A076309 a(n) = floor(n/10) - 2*(n mod 10).

0, -2, -4, -6, -8, -10, -12, -14, -16, -18, 1, -1, -3, -5, -7, -9, -11, -13, -15, -17, 2, 0, -2, -4, -6, -8, -10, -12, -14, -16, 3, 1, -1, -3, -5, -7, -9, -11, -13, -15, 4, 2, 0, -2, -4, -6, -8, -10, -12, -14, 5, 3, 1, -1, -3, -5, -7, -9, -11, -13, 6, 4, 2, 0, -2, -4, -6, -8, -10

Offset

0,2

Comments

Delete the last digit from n and subtract twice this digit from the shortened number. - N. J. A. Sloane, May 25 2019

(n==0 modulo 7) iff (a(n)==0 modulo 7); applied recursively, this property provides a useful test for divisibility by 7.

References

Erdős, Paul, and János Surányi. Topics in the Theory of Numbers. New York: Springer, 2003. Problem 6, page 3.

Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Eric Weisstein's World of Mathematics, Divisibility Tests.

Wikipedia, Divisibility rule

Formula

From R. J. Mathar, Nov 23 2010: (Start)

a(n) = a(n-1) + a(n-10) - a(n-11).

G.f.: x*(-2 -2*x -2*x^2 -2*x^3 -2*x^4 -2*x^5 -2*x^6 -2*x^7 -2*x^8 +19*x^9)/((1+x)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)*(x-1)^2). (End)

Example

695591 is not a multiple of 7, as 695591 -> 69559-2*1=69557 -> 6955-2*7=6941 -> 694-2*1=692 -> 69-2*2=65=7*9+2, therefore the answer is NO.
Is 3206 divisible by 7? 3206 -> 320-2*6=308 -> 30-2*8=14=7*2, therefore the answer is YES, indeed 3206=2*7*229.

Mathematica

Table[Floor[n/10] - 2*Mod[n, 10], {n, 0, 100}] (* G. C. Greubel, Apr 07 2016 *)

Prog

(Haskell)
a076309 n =  n' - 2 * m where (n', m) = divMod n 10
-- Reinhard Zumkeller, Jun 01 2013
(PARI) a(n) = n\10 - 2*(n % 10); \\ Michel Marcus, Apr 07 2016

Crossrefs

Cf. A008589, A076310, A076311, A076312.

Sequence in context: A088116 A100817 A074157 * A088133 A115299 A076312

Adjacent sequences: A076306 A076307 A076308 * A076310 A076311 A076312

Keyword

sign

Author

Reinhard Zumkeller, Oct 06 2002

Status

approved