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A073117 a(n+1) = a(n) + a(n) mod n; a(1) = 1. 4

%I

%S 1,1,2,4,4,8,10,13,18,18,26,30,36,46,50,55,62,73,74,91,102,120,130,

%T 145,146,167,178,194,220,237,264,280,304,311,316,317,346,359,376,401,

%U 402,435,450,470,500,505,550,583,590,592,634,656,688,740,778

%N a(n+1) = a(n) + a(n) mod n; a(1) = 1.

%C 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

%e 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.

%t s=1;lst={s};Do[s+=Mod[s, n];AppendTo[lst, s], {n, 1, 6!, 1}];lst [From _Vladimir Joseph Stephan Orlovsky_, Nov 07 2008]

%K nonn

%O 1,3

%A _Reinhard Zumkeller_, Aug 19 2002

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Last modified November 28 09:42 EST 2014. Contains 250308 sequences.