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

Author

M. F. Hasler, Jan 14 2014

Keywords

• nonn
• easy

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

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Crossrefs

Cf. A001651, A007612, A047235, A047350, A102039.

Programs

Mathematica

NestList[#+Mod[#,5]&,1,80] (* Harvey P. Dale, Oct 20 2024 *)

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

Formula

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

From Colin Barker, Jan 16 2014: (Start)

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)). (End)

E.g.f.: (4 + 5*exp(x)*(x - 1) + cos(x) + 2*sin(x))/2. - Stefano Spezia, Feb 22 2025