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

1, 2, 4, 8, 11, 12, 14, 18, 21, 22, 24, 28, 31, 32, 34, 38, 41, 42, 44, 48, 51, 52, 54, 58, 61, 62, 64, 68, 71, 72, 74, 78, 81, 82, 84, 88, 91, 92, 94, 98, 101, 102, 104, 108, 111, 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

Offset

1,2

Comments

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.

Links

Table of n, a(n) for n=1..77.

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Formula

a(n) = 2^(n-1 mod 4) + 10*floor((n-1)/4).

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

Prog

(PARI) is_A235700(n) = bittest(278, n%10) \\ 278=2^1+2^2+2^4+2^8
(PARI) A235700 = n -> 2^((n-1)%4)+(n-1)\4*10
(PARI) print1(a=1); for(i=1, 99, print1(", "a+=a%5))
(PARI) Vec(x*(2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ Colin Barker, Jan 16 2014

Crossrefs

Sequence in context: A256980 A014425 A028889 * A174781 A028846 A274924

Adjacent sequences: A235697 A235698 A235699 * A235701 A235702 A235703

Keyword

nonn,easy

Author

M. F. Hasler, Jan 14 2014

Status

approved