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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

Table of n, a(n) for n=0..108.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

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: A137269 A112201 A112203 * A132798 A080425 A048141

Adjacent sequences:  A196276 A196277 A196278 * A196280 A196281 A196282

KEYWORD

sign,easy

AUTHOR

Philippe Deléham, Sep 30 2011

STATUS

approved

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Last modified March 6 20:37 EST 2021. Contains 341850 sequences. (Running on oeis4.)