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Search: id:A103312
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| A103312 |
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A transform of the Jacobsthal numbers. |
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+0
1
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| 0, 1, 1, 1, 0, -3, -9, -18, -27, -27, 0, 81, 243, 486, 729, 729, 0, -2187, -6561, -13122, -19683, -19683, 0, 59049, 177147, 354294, 531441, 531441, 0, -1594323, -4782969, -9565938, -14348907, -14348907, 0, 43046721, 129140163, 258280326, 387420489, 387420489, 0, -1162261467, -3486784401
(list; graph; listen)
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OFFSET
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0,6
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COMMENT
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Apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) (expressed as Riordan arrays) to -A001045(n). All elements are multiples of a power of 3. - Ralf Stephan, Jan 28 2005
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FORMULA
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G.f.: x(1-x)^2/(1-3x+3x^2); a(n)=-sum{j=0..n, (-1)^j*C(n, j)*sum{k=0..floor(j/2), (-1)^k*C(n-k, k)A001045(j-2k)}}.
Recurrence: a(n+2) = 3a(n-1) - 3a(n), starting with 0, 1, 1, 1. - Ralf Stephan, Jan 28 2005
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PROGRAM
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(PARI) {a(n) = if( n<0, 0, polcoeff( x * (1 - x)^2 / (1 - 3*x + 3*x^2) + x * O(x^n), n))} /* Michael Somos Sep 29 2007 */
(PARI) {a(n) = if(n<2, n>0, 3^(n\2-1) * (-1)^((n+1)\6) * (1 + (-1)^((n-1)\3) * (n%3==1)))} /* Michael Somos Sep 29 2007 */
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CROSSREFS
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Cf. A057681.
Sequence in context: A030784 A123877 A057681 this_sequence A100967 A134479 A045943
Adjacent sequences: A103309 A103310 A103311 this_sequence A103313 A103314 A103315
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KEYWORD
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easy,sign
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), Jan 30 2005
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