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A107659
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a(n) = Sum_{k=0..n} 2^max(k, n-k).
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3
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1, 4, 10, 24, 52, 112, 232, 480, 976, 1984, 4000, 8064, 16192, 32512, 65152, 130560, 261376, 523264, 1047040, 2095104, 4191232, 8384512, 16771072, 33546240, 67096576, 134201344, 268410880, 536838144, 1073692672, 2147418112
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OFFSET
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0,2
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COMMENTS
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Define an infinite array by m(n,k) = 2^n-n+k for n>=k>=0 (in the lower left triangle) and by m(n,k) = 2^k+k-n for k>=n>=0 (in the upper right triangle). The antidiagonal sums of this array are a(n) = sum_{k=0..n} m(n-k,k). - J. M. Bergot, Aug 16 2013
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LINKS
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FORMULA
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a(2n) = 2^n(2^(n+2)-3), a(2n+1) = 2^n(2^(n+3)-4).
G.f.: (1+2*x)/[(1-2*x)*(1-2*x^2)].
a(0)=1, a(1)=4, a(2)=10, a(n)=2*a(n-1)+2*a(n-2)-4*a(n-3). - Harvey P. Dale, Nov 10 2013
a(n) = 2^(n+2) - (2 + mod(n+1, 2)) * 2^floor((n+1)/2). - Michael Somos, Jun 24 2018
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EXAMPLE
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G.f. = 1 + 4*x + 10*x^2 + 24*x^3 + 52*x^4 + 112*x^5 + 232*x^6 + 480*x^7 + ... - Michael Somos, Jun 24 2018
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MATHEMATICA
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Table[Sum[2^Max[k, n-k], {k, 0, n}], {n, 0, 30}] (* or *) LinearRecurrence[ {2, 2, -4}, {1, 4, 10}, 30] (* Harvey P. Dale, Nov 10 2013 *)
a[ n_] := 2^(n + 2) - (2 + Mod[n + 1, 2]) 2^Quotient[n + 1, 2]; (* Michael Somos, Jun 24 2018 *)
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PROG
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(PARI) {a(n) = 2^(n+2) - (2 + (n+1)%2) * 2^((n+1)\2)}; /* Michael Somos, Jun 24 2018 */
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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