OFFSET
1,1
COMMENTS
If a(n) = 0, a(n*k) = 0 for any positive k. - Franklin T. Adams-Watters, Nov 03 2009
25 is impossible; its multiples end either with the digits 00 or 50.
Multiples of 16 except 16 and 80 are impossible. Of the 625 multiples of 16 mod 10000, none are 246, 2468, 4680, 6802, or 8024. That leaves only 80 as a possible value for multiples of 16. It appears that the multiples of 16 and 25 are the only numbers for which a(n) = 0 - Franklin T. Adams-Watters, Nov 03 2009
LINKS
Franklin T. Adams-Watters, Table of n, a(n) for n = 1..1000.
EXAMPLE
a(7) = 7*6686 = 46802 and this number has increasing larger even digits (mod 10). a(12) = 24 = 12*2 has increasing even digits.
MATHEMATICA
f[n_] := Block[{x = 0, a = IntegerDigits[n], i = 1}, l =Length[a]; While < l, If[ Mod[ a[[i]] + 2, 10] != a + 1]], x = 1]; i++ ]; Return[x]]; Dock = n; While[ Union[ even[ IntegerDigits[k]]] != {True} || Fmk] == 1, k += n]; Print[k], {n, 1, 20}]
PROG
(PARI) evenincr(n)=local(d, r); d=n%4*2+2; n\=4; r=0; for(k=0, n, r=r*10+(d+2*k)%10); r
a(n)=if(n%25==0 || (n%16==0&80%n!=0), 0, k=0; while(evenincr(k)%n!=0, k++); evenincr(k)) /* This program will loop if the conjecture above is incorrect. */
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Amarnath Murthy, Jun 28 2001
EXTENSIONS
Edited and extended by Robert G. Wilson v, Feb 22 2002
Edited and extended by Franklin T. Adams-Watters, Nov 03 2009
STATUS
approved