Original entry on oeis.org
1, -1, 5, 11, -23, -41, 45, -29, 337, 1199, -3115, -6469, 10297, 8839, 16125, 108691, -354143, -873121, 1721765, 2521451, -1476983, 6699319, -34182195, -103232189, 242017777, 451910159, -597551755, -130656229, -2465133863, -10513816601, 29729597085, 66349305331, -116749235903
Offset: 0
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LinearRecurrence[{0, -5, 0, -19, 0, 25}, {1, -1, 5, 11, -23, -41}, 33] (* Jean-François Alcover, Apr 08 2024 *)
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{a(n)=polcoeff((1-x+10*x^2+6*x^3+21*x^4-5*x^5)/(1-x^2)/(1+6*x^2+25*x^4+x*O(x^n)),n)}
Original entry on oeis.org
1, 0, 4, 0, 48, 160, -256, 0, -1792, -7680, 16384, 0, 135168, 532480, -1048576, 0, -8323072, -33423360, 67108864, 0, 537919488, 2149580800, -4294967296, 0, -34342961152, -137405399040, 274877906944, 0, 2199291691008, 8796629893120, -17592186044416, 0, -140733193388032, -562941363486720
Offset: 0
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CoefficientList[Series[(1-2x+8x^2-16x^3+128x^4-96x^5+128x^6-256x^7)/ (1-2x)/(1+4x^2)/(1+64x^4),{x,0,50}],x] (* or *) Join[{1}, LinearRecurrence[ {2,-4,8,-64,128,-256,512},{0,4,0,48,160,-256,0},50]] (* Harvey P. Dale, Apr 30 2011 *)
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{a(n)=polcoeff((1-2*x+8*x^2-16*x^3+128*x^4-96*x^5+128*x^6-256*x^7)/(1-2*x)/(1+4*x^2)/(1+64*x^4+x*O(x^n)),n)}
A118435
Triangle T, read by rows, equal to the matrix product T = H*[C^-1]*H, where H is the self-inverse triangle A118433 and C is Pascal's triangle.
Original entry on oeis.org
1, 1, 1, -3, 2, 1, -11, 15, 3, 1, 25, -44, -18, 4, 1, 41, -115, -110, 50, 5, 1, -43, 246, 375, -220, -45, 6, 1, 29, 315, 861, -805, -385, 105, 7, 1, -335, 232, -1204, 2296, 1750, -616, -84, 8, 1, -1199, 3033, 1044, 3780, 5166, -2898, -924, 180, 9, 1
Offset: 0
Triangle begins:
1;
1, 1;
-3, 2, 1;
-11, 15, 3, 1;
25,-44,-18, 4, 1;
41,-115,-110, 50, 5, 1;
-43, 246, 375,-220,-45, 6, 1;
29, 315, 861,-805,-385, 105, 7, 1;
-335, 232,-1204, 2296, 1750,-616,-84, 8, 1;
-1199, 3033, 1044, 3780, 5166,-2898,-924, 180, 9, 1;
...
The matrix log, log(T) = A118441, starts:
0;
1, 0;
-4, 2, 0;
-12, 12, 3, 0;
32,-48,-24, 4, 0;
80,-160,-120, 40, 5, 0;
...
where matrix square, log(T)^2, is a single diagonal:
0;
0,0;
2,0,0;
0,6,0,0;
0,0,12,0,0;
0,0,0,20,0,0;
...
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nmax = 12;
h[n_, k_] := Binomial[n, k]*(-1)^(Quotient[n+1, 2] - Quotient[k, 2]+n-k);
H = Table[h[n, k], {n, 0, nmax}, {k, 0, nmax}];
Cn = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, nmax}];
Tn = H.Inverse[Cn].H;
T[n_, k_] := Tn[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 08 2024 *)
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{T(n,k)=local(M=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1)*(-1)^(r\2- (c-1)\2+r-c))),C=matrix(n+1,n+1,r,c,if(r>=c,binomial(r-1,c-1))));(M*C^-1*M)[n+1,k+1]}
Showing 1-3 of 3 results.
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