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FORMULA
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Construct an array of rows such that n-th row = partial sums of (n-1)-th row
of A010815: (1, -1, -1, 0, 0, 1, 0, 1,...).
A152537 = sums of antidiagonal terms of the array.
The sequence may be obtained directly from the following set of operations:
Our given sequence = A000041: (1, 1, 2, 3, 5, 7, 11,...). Delete the first
"1" then consider (1, 2, 3, 5, 7, 11,...) as an operator Q which we write in reverse with 1,2,3,...terms for each operation. Letting R = the target sequence (1,2,4,8,...); we begin a(0) = 1, a(1) = 1, then perform successive
operations of: "next term in (1,2,4,...) - dot product of Q*R" where Q is
written right to left and R (the ongoing result) written left to right).
Examples: Given 4 terms Q, R, we have: (5,3,2,1) dot (1,1,1,2) = (5+3+2+2) =
12, which we subtract from 16, = 4.
Given 5 terms of Q,R and A152537, we have (7,5,3,2,1) dot (1,1,1,2,4) = 23
which is subtracted from 32 giving 9. Continue with analogous operations to generate the series.
a(n)=sum_{j=0..n} A010815(j)*2^(n-j). G.f.: A000079(x)/A000041(x) = A010815(x)/(1-2x), where A......(x) denotes the g.f. of the associated sequence. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 09 2008]
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EXAMPLE
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a(5) = 9 = 32 - 23 = (32 - ((7,5,3,2,1) dot (1,1,1,2,4)))
(1,1,2,3) convolved with (1,1,1,2) = 8, where (1,1,2,3...) = the first four partition numbers.
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