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Search: id:A156346
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| A156346 |
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A154811=1,2,5,4,7,8,8,7,4,5,2,1 mod 9 leading to (positive or negative) 1's,2's,4's. Palindrom of period 12:repeat 1,2,-4,4,-2,-1,-1,-2,4,-4,2,1. |
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+0
3
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| 1, 2, -4, 4, -2, -1, -1, -2, 4, -4, 2, 1, 1, 2, -4, 4, -2, -1, -1, -2, 4, -4, 2, 1, 1, 2, -4, 4, -2, -1, -1, -2, 4, -4, 2, 1, 1, 2, -4, 4, -2, -1, -1, -2, 4, -4, 2, 1
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Also by (2*A154811=2,4,10,8,14,16,16,14,8,10,4,2 mod 9=2,4,1,8,5,7,7,5,8,1,4,2) - (A154811=1,2,5,4,7,8,8,7,4,5,2,1). See A156283=1,2,4,-4,-2,-1 from 1,2,4,5,7,8=A141425.
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FORMULA
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a(n)=(1/12)*{[(n+1) mod 12]-6*[(n+2) mod 12]+8*[(n+3) mod 12]-6*[(n+4) mod 12]+[(n+5) mod 12]-[(n+7) mod 12]+6*[(n+8) mod 12]-8*[(n+9) mod 12]+6*[(n+10) mod 12]-[(n+11) mod 12]}, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Feb 13 2009]
a(n)=(1/12)*{[(n+1) mod 12]-6*[(n+2) mod 12]+8*[(n+3) mod 12]-6*[(n+4) mod 12]+[(n+5) mod 12]-[(n+7) mod 12]+6*[(n+8) mod 12]-8*[(n+9) mod 12]+6*[(n+10) mod 12]-[(n+11) mod 12]}, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Feb 25 2009]
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CROSSREFS
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Sequence in context: A108620 A070512 A156283 this_sequence A126123 A096832 A016588
Adjacent sequences: A156343 A156344 A156345 this_sequence A156347 A156348 A156349
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KEYWORD
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uned,sign
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AUTHOR
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Paul Curtz (bpcrtz(AT)free.fr), Feb 08 2009
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