A370261 -id:A370261 - cp's OEIS frontend

A342205 a(n) = T(n,n+1) where T(n,x) is a Chebyshev polynomial of the first kind.

1, 2, 17, 244, 4801, 120126, 3650401, 130576328, 5374978561, 250283080090, 13007560326001, 746411226303612, 46873096812360001, 3197490648645613334, 235451028081583642049, 18614381236112230383376, 1572584048032918633353217

Offset: 0

Seiichi Manyama, Mar 05 2021

Seiichi Manyama, Table of n, a(n) for n = 0..351
Wikipedia, Chebyshev polynomials.

Cf. A115066, A323117, A342206, A342207, A370259, A370260, A370261.

Table[ChebyshevT[n, n + 1], {n, 0, 16}] (* Amiram Eldar, Mar 05 2021 *)

PARI
```
a(n) = polchebyshev(n, 1, n+1);
```
PARI
```
a(n) = round(cos(n*acos(n+1)));
```

a(n) = if(n==0, 1, n*sum(k=0, n, (2*n)^k*binomial(n+k, 2*k)/(n+k)));

a(n) = cos(n*arccos(n+1)).
a(n) = n * Sum_{k = 0..n} (2*n)^k * binomial(n+k,2*k)/(n+k) for n > 0.
From Peter Bala, Mar 11 2024: (Start)
a(2*n+1) == 1 (mod (2*n + 1)^3); a(2*n) == 1 (mod (n + 1)*(2*n)^3).
a(n) = hypergeom([n, -n], [1/2], -n/2). (End)
a(n) ~ exp(1) * 2^(n-1) * n^n. - Vaclav Kotesovec, Mar 12 2024

A370259 a(n) = (T(n,n+1) - 1)/n^3 for n >= 1, where T(n,x) is the n-th Chebyshev polynomial of the first kind.

Original entry on oeis.org

1, 2, 9, 75, 961, 16900, 380689, 10498005, 343323841, 13007560326, 560789801881, 27125634729375, 1455389462287489, 85805768251305992, 5515372218107327521, 383931652351786775721, 28778117694539885440129, 2311202255914842794592010, 198009919900727928789497641, 18027589454633803742596931571

Offset: 1

Peter Bala, Mar 11 2024

It appears that a(2*n+1) is always a square, while a(2*n) = (n + 1) * a square. See A370260 and A370261.

Paolo Xausa, Table of n, a(n) for n = 1..350

Cf. A008310, A342205, A370260, A370261.

seq( simplify( (ChebyshevT(n, n+1) - 1)/n^3 ), n = 1..20);

Array[(ChebyshevT[#, #+1]-1)/#^3 &, 20] (* Paolo Xausa, Mar 14 2024 *)

from sympy import chebyshevt
def A370259(n): return (chebyshevt(n,n+1)-1)//n**3 # Chai Wah Wu, Mar 13 2024

a(n) = Sum_{k = 1..n} (2^k)*n^(k-2)*binomial(n+k, 2*k)/(n + k) (shows that a(n) is an integer).
a(n) = (cos(n*arccos(n+1)) - 1)/n^3.
a(n) = (A342205(n) - 1)/n^3.
a(n) = ( (n + 1 + sqrt(n*(n+2)))^n + (n + 1 - sqrt(n*(n+2)))^n - 2 )/(2*n^3).

A370260 a(n) = sqrt(A370259(2*n+1)).

Original entry on oeis.org

1, 3, 31, 617, 18529, 748859, 38149567, 2348482961, 169641143873, 14071599763379, 1318414335714015, 137720427724123513, 15871136311527376801, 2000355821099358166891, 273735526097742996298111, 40419227378551955037029921, 6405616571975691389276400257

Offset: 0

Peter Bala, Mar 11 2024

The sequence is conjectured to be integral [added 03 Mar 2024: now confirmed - see the Formula section].

Paolo Xausa, Table of n, a(n) for n = 0..300

Cf. A008310, A342205, A370259, A370261, A370262.

A370259 := n -> simplify( (ChebyshevT(n, n+1) - 1)/n^3 ):
seq(sqrt(A370259(2*n+1)), n = 0..20);

Table[Sqrt[(ChebyshevT[k, k + 1] - 1)/k^3], {k, 1, 40, 2}] (* Paolo Xausa, Jul 24 2024 *)

a(n) = sqrt( (T(2*n+1, 2*n+2) - 1)/(2*n+1)^3 ), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
a(n) = sqrt( Sum_{k = 1..2*n+1} (2^k)*(2*n + 1)^(k-2)*binomial(2*n + k + 1, 2*k)/(2*n + k + 1) ).
a(n) = Sum_{k = 0..n} binomial(n+k, n-k)/(2*k + 1) * (4*n + 2)^k (shows the sequence to be integral) = R(n,2), where R(n, x) is the n-th row polynomial of A370262. - Peter Bala, Apr 03 2024

cp's OEIS Frontend

A342205 a(n) = T(n,n+1) where T(n,x) is a Chebyshev polynomial of the first kind.

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A370259 a(n) = (T(n,n+1) - 1)/n^3 for n >= 1, where T(n,x) is the n-th Chebyshev polynomial of the first kind.

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A370260 a(n) = sqrt(A370259(2*n+1)).

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