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A076311
a(n) = floor(n/10) - 5*(n mod 10).
8
0, -5, -10, -15, -20, -25, -30, -35, -40, -45, 1, -4, -9, -14, -19, -24, -29, -34, -39, -44, 2, -3, -8, -13, -18, -23, -28, -33, -38, -43, 3, -2, -7, -12, -17, -22, -27, -32, -37, -42, 4, -1, -6, -11, -16, -21, -26, -31, -36, -41, 5, 0, -5, -10, -15, -20, -25, -30, -35
OFFSET
0,2
COMMENTS
(n==0 modulo 17) iff (a(n)==0 modulo 17); applied recursively, this property provides a divisibility test for numbers given in base 10 notation.
REFERENCES
Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.
LINKS
Eric Weisstein's World of Mathematics, Divisibility Tests.
Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).
FORMULA
a(n)= +a(n-1) +a(n-10) -a(n-11). G.f. x *(-5-5*x-5*x^2-5*x^3-5*x^4-5*x^5-5*x^6-5*x^7-5*x^8+46*x^9) / ( (1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) *(x-1)^2 ). - R. J. Mathar, Feb 20 2011
EXAMPLE
12808 is not a multiple of 17, as 12808 -> 1280-5*8=1240 -> 124-5*0=124 -> 12-5*4=-8=17*(-1)+9, therefore the answer is NO.
Is 9248 divisible by 17? 9248 -> 924-5*8=884 -> 88-5*4=68=17*4, therefore the answer is YES.
MATHEMATICA
Table[Floor[n/10]-5Mod[n, 10], {n, 0, 60}] (* or *) LinearRecurrence[ {1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, -5, -10, -15, -20, -25, -30, -35, -40, -45, 1}, 60] (* Harvey P. Dale, Dec 21 2014 *)
PROG
(Haskell)
a076311 n = n' - 5 * m where (n', m) = divMod n 10
-- Reinhard Zumkeller, Jun 01 2013
(Magma) [Floor(n/10)-5*(n mod 10): n in [0..50]]; // Vincenzo Librandi, Jun 23 2015
(PARI) a(n)=n\10 - n%10*5 \\ Charles R Greathouse IV, Oct 07 2015
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Reinhard Zumkeller, Oct 06 2002
STATUS
approved