A235699 a(n+1) = a(n) + (a(n) mod 10) + 1, a(0) = 0.

0, 1, 3, 7, 15, 21, 23, 27, 35, 41, 43, 47, 55, 61, 63, 67, 75, 81, 83, 87, 95, 101, 103, 107, 115, 121, 123, 127, 135, 141, 143, 147, 155, 161, 163, 167, 175, 181, 183, 187, 195, 201, 203, 207, 215, 221, 223, 227, 235, 241, 243, 247, 255, 261, 263, 267, 275, 281, 283, 287, 295, 301, 303, 307, 315, 321, 323, 327, 335, 341, 343, 347, 355

Offset

0,3

Comments

Instead of (a(n) mod 10) one might say "the last (decimal) digit of a(n)".

Apart from the initial term, the first differences form the periodic sequence (2,4,8,6)[repeated].

Without the final "+ 1" and starting with 1, one gets A102039: Indeed, the last digit cycles through 2,4,8,6 and therefore the sequence never becomes constant.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

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

Formula

a(n) = 5*n-6+cos(n*Pi/2)+2*sin(n*Pi/2), for n>0. - Giovanni Resta, Jan 15 2014

From Colin Barker, Jan 16 2014: (Start)

a(n) = -6+(1/2+i)*(-i)^n+(1/2-i)*i^n+5*n for n>0 where i=sqrt(-1).

a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>4.

G.f.: x*(5*x^3+3*x^2+x+1) / ((x-1)^2*(x^2+1)). (End)

Mathematica

NestList[#+Mod[#, 10]+1&, 0, 80] (* or *) Join[{0}, LinearRecurrence[{2, -2, 2, -1}, {1, 3, 7, 15}, 80]] (* Harvey P. Dale, Dec 21 2014 *)

Prog

(PARI) print1(a=0); for(i=1, 99, print1(", "a+=a%10+1))

Crossrefs

Cf. A102039, A235698, A045844.

Sequence in context: A350696 A144751 A138847 * A077777 A153829 A153830

Adjacent sequences: A235696 A235697 A235698 * A235700 A235701 A235702

Keyword

nonn,base,easy

Author

M. F. Hasler, Jan 14 2014

Status

approved