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A094405
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a(1) = 1; a(n) = (sum of previous terms) mod n.
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1
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1, 1, 2, 0, 4, 2, 3, 5, 0, 8, 4, 6, 10, 4, 5, 7, 11, 1, 17, 11, 18, 10, 15, 1, 21, 11, 16, 26, 17, 27, 16, 24, 7, 5, 1, 29, 13, 17, 25, 1, 33, 15, 20, 30, 5, 45, 33, 7, 2, 42, 22, 32, 52, 38, 8, 2, 47, 23, 32, 50, 25, 35, 55, 31, 46, 10, 3, 57, 29, 41, 65, 41, 64, 36, 53, 11, 2, 62, 26
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Theorem. For all values of n>=397, a(n)=97. Proof. Let s(n) denote Sum[a(i), i=1..n-1]. Calculation shows that s(397)=38606=397*97+97. Thus a(397)=397*97+97 mod 397=97. Then s(398)=s(397)+97=398*97+97, giving a(398)=97. A simple inductive argument shows that a(397+k)=97 for all integers k>=0. - John W. Layman (layman(AT)math.vt.edu), Jun 07 2004
Conjecture: For any seed a(1) the sequence "a(n) = (sum of previous terms) mod n" ends with repeating constant. This is true for a(1) = 1,...,941. - Zak Seidov (zakseidov(AT)yahoo.com), Feb 24 2006
Essentially the same as A066910. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 05 2008]
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EXAMPLE
| a(4) = 0 because the previous terms 1, 1, 2 sum to 4 and 4 mod 4 is 0. a(5) = 4 because the previous terms 1, 1, 2, 0 sum to 4 and 4 mod 5 is 4.
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MAPLE
| L := [1]; s := 1; p := 2; while (nops(L) < 90) do; if 1>0 then; t := s mod p; L := [op(L), t]; s := s+t; p := p+1; fi; od; L;
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CROSSREFS
| Sequence in context: A001100 A136265 A066910 * A155984 A028609 A107490
Adjacent sequences: A094402 A094403 A094404 * A094406 A094407 A094408
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KEYWORD
| nonn
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AUTHOR
| Chuck Seggelin (seqfan(AT)plastereddragon.com), Jun 03 2004
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