cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A320534 a(n) = ((1 + sqrt(4*n^2 + 1))^n + (1 - sqrt(4*n^2 + 1))^n)/2^n.

Original entry on oeis.org

2, 1, 9, 28, 577, 3251, 105193, 857501, 37831169, 403541596, 22550351001, 297238464799, 20106709638337, 315569447182601, 25059144736026633, 456277507970965876, 41600491470425952257, 862007599260004863571, 88733427132980061934777, 2061632980592377284802309
Offset: 0

Views

Author

Vladimir Reshetnikov, Oct 14 2018

Keywords

Comments

a(0) = 2 assuming 0^0 = 1, or using the limit for n -> 0 (assuming n is a real variable); the same value for a(0) arises from other formulae for this sequence.

Links

Crossrefs

Cf. A084844, A320519.

Programs

  • Mathematica
    Table[2^(1 - n) Hypergeometric2F1[(1 - n)/2, -n/2, 1/2, 4 n^2 + 1], {n, 0, 19}]
(* or *)
a[0] = Limit[n^n LucasL[n, 1/n], n -> 0]; (* a[0] = 2 *)
a[n_] := a[n] = n^n LucasL[n, 1/n];
Table[a[n], {n, 0, 19}]

Formula

a(n) = 2^(1 - n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k)*(4*n^2 + 1)^k.
a(n) = 2^(1 - n) * hypergeom([(1 - n)/2, -n/2], [1/2], 4*n^2 + 1).
For n > 0, a(n) = n^n * L_n(1/n), where L_n(x) is the Lucas polynomial.
For n > 0, a(n) = 2*(-i*n)^n*cos(n*arcsin(sqrt(4*n^2+1)/(2*n))). - Peter Luschny, Oct 14 2018

A320565 a(n) = ((1 + sqrt(4*n^2 + 1))^n - (1 - sqrt(4*n^2 + 1))^n)/(2^n * sqrt(4*n^2 + 1)).

Original entry on oeis.org

0, 1, 1, 10, 33, 701, 4033, 132301, 1089921, 48460114, 520210801, 29215223489, 386721507745, 26250621340841, 413242502386337, 32899021525375426, 600383148312628737, 54846079150716441949, 1138470675779123657425, 117372939125452004885621
Offset: 0

Views

Author

Vladimir Reshetnikov, Oct 15 2018

Keywords

Comments

a(0) = 0 assuming 0^0 = 1, or using the limit for n -> 0 (assuming n is a real variable); the same value for a(0) arises from other formulae for this sequence.

Links

Crossrefs

Cf. A084844, A320519, A320534.

Programs

  • Magma
    [2^(1-n)*(&+[Binomial(n,2*k+1)*(4*n^2+1)^k: k in [0..Floor(n/2)]]): n in [0..20]]; // G. C. Greubel, Oct 15 2018
  • Mathematica
    a[0] = Limit[n^(n - 1) Fibonacci[n, 1/n], n -> 0]; (* a[0] = 0 *)
a[n_] := a[n] = n^(n - 1) Fibonacci[n, 1/n];
Table[a[n], {n, 0, 19}]
  • PARI
    for(n=0,20, print1( 2^(1-n)*sum(k=0,floor(n/2), binomial(n,2*k+1)*(4*n^2+1)^k) , ", ")) \\ G. C. Greubel, Oct 15 2018

Formula

a(n) = 2^(1 - n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k + 1)*(4*n^2 + 1)^k.
a(n) = 2 * (i*n)^n * sinh(n*arctanh(sqrt(4*n^2 + 1)))/sqrt(4*n^2 + 1), assuming 0^0 = 1 for n = 0.
For n > 0, a(n) = n^(n - 1) * F_n(1/n), where F_n(x) is the Fibonacci polynomial.
For n > 0, a(n) = sqrt(Pi/4)*i*(-i*n)^n*LegendreP((n - 1)/2, -1/2, -1/(2*n^2) - 1) / (4*n^2 + 1)^(1/4). - Peter Luschny, Oct 15 2018
Showing 1-2 of 2 results.

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