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A229818
Even bisection gives sequence a itself, n->a(2*(3*n+k)-1) gives k-th differences of a for k=1..3 with a(n)=n for n<2.
9
0, 1, 1, -1, 1, -1, -1, 0, 1, -2, -1, 6, -1, -2, 0, 4, 1, -8, -2, 2, -1, -4, 6, 6, -1, -2, -2, 2, 0, -1, 4, 0, 1, 1, -8, -1, -2, 1, 2, 0, -1, -4, -4, 1, 6, -4, 6, 8, -1, -3, -2, 4, -2, 2, 2, 1, 0, 6, -1, -20, 4, 7, 0, -14, 1, 20, 1, -7, -8, 6, -1, -3, -2, -1
OFFSET
0,10
LINKS
FORMULA
a(2*n) = a(n),
a(6*n+1) = a(n+1) - a(n),
a(6*n+3) = a(n+2) - 2*a(n+1) + a(n),
a(6*n+5) = a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n).
MAPLE
a:= proc(n) option remember; local m, q, r;
m:= (irem(n, 6, 'q')+1)/2;
`if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),
add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))
end:
seq(a(n), n=0..100);
MATHEMATICA
a[n_] := a[n] = Module[{m, q, r, q2, r2}, {q, r} = QuotientRemainder[n, 6]; m = (r+1)/2; If[n<2, n, {q2, r2} = QuotientRemainder[n, 2]; If[r2 == 0, a[q2], Sum[a[q+m-j]*(-1)^j*Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Mar 08 2017, translated from Maple *)
KEYWORD
sign,look,hear,eigen,changed
AUTHOR
Alois P. Heinz, Sep 30 2013
STATUS
approved