A073117 a(n+1) = a(n) + a(n) mod n; a(1) = 1.

1, 1, 2, 4, 4, 8, 10, 13, 18, 18, 26, 30, 36, 46, 50, 55, 62, 73, 74, 91, 102, 120, 130, 145, 146, 167, 178, 194, 220, 237, 264, 280, 304, 311, 316, 317, 346, 359, 376, 401, 402, 435, 450, 470, 500, 505, 550, 583, 590, 592, 634, 656, 688, 740, 778

Offset: 1

Author

Reinhard Zumkeller, Aug 19 2002

Keywords

• nonn

Comments

Conjecture (seems provable): More generally let a and b(1) be integers. If b(n+1) = b(n) + b(n) (mod(n+a)) there is an integer x(a,b(1)) such that b(n+1) = b(n) + x(a,b(1)) for n sufficiently large. We have x(0,1) = x(1,1) = x(2,1) = 97, x(3,1) = 1, x(4,1) = 3, x(5,1) = 3, x(6,1) = 6, ..., x(97,1) = 43, x(0,11) = 2, etc. - Benoit Cloitre, Aug 20 2002

Examples

			a(397) = 38606 = 2*97*199 = (2*199)*97 = 398*97 = (397+1)*97; a(397) mod 397 = (397*97 + 97) mod 397 = 97, a(398) = a(397) + a(397) mod 397 = (397+1)*97 + 97 = (398+1)*97, etc.: a(n+1) = a(n) + 97 for n >= 397.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A066910. - Rémy Sigrist, Mar 24 2017

Programs

Mathematica

s=1;lst={s};Do[s+=Mod[s, n];AppendTo[lst, s], {n, 1, 6!, 1}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 07 2008 *)