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%I #26 Feb 10 2024 09:23:10
%S 1,2,4,8,11,12,14,18,21,22,24,28,31,32,34,38,41,42,44,48,51,52,54,58,
%T 61,62,64,68,71,72,74,78,81,82,84,88,91,92,94,98,101,102,104,108,111,
%U 112,114,118,121,122,124,128,131,132,134,138,141,142,144,148,151,152,154,158,161,162,164,168,171,172,174,178,181,182,184,188,191
%N a(n+1) = a(n) + (a(n) mod 5), a(1)=1.
%C Although the present sequence has not been thought of via "writing a(n) in base b", this could be seen as "base 5" version of A102039 (base 10) and A001651 (base 3), A047235 (base 6), A047350 (base 7) and A007612 (base 9). For 4 or 8 one would get a sequence constant from that (3rd resp. 4th) term on.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-1).
%F a(n) = 2^(n-1 mod 4) + 10*floor((n-1)/4).
%F a(n) = (-10+(1+2*i)*(-i)^n+(1-2*i)*i^n+10*n)/4 where i=sqrt(-1). a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4). G.f.: x*(2*x^3+2*x^2+1) / ((x-1)^2*(x^2+1)). - _Colin Barker_, Jan 16 2014
%o (PARI) is_A235700(n) = bittest(278,n%10) \\ 278=2^1+2^2+2^4+2^8
%o (PARI) A235700 = n -> 2^((n-1)%4)+(n-1)\4*10
%o (PARI) print1(a=1);for(i=1,99,print1(","a+=a%5))
%o (PARI) Vec(x*(2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ _Colin Barker_, Jan 16 2014
%K nonn,easy
%O 1,2
%A _M. F. Hasler_, Jan 14 2014
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