A143629 Define E(n) = Sum_{k>=0} (-1)^floor(k/3)*k^n/k! for n = 0,1,2,... . Then E(n) is an integral linear combination of E(0), E(1) and E(2). This sequence lists the coefficients of E(1).
0, 1, 0, -2, -7, -23, -80, -271, -750, -647, 13039, 152011, 1232583, 8750796, 57405464, 349329354, 1899818951, 8008845556, 5981853002, -425732481925, -7285403175563, -89895756043392, -970910901819211, -9663021449412616
Offset: 0
Examples
E(n) as linear combination of E(i), i = 0..2. ==================================== ..E(n)..|.....E(0).....E(1)....E(2). ==================================== ..E(3)..|......-1......-2........3.. ..E(4)..|......-6......-7........7.. ..E(5)..|.....-25.....-23.......14.. ..E(6)..|.....-89.....-80.......16.. ..E(7)..|....-280....-271......-77.. ..E(8)..|....-700....-750.....-922.. ..E(9)..|....-380....-647....-6660.. ..E(10).|...13452...13039...-41264.. ... a(5) = -23 because E(5) = -25*E(0) - 23*E(1) + 14*E(2). a(6) = -80 because E(6) = -89*E(0) - 80*E(1) + 16*E(2).
Programs
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Maple
# Compare with A143818 M:=24: a:=array(0..100): b:=array(0..100): c:=array(0..100): a[0]:=1: b[0]:=0: c[0]:=0: for n from 1 to M do a[n]:= -add(binomial(n-1,k)*c[k], k=0..n-1); b[n]:= add(binomial(n-1,k)*a[k], k=0..n-1); c[n]:= add(binomial(n-1,k)*b[k], k=0..n-1); end do: A143629:=[seq(b[n]-c[n], n=0..M)];
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Mathematica
m = 23; a[0] = 1; b[0] = 0; c[0] = 0; For[n = 1, n <= m, n++, a[n] = -Sum[ Binomial[n - 1, k]*c[k], {k, 0, n - 1}]; b[n] = Sum[ Binomial[n - 1, k]*a[k], {k, 0, n - 1}]; c[n] = Sum[ Binomial[n - 1, k]*b[k], {k, 0, n - 1}] ]; A143629 = Table[b[n] - c[n], {n, 0, m}] (* Jean-François Alcover, Mar 06 2013, after Maple *)
Formula
Define three sequences A(n), B(n) and C(n) by the relations: A(n+1) = - Sum_{i = 0..n} binomial(n,i)*C(i), B(n+1) = Sum_{i = 0..n} binomial(n,i)*A(i), C(n+1) = Sum_{i = 0..n} binomial(n,i)*B(i), with initial conditions A(0) = 1, B(0) = C(0) = 0. Then a(n) = B(n) - C(n). The other sequences are A(n) = A143628(n) and C(n) = A143630(n). The values of B(n) are recorded in A143631. Compare with A143818. Also a(n) = A143628(n) - A000587(n).
Comments