A112678 Sum of digits of previous 5 terms.

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, 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, 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

Offset

0,6

Comments

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.

Links

Table of n, a(n) for n=0..89.

Formula

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)).

Example

a(0)=a(1)=a(2)=a(3)=a(4)=1.
a(5) = SOD(1+1+1+1+1) = SOD(5) = 5.
a(6) = SOD(1+1+1+1+5) = SOD(9) = 9.
a(7) = SOD(1+1+1+5+9) = SOD(17) = 8.
a(8) = SOD(1+1+5+9+8) = SOD(24) = 6.
a(9) = SOD(1+5+9+8+6) = SOD(29) = 11, note that we do not iterate SOD to reduce 11 to 2.

Crossrefs

Cf. A000322, A001591, A004090, A007953, A010888, A030132, A112661.

Sequence in context: A308577 A217249 A091812 * A021171 A011493 A254345

Adjacent sequences: A112675 A112676 A112677 * A112679 A112680 A112681

Keyword

base,easy,nonn

Author

Jonathan Vos Post, Dec 30 2005

Status

approved