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
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).
Programs
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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
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Mathematica
Select[Range[130], MemberQ[{0, 1, 4, 7, 9, 10, 13, 16, 17}, Mod[#, 18]] &] (* Amiram Eldar, Nov 12 2022 *)
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PARI
isok(k) = k == Mod(k,9)^k; \\ Michel Marcus, Nov 22 2022
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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
Comments