OFFSET
0,3
COMMENTS
Starting on the right, sum digits after doubling alternating digits beginning with the second. If doubled digit >9, reduce by 9 (sum of digits).
a(n) = A007953(A249873(n)); A093019(n) = 10 - a(10*n) mod 10 if less than 10, otherwise 0. - Reinhard Zumkeller, Nov 08 2014
First differences are b(n) defined for n>0 as follows. Take the prime factorization of n and let x be the number of 2's, y be the number of 5's, and z be min(x,y). If z is even, b(n) = 1 - 9*z. If z is odd and y=z, b(n) = 2 - 9*z. If z is odd and y>z, b(n) = -7 - 9*z. Now a(n) = a(n-1) + b(n). - Mathew Englander, Aug 04 2021
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
John Kilgo, Using the Luhn Algorithm, DotNetJohn.com.
Webopedia, Luhn formula
Wikipedia, Luhn algorithm
FORMULA
a(0)=0; for n not divisible by 10, a(n)=1+a(n-1); for n divisible by 10 but not 50, a(n)=2+a(n-10); for n divisible by 50 but not 100, a(n)=1+a(n-50); for n divisible by 100, a(n)=a(n/100). - Mathew Englander, Aug 04 2021
EXAMPLE
a(18) = 2*1 + 8 = 10.
a(59) = (1+0) + 9 = 10 (1 and 0 are the digits in 10 = 2*5).
PROG
(Haskell)
a093017 n = if n == 0 then 0 else a093017 n' + a007953 (2 * t) + d
where (n', td) = divMod n 100; (t, d) = divMod td 10
-- Reinhard Zumkeller, Nov 08 2014
(Python)
def a(n):
s = str(n)
r = s[::-1]
x = sum(int(d) for d in r[::2])
x += sum(q if (q:=2*int(d)) < 10 else q-9 for d in r[1::2])
return x
print([a(n) for n in range(87)]) # Michael S. Branicky, Jul 23 2024
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
Ray Chandler, Apr 03 2004
STATUS
approved