A100402 Digital root of 4^n.

1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7

Offset: 0

Author

Cino Hilliard, Dec 31 2004

Keywords

• easy
• nonn
• base

Comments

Equals A141725 mod 9. - Paul Curtz, Sep 15 2008

Sequence is the digital root of A016777. - Odimar Fabeny, Sep 13 2010

Digital root of the powers of any number congruent to 4 mod 9. - Alonso del Arte, Jan 26 2014

Period 3: repeat [1, 4, 7]. - Wesley Ivan Hurt, Aug 26 2014

From Timothy L. Tiffin, Dec 02 2023: (Start)

The period 3 digits of this sequence are the same as those of A070403 (digital root of 7^n) but the order is different: [1, 4, 7] vs. [1, 7, 4].

The digits in this sequence appear in the decimal expansions of the following rational numbers: 49/333, 490/333, 4900/333, .... (End)

Examples

			4^2 = 16, digitalroot(16) = 7, the third entry.
		

References

• Cecil Balmond, Number 9: The Search for the Sigma Code. Munich, New York: Prestel (1998): 203.

Links

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

Crossrefs

Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 5, A070366; c = 7, A070403; c = 8, A010689.

Cf. A000302, A001022, A009966, A009975, A009984, A010872, A010888, A070403, A087752, A121013, A141725, A016777.

Programs

Mathematica

Table[PowerMod[4, n, 9], {n, 0, 100}] (* Timothy L. Tiffin, Dec 03 2023 *)
StringRepeat["1, 4, 7, ", 100] (* Timothy L. Tiffin, Dec 03 2023 *)

PARI

a(n)=[1,4,7][1+n%3]; \\ Joerg Arndt, Aug 26 2014

Sage

[power_mod(4, n, 9) for n in range(0, 105)] # Zerinvary Lajos, Nov 25 2009

Formula

a(n) = 4^n mod 9. - Zerinvary Lajos, Nov 25 2009

From R. J. Mathar, Apr 13 2010: (Start)

a(n) = a(n-3) for n>2.

G.f.: (1+4*x+7*x^2)/ ((1-x)*(1+x+x^2)). (End)

a(n) = A010888(A000302(n)). - Michel Marcus, Aug 25 2014

a(n) = 3*A010872(n) + 1. - Robert Israel, Aug 25 2014

a(n) = 4 - 3*cos(2*n*Pi/3) - sqrt(3)*sin(2*n*Pi/3). - Wesley Ivan Hurt, Jun 30 2016

a(n) = A153130(2n). - Timothy L. Tiffin, Dec 01 2023

a(n) = A010888(A001022(n)) = A010888(A009966(n)) = A010888(A009975(n)) = A010888(A009984(n)) = A010888(A087752(n)) = A010888(A121013(n)). - Timothy L. Tiffin, Dec 02 2023

a(n) = A010888(4*a(n-1)). - Stefano Spezia, Mar 20 2025