A176233
Determinant of n X n matrix with rows (n^2,-1,0,...,0), (1,n^2,-1, 0,...,0), (0,1,n^2,-1,0,...,0), ...,(0,0,...,1,n^2).
Original entry on oeis.org
1, 17, 747, 66305, 9828200, 2185188193, 679919101029, 281956264747009, 150277722869740455, 100090028003500150001, 81458362232421250207824, 79539026883848399173231873, 91771878445323959814042316673
Offset: 1
For n = 1, det[1] = 1.
For n = 2, det([[4,-1],[1,4]]) = 17, and the continued fraction expansion is 17/4 = [2^2,2^2].
For n = 3, det([[9,-1,0],[1,9,-1],[0,1,9]]) = 747, and the continued fraction expansion is 747/det([[9,-1],[1,9]]) = 747/82 = [3^2,3^2,3^2].
- J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Collection ellipses (2004), p. 115.
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for n from 15 by -1 to 1 do x0:=n^2: for p from n by -1 to 2 do : x0:= n^2 + 1/x0 :od: print(x0): od :
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nmax = 20; Do[x0 = n^2; Do[x0 = n^2 + 1/x0, {p, n, 2, -1}]; a[n] = Numerator[x0];, {n, nmax, 1, -1}]; Table[a[n], {n, 1, nmax}] (* Vaclav Kotesovec, Dec 29 2019 *)
A368892
a(n) = Sum_{k=0..floor(n/3)} n^(n-3*k) * binomial(n-2*k,k).
Original entry on oeis.org
1, 1, 4, 28, 264, 3200, 47521, 835569, 16974208, 391147867, 10080150040, 287244283821, 8967781893889, 304393809948904, 11160668048222588, 439582708115133751, 18509867068477014112, 829768603643818659302, 39454459640462073466945
Offset: 0
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Join[{1}, Table[n^n * HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n/3}, {1/2 - n/2, -n/2}, -27/(4*n^3)], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 09 2024 *)
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a(n) = sum(k=0, n\3, n^(n-3*k)*binomial(n-2*k, k));
A352798
a(n) = 1/(cf[0;n,n,n,...,n] - cf[0;n,n,...,n]) where the first continued fraction has n+1 terms and the second has n terms.
Original entry on oeis.org
1, -10, 330, -21960, 2551640, -461930274, 120572270007, -42930583856160, 20008932768992430, -11825788272679695050, 8643081649999714376976, -7654102744143874729100040, 8076084821027629176909996013, -10010473694454865001226770534530, 14402393216408406872433735669683370
Offset: 1
a(2) = -10 because the two continued fractions are cf[0;2,2] = 0 + 1/(2 + 1/2) = 2/5 and cf[0;2] = 0 + 1/2 = 1/2 and the reciprocal of their difference is 1/(2/5 - 1/2) = -10.
a(3) = 330 because the two continued fractions are cf[0;3,3,3] = 0 + 1/(3 + 1/(3 + 1/3)) = 10/33 and cf[0;3,3] = 0 + 1/(3 + 1/3) = 3/10, and 1/(10/33 - 3/10) = 330.
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a:= n-> (f-> -(-1)^n*f(n,n)*f(n+1,n))(combinat[fibonacci]):
seq(a(n), n=1..15); # Alois P. Heinz, Jul 06 2022
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a(n) = (-1)^(n+1) * vecprod(Vec(lift(Mod('x,'x^2-n*'x-1)^(n+1)))); \\ Kevin Ryde, Apr 18 2022
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from sympy.ntheory.continued_fraction import continued_fraction_reduce
def A352798(n): return int(1/(continued_fraction_reduce([0]+[n]*n)-continued_fraction_reduce([0]+[n]*(n-1)))) # Chai Wah Wu, Jul 06 2022
A368890
a(n) = Sum_{k=0..floor(n/2)} n^(3*(n-2*k)) * binomial(n-k,k).
Original entry on oeis.org
1, 1, 65, 19737, 16789505, 30525391000, 101570840860033, 558574349855881107, 4722492584690006360065, 58150612359276833311664895, 1000009000028000035000015000001, 23225285520096132372224712190010064, 708804486128121003209727133170234347521
Offset: 0
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Join[{1}, Table[n^(3*n) * Hypergeometric2F1[1/2 - n/2, -n/2, -n, -4/n^6], {n, 1, 15}]] (* Vaclav Kotesovec, Jan 09 2024 *)
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a(n) = sum(k=0, n\2, n^(3*(n-2*k))*binomial(n-k, k));
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