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A235700 a(n+1) = a(n) + (a(n) mod 5), a(1)=1.

%I A235700 #37 Feb 22 2025 14:53:20
%S A235700 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 A235700 61,62,64,68,71,72,74,78,81,82,84,88,91,92,94,98,101,102,104,108,111,
%U A235700 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 A235700 a(n+1) = a(n) + (a(n) mod 5), a(1)=1.
%C A235700 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 A235700 Harvey P. Dale, <a href="/A235700/b235700.txt">Table of n, a(n) for n = 1..1000</a>
%H A235700 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-2,2,-1).
%F A235700 a(n) = 2^(n-1 mod 4) + 10*floor((n-1)/4).
%F A235700 From _Colin Barker_, Jan 16 2014: (Start)
%F A235700 a(n) = (-10+(1+2*i)*(-i)^n+(1-2*i)*i^n+10*n)/4 where i=sqrt(-1).
%F A235700 a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4).
%F A235700 G.f.: x*(2*x^3+2*x^2+1) / ((x-1)^2*(x^2+1)). (End)
%F A235700 E.g.f.: (4 + 5*exp(x)*(x - 1) + cos(x) + 2*sin(x))/2. - _Stefano Spezia_, Feb 22 2025
%t A235700 NestList[#+Mod[#,5]&,1,80] (* _Harvey P. Dale_, Oct 20 2024 *)
%o A235700 (PARI) is_A235700(n) = bittest(278,n%10) \\ 278=2^1+2^2+2^4+2^8
%o A235700 (PARI) A235700 = n -> 2^((n-1)%4)+(n-1)\4*10
%o A235700 (PARI) print1(a=1);for(i=1,99,print1(","a+=a%5))
%o A235700 (PARI) Vec(x*(2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ _Colin Barker_, Jan 16 2014
%Y A235700 Cf. A001651, A007612, A047235, A047350, A102039.
%K A235700 nonn,easy
%O A235700 1,2
%A A235700 _M. F. Hasler_, Jan 14 2014