A155754 A variation on 10^n mod 19.

1, -9, 5, -7, 6, 3, -8, -4, -2, -1, 9, -5, 7, -6, -3, 8, 4, 2, 1, -9, 5, -7, 6, 3, -8, -4, -2, -1, 9, -5, 7, -6, -3, 8, 4, 2, 1, -9, 5, -7, 6, 3, -8, -4, -2, -1, 9, -5, 7, -6, -3, 8, 4, 2, 1, -9, 5, -7, 6, 3, -8, -4, -2, -1, 9, -5, 7, -6, -3, 8, 4, 2

Offset: 0

Author

Ferruccio Guidi (fguidi(AT)cs.unibo.it), Jan 26 2009

Keywords

• easy
• sign

Comments

This sequence can be employed in a test for divisibility by 19 and works like A033940 works for 7.

The use of negative coefficients ensures the termination of the test because the modulus of the intermediate sum at each step of the test decreases strictly.

The test is successful if the final sum is 0.

The negative coefficients have the form (10^n mod 19) - 19 when 10^n mod 19 > 9.

Example: 8284 is divisible by 19 since |4*1 + 8*(-9) + 2*5 + 8*(-7)| = 114 and 4*1 + 1*(-9) + 1*5 = 0.

Links

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

Crossrefs

Cf. A033940, A119910, A117378.

Formula

a(n) = -a(n-9). G.f.: (-2*x^8-4*x^7-8*x^6+3*x^5+6*x^4-7*x^3+5*x^2-9*x+1) / (x^9+1). [Colin Barker, Feb 14 2013]