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A103196
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a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).
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3
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1, 2, 3, 8, 13, 30, 55, 116, 225, 458, 907, 1824, 3637, 7286, 14559, 29132, 58249, 116514, 233011, 466040, 932061, 1864142, 3728263, 7456548, 14913073, 29826170, 59652315, 119304656, 238609285, 477218598, 954437167
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OFFSET
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0,2
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COMMENTS
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A floretion-generated sequence relating to the Jacobsthal sequence A001045 as well as to A095342 (Number of elements in n-th string generated by a Kolakoski(5,1) rule starting with a(1)=1). (a(n)) may be seen as the result of a certain transform of the natural numbers (see program code).
Floretion Algebra Multiplication Program, FAMP Code: 4jesleftforseq[A*B] with A = + 'i + 'j + i' + j' + 'ii' + 'jj' + 'ij' + 'ji' + e and B = - .25'i + .25'j + .25'k + .25i' - .25j' + .25k' - .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' - .25'ki' - .25'kj' - .25e; 1vesforseq[A*B](n) = n, ForType: 1A.
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LINKS
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FORMULA
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a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)phi(phi(3^k))}; a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)(2*3^k/9+C(1, k)/3+4*C(0, k)/9)}; a(n)=sum{k=0..n+2, J(n-k+3)((-1)^(k+1)-2C(1, k)+4C(0, k))} where J(n)=A001045(n); a(n)=A113954(n+2). - Paul Barry, Nov 09 2005
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MATHEMATICA
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Table[(2^(n+3)-(-1)^n (3n-1))/9, {n, 0, 30}] (* or *) LinearRecurrence[ {0, 3, 2}, {1, 2, 3}, 40] (* Harvey P. Dale, Jul 09 2018 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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