A358348 Numbers k such that k == k^k (mod 9).

1, 4, 7, 9, 10, 13, 16, 17, 18, 19, 22, 25, 27, 28, 31, 34, 35, 36, 37, 40, 43, 45, 46, 49, 52, 53, 54, 55, 58, 61, 63, 64, 67, 70, 71, 72, 73, 76, 79, 81, 82, 85, 88, 89, 90, 91, 94, 97, 99, 100, 103, 106, 107, 108, 109, 112, 115, 117, 118, 121, 124, 125, 126

Offset: 1

Author

Ivan Stoykov, Nov 11 2022

Keywords

• nonn
• base
• easy

Comments

Each multiple of 9 is in the sequence. Additionally, the squares are also present.

Examples

			4 is a term since 4^4 = 256 == 4 (mod 9).
		

References

• M. Fujiwara and Y. Ogawa, Introduction to Truly Beautiful Mathematics. Tokyo: Chikuma Shobo, 2005.

Links

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

Crossrefs

Cf. A007953, A010888, A082576, A189510.

Programs

Maple

A358348 := proc(n)
    2*(n+1)-op(modp(n,9)+1,[2,3,2,1,1,2,1,0,1]) ;
end proc:
seq(A358348(n),n=1..50) ; # R. J. Mathar, Mar 29 2023

Mathematica

Select[Range[130], MemberQ[{0, 1, 4, 7, 9, 10, 13, 16, 17}, Mod[#, 18]] &] (* Amiram Eldar, Nov 12 2022 *)

PARI

isok(k) = k == Mod(k,9)^k; \\ Michel Marcus, Nov 22 2022

Python

def A358348(n):
    return ((0, 1, 4, 7, 9, 10, 13, 16, 17)[m := n % 9]
         + (n - m << 1))  # Chai Wah Wu, Feb 09 2023

Formula

G.f.: x*(x+1)*(x^7+3*x^5+x^3+x^2+2*x+1)/((1-x)^2*(1+x^3+x^6)*(1+x+x^2)). - Alois P. Heinz, Feb 08 2023

a(n) = 2*(n+1) - b(n) where b(n>=0) = 2,3,2,1,1,2,1,0,1,2,3,2,... has period 9. - Kevin Ryde, Mar 26 2023