A033115 Base-5 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

1, 5, 26, 130, 651, 3255, 16276, 81380, 406901, 2034505, 10172526, 50862630, 254313151, 1271565755, 6357828776, 31789143880, 158945719401, 794728597005, 3973642985026, 19868214925130, 99341074625651, 496705373128255

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• base
• easy

Comments

Partial sums of A015531. - Mircea Merca, Dec 28 2010

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (5,1,-5).

Crossrefs

Cf. A015531.

Programs

Magma

[Round((5*5^n-3)/24): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011

Maple

seq(1/3*floor(5^(n+1)/8),n=1..32); # Mircea Merca, Dec 26 2010

Mathematica

Table[FromDigits[PadRight[{},n,{1,0}],5],{n,30}] (* or *) LinearRecurrence[ {5,1,-5},{1,5,26},30] (* Harvey P. Dale, Jan 28 2017 *)

Formula

a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3). - Joerg Arndt, Jan 08 2011

From Paul Barry, Nov 12 2003: (Start)

a(n) = floor(5^(n+2)/24);

a(n) = Sum_{k=0..floor(n/2)} 5^(n-2*k);

a(n) = Sum_{k=0..n} Sum_{j=0..k} (-1)^(j+k)*5^j.

Partial sums of A083425.

G.f.: 1/((1-x)*(1+x)*(1-5*x));

a(n) = 4*a(n-1) + 5*a(n-2) + 1. (End)

From Mircea Merca, Dec 28 2010: (Start)

a(n) = (1/3)*floor(5^(n+1)/8) = floor((5*5^n - 1)/24) = round((5*5^n - 3)/24) = round((5*5^n - 5)/24) = ceiling((5*5^n - 5)/24);

a(n) = a(n-2) + 5^(n-1), n > 1. (End)