 9, 0, 1, 96, 7, 6, 5, 6, 3, 96, 9, 0, 1, 76, 7, 6, 5, 6, 3, 16, 9, 0, 1, 56, 7, 6, 5, 6, 3, 36, 9, 0

Offset: 1

Author

Bernard Schott and Robert G. Wilson v, Jun 28 2017

Keywords

• easy
• nonn
• base

Comments

Definition: For positive integers b (the base) and n, the positive integer (allowing initial zeros) a(n) is expomorphic relative to base b if a(n) has exactly n decimal digits and if b^a(n) == a(n) (mod 10^n) or, equivalently, b^a(n) ends in a(n). [See Crux Mathematicorum link.]

The only twelve values a(n) can take are 0, 1, 3, 5, 6, 7, 9, 16, 36, 56, 76 and 96;

and the percentages of the time these occur are 10, 10, 10, 10, 20, 10, 10, 4, 4, 4, 4 and 4, respectively.

The bases, n, for which k is:

0: n == 0 (mod 10)

1: n == 1 (mod 10)

3: n == 7 (mod 10)

5: n == 5 (mod 10)

6: n == 4 or 6 (mod 10)

7: n == 3 (mod 10)

9: n == 9 (mod 10)

16: n == +/- 12 (mod 50)

36: n == +/- 2 (mod 50)

56: n == +/- 8 (mod 50)

76: n == +/- 18 (mod 50)

96: n == +/- 22 (mod 50).

Periodicity is 50.

Examples

			a(4) is 6 since 4^6 = 4096 which ends in 6.
		

Links

• Charles W. Trigg, Problem 559, Crux Mathematicorum, page 192, Vol. 7, Jun. 1981.
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Crossrefs

Cf. A288845.

Programs

Maple

f:= proc(n) local k;
   if n mod 10 = 0 then return 0 fi;
   for k from 1 do if n^k - k mod 10^(1+ilog10(k)) = 0 then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Jul 07 2017

Mathematica

f[n_] := If[ Mod[n, 10] > 0, Block[{k = 1}, While[ PowerMod[n, k, 10^IntegerLength[k]] != k, k++]; k], 0]; Array[f, 88]

Python

def a(n):
    if n%10==0: return 0
    k=1
    while pow(n, k, 10**len(str(k)))!=k: k+=1
    return k
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 29 2017