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%I #5 Mar 30 2012 18:40:30
%S 1,1,1,1,1,5,9,8,6,11,12,10,11,5,13,6,9,8,5,5,6,6,3,7,9,4,11,7,11,6,
%T 12,11,11,6,10,5,7,12,4,11,12,10,13,5,6,10,8,6,8,11,7,4,9,12,7,12,8,
%U 12,6,9,11,10,12,12,9,9,7,13,5,7,5,10,4,4,3,8,11,3,11,9,6,4,6,9,7,5,4,4,11,4
%N Sum of digits of previous 5 terms.
%C This is to the pentanacci sequence A001591 as A112661 is to the tribonacci and as A030132 is to Fibonacci. A000322 is the pentanacci sequence (A001591) but starting with values (1,1,1,1,1). Andrew Carmichael Post (andrewpost(AT)gmail.com) wrote the program that generated this sequence and showed that for any 5 initial integers a(0),a(1),a(2),a(3),a(4) the length of the cycle eventually entered is a factor of 2184. For the SOD(teranacci) the limit cycle length is always a factor of 312. For the SOD(tribonacci) which is A112661, the length of any cycle eventually entered is a factor of 78.
%F a(0)=a(1)=a(2)=a(3)=a(4)=1. a(n) = SumDigits(a(n-1)+a(n-2)+a(n-3)+a(n-4)+a(n-5)). a(n) = SumDigits(A000322(n)).
%e a(0)=a(1)=a(2)=a(3)=a(4)=1.
%e a(5) = SOD(1+1+1+1+1) = SOD(5) = 5.
%e a(6) = SOD(1+1+1+1+5) = SOD(9) = 9.
%e a(7) = SOD(1+1+1+5+9) = SOD(17) = 8.
%e a(8) = SOD(1+1+5+9+8) = SOD(24) = 6.
%e a(9) = SOD(1+5+9+8+6) = SOD(29) = 11, note that we do not iterate SOD to reduce 11 to 2.
%Y Cf. A000322, A001591, A004090, A007953, A010888, A030132, A112661.
%K base,easy,nonn
%O 0,6
%A _Jonathan Vos Post_, Dec 30 2005
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