A098443 -id:A098443 - cp's OEIS frontend

A073000 Decimal expansion of arctangent of 1/2.

4, 6, 3, 6, 4, 7, 6, 0, 9, 0, 0, 0, 8, 0, 6, 1, 1, 6, 2, 1, 4, 2, 5, 6, 2, 3, 1, 4, 6, 1, 2, 1, 4, 4, 0, 2, 0, 2, 8, 5, 3, 7, 0, 5, 4, 2, 8, 6, 1, 2, 0, 2, 6, 3, 8, 1, 0, 9, 3, 3, 0, 8, 8, 7, 2, 0, 1, 9, 7, 8, 6, 4, 1, 6, 5, 7, 4, 1, 7, 0, 5, 3, 0, 0, 6, 0, 0, 2, 8, 3, 9, 8, 4, 8, 8, 7, 8, 9, 2, 5, 5, 6, 5, 2, 9

Offset: 0

Robert G. Wilson v, Aug 03 2002

The angle at which you must shoot a cue ball on a standard pool table so that it will strike all four sides and return to its origin. [Barrow] - Robert G. Wilson v, Nov 29 2015

			Arctan(1/2)
=0.463647609000806116214256231461214402028537054286120263810933088720197864165... radians
=26°.56505117707798935157219372045329467120421429964522102798601631528806582148474...
=26°33'.9030706246793610943316232271976802722528579787132616791609789172839492890...
=26°33'54".184237480761665659897393631860816335171478722795700749658735037036957...
complement = 63°.43494882292201064842780627954670532879578570035477897201398368471...
supplement = 153°.4349488229220106484278062795467053287957857003547789720139836847...

John D. Barrow, One Hundred Essential Things You Didn't Know You Didn't Know, W. W. Norton & Co., NY & London, 2008.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 242.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Peter Bala, New series for old functions.
R. J. Mathar, Hierarchical Subdivision of the Simple Cubic Lattice, arXiv preprint arXiv:1309.3705 [math.MG], 2013.
Simon Plouffe, arctan(1/2).
Index entries for transcendental numbers

Cf. A001622, A002162, A002391, A105531, A254619.

evalf(arctan(0.5)) ; # R. J. Mathar, Aug 22 2013

Mathematica
```
RealDigits[ ArcTan[1/2], 10, 110] [[1]]
```

default(realprecision,2000); atan(1/2) \\ Anders Hellström, Nov 30 2015

Equals Pi/2 - A105199 = A019669-A105199. - R. J. Mathar, Aug 21 2013
From Peter Bala, Feb 04 2015: (Start)
Arctan(1/2) = 1/2*Sum_{k >= 0} (-1)^k/((2*k + 1)*4^k).
Define a pair of integer sequences A(n) = 4^n*(2*n + 1)!/n! and B(n) = A(n)*Sum_{k = 0..n} (-1)^k/((2*k + 1)*4^k). Both sequences satisfy the same second order recurrence equation u(n) = (12*n + 10)*u(n-1) + 16*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 2*arctan(1/2) = 1 - 2/(24 + 16*3^2/(34 + 16*5^2/(46 + ... + 16*(2*n - 1)^2/((12*n + 10) + ...)))). See A002391, A105531 and A002162 for similar expansions.
Arctan(1/2) = 2/5 * Sum_{k >= 0} (4/5)^k/((2*k + 1)*binomial(2*k,k)).
Define a pair of integer sequences C(n) = 5^n*(2*n + 1)!/n! and D(n) = C(n)*Sum_{k = 0..n} (4/5)^k/((2*k + 1)*binomial(2*k,k)). Both sequences satisfy the same second order recurrence equation u(n) = (24*n + 10)*u(n-1) - 40*n*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion 5/2*arctan(1/2) = 1 + 4/(30 - 240/(58 - 600/(82 - ... - 40*n*(2*n - 1)/((24*n + 10) - ... )))).
Arctan(1/2) = 2/25 * Sum_{k >= 0} (24*k + 17)*(4/5)^(2*k)/( (4*k + 1)*(4*k + 3)*binomial(4*k,2*k) ).
Arctan(1/2) = 2/125 * Sum_{k >= 0} (1116*k^2 + 1446*k + 433)*(4/5)^(3*k)/( (6*k + 1)*(6*k + 3)*(6*k + 5)*binomial(6*k,3*k) ). (End)
Equals Integral_{x = 0..oo} exp(-2*x)*sin(x)/x dx. - Peter Bala, Nov 05 2019
Equals 2 * arccot(phi^3), where phi is the golden ratio (A001622). - Amiram Eldar, Jul 06 2023
Equals Sum_{n >= 1} i/(n*P(n, 2*i)*P(n-1, 2*i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A098443(n)*A098443(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(2))^n ). - Peter Bala, Mar 16 2024

A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 6, 5, 1, 4, 12, 20, 14, 1, 5, 20, 50, 70, 42, 1, 6, 30, 100, 210, 252, 132, 1, 7, 42, 175, 490, 882, 924, 429, 1, 8, 56, 280, 980, 2352, 3696, 3432, 1430, 1, 9, 72, 420, 1764, 5292, 11088, 15444, 12870, 4862, 1, 10, 90, 600, 2940, 10584, 27720

Offset: 0

Paul Barry, Sep 09 2004

A Catalan scaled binomial matrix.
From Philippe Deléham, Sep 01 2005: (Start)
Table U(n,k), k >= 0, n >= 0, read by antidiagonals, begins:
  row k = 0: 1, 1,  2,   5,   14, ... is A000108
  row k = 1: 1, 2,  6,  20,   70, ... is A000984
  row k = 2: 1, 3, 12,  50,  280, ... is A007854
  row k = 3: 1, 4, 20, 104,  548, ... is A076035
  row k = 4: 1, 5, 30, 185, 1150, ... is A076036
G.f. for row k: 1/(1-(k+1)*x*C(x)) where C(x) is the g.f. = for Catalan numbers A000108.
U(n,k) = Sum_{j=0..n} A106566(n,j)*(k+1)^j. (End)
This sequence gives the coefficients (increasing powers of x) of the Jensen polynomials for the Catalan sequence A000108 of degree n and shift 0. For the definition of Jensen polynomials for a sequence see a comment in A094436. - Wolfdieter Lang, Jun 25 2019

			Rows begin:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  6,   5;
  1, 4, 12,  20,  14;
  1, 5, 20,  50,  70,  42;
  1, 6, 30, 100, 210, 252, 132;
  ...
Row 3: t*(1 - 3*t + 6*t^2 - 5*t^3)/(1 - 4*t)^(9/2) = 1/2*Sum_{k >= 1} k*(k+1)*(k+2)*(k+3)/4!*binomial(2*k,k)*t^k. - _Peter Bala_, Jun 13 2016

Indranil Ghosh, Rows 0..125, flattened

Row sums are A007317.
Antidiagonal sums are A090344.
Principal diagonal is A000108.
Mirror image of A124644.
Cf. A000984, A007854, A011973, A076035, A076036, A098443, A098473, A106566, A267633.

p := proc(n) option remember; if n = 0 then 1 else normal((x*(1 + 4*x)*diff(p(n-1, x), x) + (2*x + n + 1)*p(n-1, x))/(n + 1)) fi end:
row := n -> local k; seq(coeff(p(n), x, k), k = 0..n):
for n from 0 to 6 do row(n) od;  # Peter Luschny, Jun 21 2023

Table[Binomial[n, k] Binomial[2 k, k]/(k + 1), {n, 0, 10}, {k, 0, n}] // Flatten (* or *)
Table[(-1)^k*CatalanNumber[k] Pochhammer[-n, k]/k!, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 17 2017 *)

from functools import cache
@cache
def A098474row(n: int) -> list[int]:
    if n == 0: return [1]
    a = A098474row(n - 1) + [0]
    row = [0] * (n + 1)
    row[0] = 1; row[1] = n
    for k in range(2, n + 1):
        row[k] = (a[k] * (n + k + 1) + a[k - 1] * (4 * k - 2)) // (n + 1)
    return row  # Peter Luschny, Jun 22 2023

def A098474(n,k):
    return (-1)^k*catalan_number(k)*rising_factorial(-n,k)/factorial(k)
for n in range(7): [A098474(n,k) for k in (0..n)] # Peter Luschny, Feb 05 2015

G.f.: 2/(1-x+(1-x-4*x*y)^(1/2)). - Vladeta Jovovic, Sep 11 2004
E.g.f.: exp(x*(1+2*y))*(BesselI(0, 2*x*y)-BesselI(1, 2*x*y)). - Vladeta Jovovic, Sep 11 2004
G.f.: 1/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-x-xy/(1-xy/(1-... (continued fraction). - Paul Barry, Feb 11 2009
Sum_{k=0..n} T(n,k)*x^(n-k) = A126930(n), A005043(n), A000108(n), A007317(n+1), A064613(n), A104455(n) for x = -2, -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Dec 12 2009
T(n,k) = (-1)^k*Catalan(k)*Pochhammer(-n,k)/k!. - Peter Luschny, Feb 05 2015
O.g.f.: [1 - sqrt(1-4tx/(1-x))]/(2tx) = 1 + (1+t) x + (1+2t+2t^2) x^2 + (1+3t+6t^2+5t^3) x^3 + ... , generating the polynomials of this entry, reverse of A124644. See A011973 for a derivation and the inverse o.g.f., connected to the Fibonacci, Chebyshev, and Motzkin polynomials. See also A267633. - Tom Copeland, Jan 25 2016
From Peter Bala, Jun 13 2016: (Start)
The o.g.f. F(x,t) = ( 1 - sqrt(1 - 4*t*x/(1 - x)) )/(2*t*x) satisfies the partial differential equation d/dx(x*(1 - x)*F) - x*t*(1 + 4*t)*dF/dt - 2*x*t*F = 1. This gives a recurrence for the row polynomials: (n + 2)*R(n+1,t) = t*(1 + 4*t)*R'(n,t) + (2*t + n + 2)*R(n,t), where the prime ' indicates differentiation with respect to t.
Equivalently, setting Q(n,t) = t^(n+2)*R(n,-t)/(1 - 4*t)^(n + 3/2) we have t^2*d/dt(Q(n,t)) = (n + 2)*Q(n+1,t).
This leads to the following expansions:
Q(0,t) = (1/2)*Sum_{k >= 1} k*binomial(2*k,k)*t^(k+1)
Q(1,t) = (1/2)*Sum_{k >= 1} k*(k+1)/2!*binomial(2*k,k)*t^(k+2)
Q(2,t) = (1/2)*Sum_{k >= 1} k*(k+1)*(k+2)/3!*binomial(2*k,k) *t^(k+3) and so on. (End)
Sum_{k=0..n} T(n,k)*x^k = A007317(n+1), A162326(n+1), A337167(n) for x = 1, 2, 3 respectively. - Sergii Voloshyn, Mar 31 2022

New name using a formula of Paul Barry by Peter Luschny, Feb 05 2015

A386621 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4kx - 4*x^2).

Original entry on oeis.org

1, 1, 0, 1, 2, 2, 1, 4, 8, 0, 1, 6, 26, 32, 6, 1, 8, 56, 184, 136, 0, 1, 10, 98, 576, 1366, 592, 20, 1, 12, 152, 1328, 6216, 10424, 2624, 0, 1, 14, 218, 2560, 18886, 68976, 80996, 11776, 70, 1, 16, 296, 4392, 45256, 276208, 779456, 637424, 53344, 0

Offset: 0

Seiichi Manyama, Aug 29 2025

			Square array begins:
   1,    1,     1,      1,       1,        1,        1, ...
   0,    2,     4,      6,       8,       10,       12, ...
   2,    8,    26,     56,      98,      152,      218, ...
   0,   32,   184,    576,    1328,     2560,     4392, ...
   6,  136,  1366,   6216,   18886,    45256,    92886, ...
   0,  592, 10424,  68976,  276208,   822800,  2020392, ...
  20, 2624, 80996, 779456, 4114004, 15235520, 44758244, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..3 give A126869, A006139, A098443, A387428.

Main diagonal gives A387430.

a(n, k) = sum(j=0, n\2, (k^2+1)^j*(2*k)^(n-2*j)*binomial(n, 2*j)*binomial(2*j, j));

A(n,k) = Sum_{j=0..n} (k-i)^j * (k+i)^(n-j) * binomial(n,j)^2, where i is the imaginary unit.
A(n,k) = Sum_{j=0..floor(n/2)} k^(n-2*j) * binomial(2*(n-j),n-j) * binomial(n-j,j).
n*A(n,k) = 2*k*(2*n-1)*A(n-1,k) + 4*(n-1)*A(n-2,k) for n > 1.
A(n,k) = Sum_{j=0..floor(n/2)} (k^2+1)^j * (2*k)^(n-2*j) * binomial(n,2*j) * binomial(2*j,j).
A(n,k) = [x^n] (1 + 2*k*x + (k^2+1)*x^2)^n.
E.g.f. of column k: exp(2*k*x) * BesselI(0, 2*sqrt(k^2+1)*x).

A098444 Expansion of 1/sqrt(1-6x-11x^2).

Original entry on oeis.org

1, 3, 19, 117, 771, 5193, 35629, 247467, 1734931, 12250953, 87006249, 620818047, 4447016781, 31959556983, 230331965379, 1664043517557, 12047551338771, 87387014213433, 634918255153369, 4619923954541247, 33661450900419001

Offset: 0

Paul Barry, Sep 07 2004

Binomial transform of A084770. Second binomial transform of A098264. Binomial transform is A098443.

Coefficient of x^n in (1 + 3 x + 5 x^2)^n = number of paths from the origin to (n,0) with steps U=(1,1), H=(1,0) and D=(1,-1); U can have 5 colors and H can have 3 colors. - N-E. Fahssi, Jan 28 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Table[SeriesCoefficient[1/Sqrt[1-6*x-11*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 15 2012 *)

x='x+O('x^66); Vec(1/sqrt(1-6*x-11*x^2)) \\ Joerg Arndt, May 11 2013

E.g.f.: exp(3x)*BesselI(0, 2*sqrt(5)*x)
D-finite with recurrence: n*a(n) = 3*(2*n-1)*a(n-1) + 11*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ sqrt(50+15*sqrt(5))*(3+2*sqrt(5))^n/(10*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 15 2012

cp's OEIS Frontend

A073000 Decimal expansion of arctangent of 1/2.

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A098474 Triangle read by rows, T(n,k) = C(n,k)*C(2*k,k)/(k+1), n >= 0, 0 <= k <= n.

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A386621 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4*k*x - 4*x^2).

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A098444 Expansion of 1/sqrt(1-6x-11x^2).

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Author

Keywords

Comments

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Programs

Mathematica

PARI

Formula

A098474 Triangle read by rows, T(n,k) = C(n,k)C(2k,k)/(k+1), n >= 0, 0 <= k <= n.

A386621 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 - 4kx - 4*x^2).