A167624 a(n) = n^6 mod 32.

0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25, 0, 1, 0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25, 0, 1, 0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25, 0, 1, 0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25, 0, 1, 0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25, 0, 1, 0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0

Offset

0,4

Comments

Equivalently: n^(8*m+6) (mod 32). - G. C. Greubel, Jun 17 2016

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Formula

a(n+16) = a(n). - G. C. Greubel, Jun 17 2016

From Chai Wah Wu, Jan 06 2020: (Start)

a(n) = a(n-2) - a(n-4) + a(n-6) - a(n-8) + a(n-10) - a(n-12) + a(n-14) for n > 13.

G.f.: (-x^13 - 24*x^11 + 15*x^9 - 32*x^7 + 15*x^5 - 24*x^3 - x)/((x - 1)*(x + 1)*(x^4 + 1)*(x^8 + 1)). (End)

Mathematica

PowerMod[Range[0, 100], 6, 32] (* Harvey P. Dale, Apr 03 2015 *)
LinearRecurrence[{0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1}, {0, 1, 0, 25, 0, 9, 0, 17, 0, 17, 0, 9, 0, 25}, 80] (* Hugo Pfoertner, Jan 28 2025 *)

Prog

(SageMath) [power_mod(n, 6, 32) for n in range(0, 100)]
(PARI) a(n)=n^6%32 \\ Charles R Greathouse IV, Apr 06 2016
(Python) def A167624(n): return pow(n, 6, 32) # Karl-Heinz Hofmann, Jan 28 2025

Crossrefs

Sequence in context: A068741 A255403 A005079 * A181614 A108321 A059062

Adjacent sequences: A167621 A167622 A167623 * A167625 A167626 A167627

Keyword

nonn,easy

Author

Zerinvary Lajos, Nov 07 2009

Status

approved