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.

Showing 1-4 of 4 results.

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

Views

Author

Peter Bala, Mar 11 2024

Keywords

Comments

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

Crossrefs

Programs

  • Maple
    A370259 := n -> simplify( (ChebyshevT(n, n+1) - 1)/n^3 ):
    seq(sqrt(A370259(2*n+1)), n = 0..20);
  • Mathematica
    Table[Sqrt[(ChebyshevT[k, k + 1] - 1)/k^3], {k, 1, 40, 2}] (* Paolo Xausa, Jul 24 2024 *)

Formula

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

A370261 a(n) = sqrt(A370259(2*n)/(n+1)) for n >= 1.

Original entry on oeis.org

1, 5, 65, 1449, 46561, 1968525, 103565057, 6531391313, 480749649601, 40482981221781, 3840053099665729, 405275779792031225, 47113209228513626017, 5982545638922153790749, 823992221632687352744961, 122360935410018418223907489, 19489013519781051891806113153
Offset: 1

Views

Author

Peter Bala, Mar 11 2024

Keywords

Comments

The sequence is conjectured to be integral.

Crossrefs

Programs

  • Maple
    A370259 := n -> simplify( (ChebyshevT(n, n+1) - 1)/n^3 ):
    seq(sqrt(A370259(2*n)/(n+1)), n = 1..20);
  • Mathematica
    Table[Sqrt[(ChebyshevT[2*n, 2*n + 1] - 1)/(2*n)^3/(n + 1)], {n, 20}] (* Paolo Xausa, Jul 24 2024 *)
  • Python
    from math import isqrt
    from sympy import chebyshevt
    def A370261(n): return isqrt((chebyshevt((m:=n<<1),m+1)-1)//((n+1)*m**3)) # Chai Wah Wu, Mar 13 2024

Formula

a(n) = sqrt( (T(2*n, 2*n+1) - 1)/((n+1)*(2*n)^3) ), where T(n, x) is the n-th Chebyshev polynomial of the first kind.

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

Original entry on oeis.org

1, 2, 17, 244, 4801, 120126, 3650401, 130576328, 5374978561, 250283080090, 13007560326001, 746411226303612, 46873096812360001, 3197490648645613334, 235451028081583642049, 18614381236112230383376, 1572584048032918633353217
Offset: 0

Views

Author

Seiichi Manyama, Mar 05 2021

Keywords

Crossrefs

Programs

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

Formula

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

A370262 Triangle read by rows: T(n, k) = binomial(n+k, n-k)/(2*k + 1) * (2*n + 1)^k.

Original entry on oeis.org

1, 1, 1, 1, 5, 5, 1, 14, 49, 49, 1, 30, 243, 729, 729, 1, 55, 847, 5324, 14641, 14641, 1, 91, 2366, 26364, 142805, 371293, 371293, 1, 140, 5670, 101250, 928125, 4556250, 11390625, 11390625, 1, 204, 12138, 324258, 4593655, 36916282, 168962983, 410338673, 410338673
Offset: 0

Views

Author

Peter Bala, Mar 12 2024

Keywords

Comments

The table entries are integers since a(n, k) := binomial(n+k, n-k)/(2*k + 1) * (2*n + 1) gives the entries of the transpose of triangle A082985.

Examples

			Triangle begins
 n\k | 0    1      2       3        4        5        6
 - - - - - - - - - - - - - - - - - - - - - - - - - - - -
   0 | 1
   1 | 1    1
   2 | 1    5      5
   3 | 1   14     49      49
   4 | 1   30    243     729      729
   5 | 1   55    847    5324    14641    14641
   6 | 1   91   2366   26364   142805   371293   371293
  ...
		

Crossrefs

A371697 (row sums), A052750 (main diagonal and subdiagonal), A000330 (column 1).

Programs

  • Maple
    seq(seq(binomial(n+k, n-k)/(2*k + 1) * (2*n + 1)^k, k = 0..n), n = 0..10);
  • Mathematica
    Table[Binomial[n + k, n - k] / (2*k + 1) * (2*n + 1)^k, {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 17 2024 *)

Formula

n-th row polynomial R(n, x) = Sum_{k = 0..n} T(n, k)*x^k = sqrt( 2* Sum_{k = 0..2*n} (2*n + 1)^(k-1) *binomial(2*n+k+2, 2*k+2)/(2*n + k + 2) * x^k ).
R(n, x)^2 = 2/(x*(2*n + 1)^3) * ( ChebyshevT(2*n+1, 1 + (2*n+1)*x/2) - 1 ).
R(n, 2) = A370260(n).
Showing 1-4 of 4 results.