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

1,2

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.

Links

Table of n, a(n) for n=1..100.

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).

Example

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

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]

Prog

(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

Crossrefs

Cf. A288845.

Sequence in context: A100252 A020340 A255868 * A181759 A280679 A343921

Adjacent sequences: A289135 A289136 A289137 * A289139 A289140 A289141

Keyword

easy,nonn,base

Author

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

Status

approved