A202543 -id:A202543 - cp's OEIS frontend

A002544 a(n) = binomial(2n+1,n)(n+1)^2.

1, 12, 90, 560, 3150, 16632, 84084, 411840, 1969110, 9237800, 42678636, 194699232, 878850700, 3931426800, 17450721000, 76938289920, 337206098790, 1470171918600, 6379820115900, 27569305764000, 118685861314020, 509191949220240, 2177742427450200, 9287309860732800

Offset: 0

N. J. A. Sloane, Simon Plouffe

Coefficients for numerical differentiation.
Take the first n integers 1,2,3..n and find all combinations with repetitions allowed for the first n of them. Find the sum of each of these combinations to get this sequence. Example for 1 and 2: 1,2,1+1,1+2,2+2 gives sum of 12=a(2). - J. M. Bergot, Mar 08 2016
Let cos(x) = 1 -x^2/2 +x^4/4!-x^6/6!.. = Sum_i (-1)^i x^(2i)/(2i)! be the standard power series of the cosine, and y = 2*(1-cos(x)) = 4*sin^2(x/2) = x^2 -x^4/12 +x^6/360 ...= Sum_i 2*(-1)^(i+1) x^(2i)/(2i)! be a closely related series. Then this sequence represents the reversion x^2 = Sum_i 1/a(i) *y^(i+1). - R. J. Mathar, May 03 2022

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..200
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables) [Annotated scanned copy]
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135. [Annotated scanned copy]
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
R. Shenton and A. W. Kemp, An S-fraction and ln^2(1+x), Journal of Computational and Applied Mathematics, 26 (1989) 367-370 North-Holland.
T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.
Mats Vermeeren, A dynamical solution to the Basel problem, arXiv preprint arXiv:1506.05288 [math.CA], 2015.

Cf. A085373, A002457, A002674, A202543.
Equals A002736/2.
A diagonal of A331430.

seq((n+1)^2*(binomial(2*n+2, n+1))/2, n=0..29); # Zerinvary Lajos, May 31 2006

Table[Binomial[2n+1,n](n+1)^2,{n,0,20}] (* Harvey P. Dale, Mar 23 2011 *)

PARI
```
a(n)=binomial(2*n+1,n)*(n+1)^2
```

x='x+O('x^99); Vec((1+2*x)/(1-4*x)^(5/2)) \\ Altug Alkan, Jul 09 2016

from sympy import binomial
def a(n): return binomial(2*n + 1, n)*(n + 1)**2 # Indranil Ghosh, Apr 18 2017

G.f.: (1 + 2x)/(1 - 4x)^(5/2).
a(n-1) = sum(i_1 + i_2 + ... + i_n) where the sum is over 0 <= i_1 <= i_2 <= ... <= i_n <= n; a(n) = (n+1)^2 C(2n+1, n). - David Callan, Nov 20 2003
a(n) = (n+1)^2 * binomial(2*n+2,n+1)/2. - Zerinvary Lajos, May 31 2006
Asymptotics: a(n)-> (1/64) * (128*n^2+176*n+41) * 4^n * n^(-1/2)/(sqrt(Pi)), for n->infinity. - Karol A. Penson, Aug 05 2013
G.f.: 2F1(3/2,2;1;4x). - R. J. Mathar, Aug 09 2015
a(n) = A002457(n)*(n+1). - R. J. Mathar, Aug 09 2015
a(n) = A000217(n)*A000984(n). - J. M. Bergot, Mar 10 2016
a(n-1) = A001791(n)*n*(n+1)/2. - Anton Zakharov, Jul 04 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
E.g.f.: ((1 + 2*x)*(1 + 8*x)*BesselI(0,2*x) + 2*x*(3 + 8*x)*BesselI(1,2*x))*exp(2*x).
Sum_{n>=0} 1/a(n) = Pi^2/9 = A100044. (End)
From Peter Bala, Apr 18 2017: (Start)
With x = y^2/(1 + y) we have log^2(1 + y) = Sum_{n >= 0} (-1)^n*x^(n+1)/a(n). See Shenton and Kemp.
Series reversion ( Sum_{n >= 0} (-1)^n*x^(n+1)/a(n) ) = Sum_{n >= 1} 2*x^n/(2*n)! = Sum_{n >= 1} x^n/A002674(n). (End)
D-finite with recurrence n^2*a(n) -2*(n+1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)^2 = A202543^2. - Amiram Eldar, May 14 2022

A202537 Decimal expansion of x satisfying e^x-e^(-2x)=1.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling, Dec 21 2011

If u>0 and v>0, there is a unique number x satisfying e^(ux)-e^(-vx)=1.  Guide to related sequences, with graphs included in Mathematica programs:
u.... v.... x
1.... 1.... A002390
1.... 2.... A202537
1.... 3.... A202538
2.... 1.... A202539
3.... 1.... A202540
2.... 2.... A202541
3.... 3.... A202542
1/2..1/2... A202543
Suppose that f(x,u,v) is a function of three real variables and that g(u,v) is a function defined implicitly by f(g(u,v),u,v)=0.  We call the graph of z=g(u,v) an implicit surface of f.  For an example related to A202537, take f(x,u,v)=e^(ux)-e^(-vx)-1 and g(u,v) = a nonzero solution x of f(x,u,v)=0.  If there is more than one nonzero solution, care must be taken to ensure that the resulting function g(u,v) is single-valued and continuous.  A portion of an implicit surface is plotted by Program 2 in the Mathematica section.

			0.382245085840035641329358499184857393759416422...

Index entries for transcendental numbers.

Cf. A002390.

(* Program 1:  A202537 *)
u = 1; v = 2;
f[x_] := E^(u*x) - E^(-v*x); g[x_] := 1
Plot[{f[x], g[x]}, {x, -2, 2}, {AxesOrigin -> {0, 0}}]
r = x /. FindRoot[f[x] == g[x], {x, .3, .4}, WorkingPrecision -> 110]
RealDigits[r]  (* A202537 *)
(* Program 2: implicit surface for e^(ux)-e(-vx)=1 *)
f[{x_, u_, v_}] := E^(u*x) - E^(-v*x) - 1;
t = Table[{u, v, x /. FindRoot[f[{x, u, v}] == 0, {x, 0, .3}]}, {v, 1, 4}, {u, 2, 20}];
ListPlot3D[Flatten[t, 1]] (* for A202537 *)
First[ RealDigits[ Log[ Root[#^3 - #^2 - 1 & , 1]], 10, 99]] (* Jean-François Alcover, Feb 26 2013 *)

solve(x=0,1,exp(x)-exp(-2*x)-1) \\ Charles R Greathouse IV, Feb 26 2013

log(polrootsreal(x^3-x^2-1)[1]) \\ Charles R Greathouse IV, Feb 07 2025

Digits from a(90) on corrected by Jean-François Alcover, Feb 26 2013

A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 2, 10, 505, 12878813, 4229958765311886322, 5876687051603582015287706866081267480733704277890

Offset: 0

Andrew Hone, Dec 12 2005

nonn

A second-order recurrence with the Laurent property. This property is satisfied by any second-order recurrence of the form a(n+2) = f(a(n+1))/a(n) with f being a polynomial of the form f(x) = x*p(x) where p is a polynomial of degree d with integer coefficients such that p(0)=1 and p has the reciprocal property x^d*p(1/x) = p(x). Hence if a(0) = a(1) = 1 then a(n) is an integer for all n.

As n tends to infinity, log(log(a(n)))/n tends to log((3+sqrt(5))/2) or about 0.962 (A202543).

Seiichi Manyama, Table of n, a(n) for n = 0..10
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.

Cf. A101879, A112373, A202543.

a[0]:=1; a[1]:=1; f(x):=x^3+x;
for n from 0 to 8 do a[n+2]:=simplify(subs(x=a[n+1],f(x))/a[n]) od;
s[3]:=ln(10); s[4]:=ln(505);
for n from 3 to 10000 do s[n+2]:=evalf(3*s[n+1]+ln(1+exp(-2*s[n+1]))-s[n]): od: print(evalf(ln(s[10002])/(10002))): evalf(ln((3+sqrt(5))/2));
# s[n]=ln(a[n]); ln(s[n])/n converges slowly to 0.962...
f:=proc(n) option remember; local i,j,k,t1,t2,t3; if n <= 1 then RETURN(1); fi; (f(n-1)^3+f(n-1))/f(n-2); end;
# N. J. A. Sloane

nxt[{a_,b_}]:={b,(b^3+b)/a}; NestList[nxt,{1,1},10][[All,1]] (* Harvey P. Dale, Jun 26 2017 *)

def A(l, m, n)
  a = Array.new(2 * m, 1)
  ary = [1]
  while ary.size < n + 1
    i = a[1..-1].inject(:*) + a[m] ** l
    break if i % a[0] > 0
    a = *a[1..-1], i / a[0]
    ary << a[0]
  end
  ary
end
def A112449(n)
  A(3, 1, n)
end # Seiichi Manyama, Nov 20 2016

a(1-n) = a(n). - Seiichi Manyama, Nov 20 2016

A318057 a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.

Original entry on oeis.org

0, -2, 3, 2, 5, 2, 6, 9, 10, 9, 13, 12, 15, 16, 19, 16, 20, 22, 24, 25, 27, 29, 28, 30, 33, 32, 36, 32, 38, 32, 41, 42, 44, 45, 46, 47, 50, 48, 52, 54, 53, 56, 53, 58, 59, 60, 64, 62, 66, 62, 67, 69, 71, 73, 75, 74, 77, 78, 80, 82, 81, 84, 81, 87, 81, 88, 90

Offset: 1

A.H.M. Smeets, Aug 14 2018

The correct binary value of the golden section is given in A068432; the continued fraction terms of the golden section is given in A000012.
For the number of correct decimal digits of the golden section see A318058.
The denominator of the k-th convergent obtained from a continued fraction tend to k*A001622; the error between the k-th convergent and the constant itself tends to 1/(2*k*A001622), or in binary digits 2*k*log(A001622)/log(2) bits after the binary point.
The sequence for quaternary digits is obtained by floor(a(n)/2), the sequence for octal digits is obtained by floor(a(n)/3), and the sequence for hexadecimal digits is obtained by floor(a(n)/4).

			   n   convergent         binary expansion       a(n)
  ==  =============  ==========================  ====
   1    1 / 1          1.0                         0
   2    2 / 1         10.0                        -2
   3    3 / 2          1.1000                      3
   4    5 / 3          1.101                       2
   5    8 / 5          1.100110                    5
   6   13 / 8          1.101                       2
   7   21 / 13         1.1001110                   6
   8   34 / 21         1.1001111001                9
   9   55 / 34         1.10011110000              10
  10   89 / 55         1.1001111001                9
  oo  lim = A068432    1.1001111000110111011110   --

Cf. A000012, A001622, A002390, A068432, A202543, A242208, A318058.

p, q, i, base = 1, 1, 0, 2
while i < 20200:
    p, q, i = p+q, p, i+1
a0, p, q = p//q, q, p
i, p, dd = 0, p*base, [0]
while i < 30000:
    d, p, i = p//q, (p%q)*base, i+1
    dd = dd+[d]
n, pn, qn = 0, 1, 0
while n < 20000:
    n, pn, qn = n+1, pn+qn, pn
    if pn//qn != a0:
        print(n, "- manual!")
    else:
        i, p, q, di = 0, (pn%qn)*base, qn, 0
        while di == dd[i]:
            i, di, p = i+1, p//q, (p%q)*base
        print(n, i-1)

Lim {n -> oo} a(n)/n = 2*log(A001622)/log(2) = 2*A002390/log(2) = A202543/log(2) = 2*A242208.

A318058 a(n) is the number of decimal places to which the n-th convergent of the continued fraction expansion of the golden section matches the correct value.

Original entry on oeis.org

0, -1, 0, 1, 1, 1, 2, 2, 2, 3, 2, 4, 4, 5, 5, 5, 6, 5, 7, 7, 8, 7, 9, 9, 9, 10, 10, 10, 11, 10, 12, 12, 13, 12, 13, 14, 15, 14, 15, 16, 16, 16, 17, 17, 17, 18, 18, 19, 18, 20, 20, 21, 21, 21, 22, 22, 23, 23, 24, 23, 24, 25, 25, 26, 26, 27, 27, 27, 28, 28, 29, 29, 29, 30, 30

Offset: 1

A.H.M. Smeets, Aug 14 2018

The correct decimal value of the golden section is given in A001622; the continued fraction terms of the golden section is given in A000012.
For the number of correct decimal digits of the golden section see A318057.
The denominator of the k-th convergent obtained from a continued fraction tend to k*A001622; the error between the k-th convergent and the constant itself tends to 1/(2*k*A001622), or in binary digits 2*k*log(A001622)/log(2) bits after the binary point.

			   n   convergent         decimal expansion     a(n)
  ==  =============  =========================  ====
   1    1 / 1         1.0                         0
   2    2 / 1         2.0                        -1
   3    3 / 2         1.5                         0
   4    5 / 3         1.66                        1
   5    8 / 5         1.60                        1
   6   13 / 8         1.62                        1
   7   21 / 13        1.615                       2
   8   34 / 21        1.619                       2
   9   55 / 34        1.617                       2
  10   89 / 55        1.6181                      3
  oo  lim = A001622   1.6180339887498948482      --

Cf. A000012, A001622, A002390, A097348, A202543, A318057.

p, q, i, base = 1, 1, 0, 10
while i < 20200:
    p, q, i = p+q, p, i+1
a0, p, q = p//q, q, p
i, p, dd = 0, p*base, [0]
while i < 30000:
    d, p, i = p//q, (p%q)*base, i+1
    dd = dd+[d]
n, pn, qn = 0, 1, 0
while n < 20000:
    n, pn, qn = n+1, pn+qn, pn
    if pn//qn != a0:
        print(n, "- manual!")
    else:
        i, p, q, di = 0, (pn%qn)*base, qn, 0
        while di == dd[i]:
            i, di, p = i+1, p//q, (p%q)*base
        print(n, i-1)

Limit_{n -> oo} a(n)/n = 2*log(A001622)/log(10) = 2*A002390/log(10) = A202543/log(10) = 2*A097348.

A182546 Decimal expansion of | log_phi(i) |, where phi is the golden ratio and i is the imaginary unit.

Original entry on oeis.org

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

Offset: 1

Volker Werner, May 04 2012

			3.26425130263649690673153367843629490781491039315880...

Cf. A000796, A001622, A202543.

RealDigits[ Im[N[Log[I]/Log[GoldenRatio], 105]]] [[1]]
RealDigits[Abs[Log[GoldenRatio,I]],10,120][[1]] (* Harvey P. Dale, Sep 22 2018 *)

Equals |Im(Log(i))/Log(A001622)|

Equals Pi/log(phi+1) = A000796/A202543. - Gleb Koloskov, Sep 28 2021

A377644 Decimal expansion of Li_3(2 - phi).

Original entry on oeis.org

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

Offset: 0

Stefano Spezia, Nov 03 2024

			0.402683962952109021159959448182511142219733807379...

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 1.6.3, p. 44.

Michael I. Shamos, A catalog of the real numbers, (2007). See p. 397.

Cf. A001622, A002117, A013661, A094214, A132338, A202543.

RealDigits[PolyLog[3,2-GoldenRatio],10,100][[1]]

Equals 4*zeta(3)/5 + Pi^2*log(A132338)/15 - log(A132338)^3/12 (see Finch).

Equals Li_3(A094214^2) = 4*zeta(3)/5 + 2*zeta(2)*log(A132338)/5 - log(A132338)^3/12 (see Shamos).

A383659 Decimal expansion of phi + 2*log(phi), where phi is the golden ratio.

Original entry on oeis.org

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

Offset: 1

Kritsada Moomuang, Jun 11 2025

			2.58045763886910174320...

Cf. A001622, A002390, A202543, A384238, A384682.

RealDigits[GoldenRatio + 2 * Log[GoldenRatio], 10, 100, 0][[1]]

Equals Integral_{x=0..1} sqrt(1/x + sqrt(1/x + sqrt(1/x + ...))) dx.
Equals Integral_{x=0..1} (1 + sqrt(1 + 4/x))/2 dx.
Equals A001622 + A202543.

cp's OEIS Frontend

A002544 a(n) = binomial(2*n+1,n)*(n+1)^2.

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A202537 Decimal expansion of x satisfying e^x-e^(-2x)=1.

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A112449 a(n+2) = (a(n+1)^3 + a(n+1))/a(n) with a(0)=1, a(1)=1.

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A318057 a(n) is the number of binary places to which n-th convergent of continued fraction expansion of the golden section matches the correct value.

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A318058 a(n) is the number of decimal places to which the n-th convergent of the continued fraction expansion of the golden section matches the correct value.

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A182546 Decimal expansion of | log_phi(i) |, where phi is the golden ratio and i is the imaginary unit.

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Mathematica

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A377644 Decimal expansion of Li_3(2 - phi).

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A002544 a(n) = binomial(2n+1,n)(n+1)^2.