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%I #16 Jun 17 2018 14:57:29
%S 0,0,0,0,1,0,0,0,1,-4,0,0,1,-3,6,0,1,-2,3,-4,1,-1,1,-1,2,0,0,0,1,-8,0,
%T 0,1,-7,22,0,1,-6,15,-28,1,-5,9,-13,18,-4,4,-4,5,-11,0,0,1,-6,24,0,1,
%U -5,18,-46,1,-4,13,-28,50,-3,9,-15,22,-33,6,-6,7
%N Quintisection a(5n+k) gives k-th differences of a for k=0..4 with a(n)=0 for n<4 and a(4)=1.
%H Alois P. Heinz, <a href="/A229655/b229655.txt">Table of n, a(n) for n = 0..15625</a>
%F a(5*n) = a(n),
%F a(5*n+1) = a(n+1) - a(n),
%F a(5*n+2) = a(n+2) - 2*a(n+1) + a(n),
%F a(5*n+3) = a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n),
%F a(5*n+4) = a(n+4) - 4*a(n+3) + 6*a(n+2) - 4*a(n+1) + a(n).
%p a:= proc(n) option remember; local m, q;
%p m:= irem(n, 5, 'q'); `if`(n<5, `if`(n=4, 1, 0),
%p add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m))
%p end:
%p seq(a(n), n=0..100);
%t a[n_] := a[n] = Module[{ m, q}, {q, m} = QuotientRemainder[n, 5]; If[n < 5, If[n == 4, 1, 0], Sum[a[q + m - j]*(-1)^j*Binomial[m, j], {j, 0, m}]]];
%t Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Jun 09 2018, from Maple *)
%Y Cf. A005590, A229653, A229654, A229656, A229657, A229658, A229659, A229660.
%K sign,look,eigen
%O 0,10
%A _Alois P. Heinz_, Sep 27 2013
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