A321959
a(n) = [x^n] ((1 - x)*x)/((1 - 2*x)^2*(2*x^2 - 2*x + 1)).
Original entry on oeis.org
0, 1, 5, 16, 42, 100, 228, 512, 1144, 2544, 5616, 12288, 26656, 57408, 122944, 262144, 556928, 1179392, 2490112, 5242880, 11010560, 23069696, 48235520, 100663296, 209713152, 436203520, 905965568, 1879048192, 3892322304, 8053080064, 16643014656, 34359738368
Offset: 0
G.f. = x + 5*x^2 + 16*x^3 + 42*x^4 + 100*x^5 + 228*x^6 + ... - _Michael Somos_, Sep 30 2022
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ogf := ((1 - x)*x)/((1 - 2*x)^2*(2*x^2 - 2*x + 1));
ser := series(ogf, x, 32): seq(coeff(ser, x, n), n=0..31);
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LinearRecurrence[{6,-14,16,-8}, {0,1,5,16}, 32] (* Georg Fischer, May 08 2021 *)
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{a(n) = if(n<0, 0, polcoeff( x*(1 - x) / ((1 - 2*x)^2*(1 - 2*x + 2*x^2)), n))}; /* Michael Somos, Sep 30 2022 */
A323346
Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).
Original entry on oeis.org
1, 2, 1, 3, 3, 1, 4, 6, 4, 2, 6, 10, 10, 6, 6, 12, 16, 20, 16, 12, 16, 28, 28, 36, 36, 28, 28, 36, 64, 56, 64, 72, 64, 56, 64, 72, 136, 120, 120, 136, 136, 120, 120, 136, 136, 272, 256, 240, 256, 272, 256, 240, 256, 272, 256, 528, 528, 496, 496, 528, 528, 496, 496, 528, 528, 496
Offset: 0
Table begins
p\q| 0 1 2 3 4 5 ...
---+-------------------------------
0 | 1, 1, 1, 2, 6, 16, ...
1 | 2, 3, 4, 6, 12, 28, ...
2 | 3, 6, 10, 16, 28, 56, ...
3 | 4, 10, 20, 36, 64, 120, ...
4 | 6, 16, 36, 72, 136, 256, ...
5 | 12, 28, 64, 136, 272, 528, ...
...
See A323100 for an example that shows T(1,3) = 6.
A323100 is the complement sequence.
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s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2, s][1 + modp(n+1, 8)]:
T := proc(n, k) option remember;
if n = 0 then return 2^k*(1 - 1/2) - 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^n*(1 - 1/2) - 2^((n - 3)/2)*h(n) fi;
T(n, k-1) + T(n-1, k) end:
for n from 0 to 9 do seq(T(n, k), k=0..9) od; # Peter Luschny, Jan 12 2019
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T[n_, k_] := 2^(n + k) - Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];
Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 19 2019 *)
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T(p,q) = sum(i=0, p, sum(j=0, q, binomial(p, i)*binomial(q, j)*!(binomial(i-j, 2)%2)))
A323225
a(n) = ((2^n*n + i*(1 - i)^n - i*(1 + i)^n))/4, where i is the imaginary unit.
Original entry on oeis.org
0, 1, 3, 7, 16, 38, 92, 220, 512, 1160, 2576, 5648, 12288, 26592, 57280, 122816, 262144, 557184, 1179904, 2490624, 5242880, 11009536, 23067648, 48233472, 100663296, 209717248, 436211712, 905973760, 1879048192, 3892305920, 8053047296, 16642981888, 34359738368
Offset: 0
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a := n -> ((2^n*n + I*(1 - I)^n - I*(1 + I)^n))/4:
seq(a(n), n=0..32);
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LinearRecurrence[{6, -14, 16, -8}, {0, 1, 3, 7}, 40] (* Jean-François Alcover, Mar 20 2019 *)
Table[((2^n n + I (1 - I)^n - I (1 + I)^n))/4, {n, 0, 29}] (* Alonso del Arte, Mar 27 2020 *)
Showing 1-3 of 3 results.
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