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A107354
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To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).
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14
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1, 1, 2, 7, 44, 516, 11622, 512022, 44588536, 7718806044, 2664170119608, 1836214076324153, 2529135272371085496, 6964321029630556852944, 38346813253279804426846032, 422247020982575523983378003936, 9298487213328788062025571134762096
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OFFSET
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0,3
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COMMENTS
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Can also be computed summing forwards:
1
1,1
1,2,2, 2
1,3,5, 7, 7, 7, 7, 7
1,4,9,16,23,30,37,44,44,44,44,44,44,44,44,44
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LINKS
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FORMULA
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a(n) = C(2^(n-1)+n-2,n-1) - Sum_{k=1..n-2} a(k)*C(2^(n-1)-2^k+n-k-1,n-k) for n>=2, with a(0)=1, a(1)=1, where C = binomial. - Paul D. Hanna, May 24 2005
The first number in row 3 is 2^(n-2)+1. - Ralf Stephan, May 24 2005
G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-1) (g.f. of subpartitions). - Paul D. Hanna, Jul 03 2006
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n+n). - Paul D. Hanna, Jul 03 2006
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EXAMPLE
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For n=4, the array looks like this:
1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
........................1..2..3..4..5..6..7..8
....................................5.11.18.26
.........................................18.44
............................................44
Therefore a(4)=44.
For n=5, we can illustrate the recurrence by:
a(5) = 516 = C(19, 4) - ( 1*C(17, 4) + 2*C(14, 3) + 7*C(9, 2) ) = C(16+4-1, 4) - ( 1*C(16-2+4-1, 4) + 2*C(16-4+3-1, 3) + 7*C(16-8+2-1, 2) ).
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MAPLE
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a:= proc(n) option remember; `if`(n=0, 1, -add(
a(j)*(-1)^(n-j)*binomial(2^j, n-j), j=0..n-1))
end:
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MATHEMATICA
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f[n_] := If[n == 0, 1, Binomial[2^(n - 1) + n - 2, n - 1] - Sum[ f[k]*Binomial[2^(n - 1) - 2^k + n - k - 1, n - k], {k, n - 2}]]; Table[ f[n], {n, 0, 15}] (* Robert G. Wilson v, May 25 2005 *)
Table[NestWhile[Accumulate[Drop[#, Ceiling[Length[#]/2]]]&, PadRight[{}, 2^n+1, 1], Length[ #]> 1&], {n, 0, 16}]//Flatten (* Harvey P. Dale, Jun 24 2018 *)
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PROG
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(PARI) {a(n)=if(n==0, 1, binomial(2^(n-1)+n-2, n-1)- sum(k=1, n-2, a(k)*binomial(2^(n-1)-2^k+n-k-1, n-k)))} \\ Paul D. Hanna, May 24 2005
(PARI) {a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-1) ), n)} \\ Paul D. Hanna, May 24 2005
(Haskell)
a107354 n = head $ snd $ until ((== 1) . fst)
f (2^n, replicate (2^n) 1) where
f (len, xs) = (len', scanl1 (+) $ drop len' xs) where
len' = len `div` 2
-- Feasible only for small n.
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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