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A112678
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Sum of digits of previous 5 terms.
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0
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,6
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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.
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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)).
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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.
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CROSSREFS
| Cf. A000322, A001591, A004090, A007953, A010888, A030132, A112661.
Sequence in context: A055566 A153610 A091812 * A021171 A011493 A077771
Adjacent sequences: A112675 A112676 A112677 * A112679 A112680 A112681
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KEYWORD
| base,easy,nonn
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AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), Dec 30 2005
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