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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

%I #26 Jul 03 2021 10:58:18

%S 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,

%T 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,

%U 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

%N 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.

%C Apparently a(9+n) = A194519(n).

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,0,0,0,1,-1).

%F 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

%F The conjecture above is correct. - _Charles R Greathouse IV_, Jan 04 2013

%e 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.

%p A196279 := proc(n)

%p r := (7*n) mod 10 ;

%p x := floor(7*n/10) ;

%p (x-2*r)/7 ;

%p end proc: # _R. J. Mathar_, Oct 04 2011

%o (PARI) a(n)=(7*n\10-7*n%10*2)/7 \\ _Charles R Greathouse IV_, Jan 04 2013

%K sign,base,easy

%O 0,2

%A _Philippe Deléham_, Sep 30 2011

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Last modified March 28 14:38 EDT 2024. Contains 371254 sequences. (Running on oeis4.)