A072968 Least k>0 such that the last digit of (n+k)^(n+k) is the same as the last digit of n^n.

10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10, 10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10, 10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10, 10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10, 10

Offset

1,1

Links

Table of n, a(n) for n=1..81.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Formula

a(n) is periodic with period (10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10, 10) of length 20

Mathematica

ld[n_]:=Module[{ldn=Mod[n^n, 10], k=1}, While[Mod[(n+k)^(n+k), 10] != ldn, k++]; k]; Array[ld, 90] (* Harvey P. Dale, Sep 07 2012 *)
LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {10, 16, 14, 2, 10, 2, 6, 4, 10, 10, 10, 2, 14, 2, 10, 8, 6, 4, 10, 10}, 81] (* Ray Chandler, Aug 26 2015 *)

Prog

(PARI) a(n)=if(n<0, 0, s=1; while(abs((n+s)^(n+s)%10-n^n%10)>0, s++); s)
(Python)
def a(n):
    k, target = 1, pow(n, n, 10)
    while pow(n+k, n+k, 10) != target: k += 1
    return k
print([a(n) for n in range(1, 82)]) # Michael S. Branicky, Oct 16 2021

Crossrefs

Sequence in context: A175335 A167788 A341012 * A072138 A109891 A104869

Adjacent sequences: A072965 A072966 A072967 * A072969 A072970 A072971

Keyword

base,easy,nonn

Author

Benoit Cloitre, Aug 13 2002

Status

approved