A102039 a(n) = a(n-1) + last digit of a(n-1), starting at 1.

1, 2, 4, 8, 16, 22, 24, 28, 36, 42, 44, 48, 56, 62, 64, 68, 76, 82, 84, 88, 96, 102, 104, 108, 116, 122, 124, 128, 136, 142, 144, 148, 156, 162, 164, 168, 176, 182, 184, 188, 196, 202, 204, 208, 216, 222, 224, 228, 236, 242, 244, 248, 256, 262, 264, 268, 276, 282

Offset: 1

Author

Samantha Stones (devilsdaughter2000(AT)hotmail.com), Dec 25 2004

Keywords

• easy
• nonn
• base

Comments

Sequence A001651 is the "base 3" version. In base 4 this rule leads to (1,2,4,4,4...), in base 5 to (1,2,4,8,11,12,14,18,21,22,24,28...) = A235700. - M. F. Hasler, Jan 14 2014

This and the following sequences (none of which is "base"!) could all be defined by a(1) = 1 and a(n+1) = a(n) + (a(n) mod b) with different values of b: A001651 (b=3), A235700 (b=5), A047235 (b=6), A047350 (b=7), A007612 (b=9). Using b=4 or b=8 yields a constant sequence from that term on. - M. F. Hasler, Jan 15 2014

Examples

			28 + 8 = 36, 36 + 6 = 42.
		

Links

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

Crossrefs

Apart from initial term, same as A002081.

Programs

Mathematica

LinearRecurrence[{2,-2,2,-1},{1,2,4,8,16},60] (* Harvey P. Dale, Jul 02 2022 *)

PARI

print1(a=1);for(i=1,99,print1(","a+=a%10)) \\ M. F. Hasler, Jan 14 2014

PARI

Vec(x*(5*x^4+2*x^3+2*x^2+1)/((x-1)^2*(x^2+1)) + O(x^100)) \\ Colin Barker, Sep 20 2014

PARI

a(n) = if(n==1, 1, (-10-(1-I/2)*(-I)^n-(1+I/2)*I^n+5*n)) \\ Colin Barker, Oct 18 2015

Formula

a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>5. G.f.: x*(5*x^4+2*x^3+2*x^2+1) / ((x-1)^2*(x^2+1)). - Colin Barker, Sep 20 2014

a(n) = (-10 - (1-i/2)*(-i)^n - (1+i/2)*i^n + 5*n) for n>1, where i = sqrt(-1). - Colin Barker, Oct 18 2015