login
The OEIS is supported by the many generous donors to the OEIS Foundation.

 

Logo
Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A196279 Let r= (7n) mod 10 and x=floor(7n/10) be the last digit and leading part of 7n. Then a(n) = (x-2r)/7. 1
0, -2, -1, 0, -2, -1, 0, -2, -1, 0, 1, -1, 0, 1, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 8, 9, 10, 8, 9 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Apparently a(9+n) = A194519(n).
LINKS
FORMULA
Conjecture: a(n)= +a(n-1) +a(n-10) -a(n-11) with G.f. x*(-2 +x -2*x^3 +x^4 -2*x^6 +x^7 +x^9 +x^2 +x^5 +x^8) / ( (1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) *(x-1)^2 ). - R. J. Mathar, Oct 04 2011
The conjecture above is correct. - Charles R Greathouse IV, Jan 04 2013
EXAMPLE
Check to see if 273 is divisible by 7 : double the last digit 3*2=6 ; subtract that from the rest of the number 27-6=21 ; check to see if the difference is divisible by 7: 21/7 is divisible by 7, therefore 273 is also divisible by 7. 273=7*39 and 21=7*3 so a(39)=3.
MAPLE
A196279 := proc(n)
r := (7*n) mod 10 ;
x := floor(7*n/10) ;
(x-2*r)/7 ;
end proc: # R. J. Mathar, Oct 04 2011
PROG
(PARI) a(n)=(7*n\10-7*n%10*2)/7 \\ Charles R Greathouse IV, Jan 04 2013
CROSSREFS
Sequence in context: A343348 A112201 A112203 * A132798 A080425 A048141
KEYWORD
sign,base,easy
AUTHOR
Philippe Deléham, Sep 30 2011
STATUS
approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified July 15 07:46 EDT 2024. Contains 374324 sequences. (Running on oeis4.)