OFFSET
0,2
COMMENTS
(n==0 modulo 13) iff (a(n)==0 modulo 13); 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
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Divisibility Tests.
Wikipedia, Divisibility rule
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*(-4-4*x-4*x^2-4*x^3-4*x^4-4*x^5-4*x^6-4*x^7-4*x^8+35*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
435598 is not a multiple of 13, as 435598 -> 43559+4*8=43591 -> 4359+4*1=4363 -> 436+4*3=448 -> 44+4*8=76 -> 7+4*6=29=13*2+3, therefore the answer is NO.
Is 8424 divisible by 13? 8424 -> 842+4*4=858 -> 85+4*8=117 -> 11+4*7=39=13*3, therefore the answer is YES.
MAPLE
MATHEMATICA
Table[Floor[n/10] + 4*Mod[n, 10], {n, 0, 100}] (* Wesley Ivan Hurt, Jan 30 2014 *)
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 1}, 80] (* Harvey P. Dale, Sep 30 2015 *)
PROG
(Haskell)
a076310 n = n' + 4 * m where (n', m) = divMod n 10
-- Reinhard Zumkeller, Jun 01 2013
(PARI) a(n) = n\10 + 4*(n % 10); \\ Michel Marcus, Jan 31 2014
(Magma) [Floor(n/10)+4*(n mod 10): n in [0..75]]; // Vincenzo Librandi, Feb 27 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Oct 06 2002
STATUS
approved