Herb Conn - cp's OEIS frontend

A185672 Let f(n) = Sum_{j>=1} j^n3^j/binomial(2j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives r_n.

4, 20, 172, 2084, 32524, 620900, 14014732, 365100644, 10781360524, 355869575780, 12984066273292, 518879340911204, 22540052170064524, 1057507154836226660, 53291594817628483852, 2870834224548449841764, 164633490033421041392524, 10013579272685278891133540, 643872718978606529940390412

Offset: 0

N. J. A. Sloane, Feb 09 2011, following a suggestion from Herb Conn

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..200
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's Interesting Series, arXiv:1009.4274 [math-ph], 2010-2011.
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's interesting series, Amer. Math. Monthly, 120 (2013), 116-130.
D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92(7) (1985), 449-457.

Cf. A181374 (s_n), A180875 and A014307 (2^j rather than 3^j).

f[n_] := Sum[j^n*3^j/Binomial[2*j, j], {j, 1, Infinity}];
a[n_] := FindIntegerNullVector[{Pi/Sqrt[3], 1, N[-f[n], 20]}][[1]];
Table[r = a[n]; Print[r]; r, {n, 0, 8}] (* Jean-François Alcover, Sep 05 2018 *)
Table[Expand[FunctionExpand[FullSimplify[Sum[j^n*3^j/Binomial[2*j, j], {j, 1, Infinity}]]]][[2]] * Sqrt[3]/Pi, {n, 0, 20}] (* Vaclav Kotesovec, May 14 2020 *)
S[k_, z_] := Sum[n!*(z/(4 - z))^n* StirlingS2[k + 1, n]*(1/n + Sum[(-1)^p*Pochhammer[1/2, p]/(p + 1)!* Binomial[n - 1, p]*(4/z)^(p + 1)*(Sqrt[z/(4 - z)]*ArcSin[Sqrt[z]/2] - 1/2*Sum[Gamma[l]/Pochhammer[1/2, l]*(z/4)^l, {l, 1, p}]), {p, 0, n - 1}]), {n, 1, k + 2}]; Table[Expand[Simplify[S[j, 3]]][[2]]*Sqrt[3]/Pi, {j, 0, 20}] (* Vaclav Kotesovec, May 15 2020 *)

a(n) ~ 2^(3/2) * n^(n+1) / (sqrt(3) * exp(n) * (log(4/3))^(n + 3/2)). - Vaclav Kotesovec, May 15 2020

More terms from Vaclav Kotesovec, May 14 2020

A185585 Let f(n) = Sum_{j>=1} j^n/binomial(2j,j) = r_nPi*sqrt(3)/3^{t_n} + s_n/3; sequence gives t_n.

Original entry on oeis.org

3, 3, 4, 5, 5, 5, 6, 5, 7, 8, 8, 9, 10, 10, 10, 10, 8, 11, 12, 12, 13, 14, 14, 13, 15, 13, 16, 17, 17, 18, 19, 19, 19, 20, 19, 21, 22, 22, 23, 24, 24, 24, 24, 23, 24, 25, 25, 26, 27, 27, 26, 28, 26, 29, 30, 30, 31, 32

Offset: 0

N. J. A. Sloane, Feb 09 2011, following a suggestion from Herb Conn

nonn

Petros Hadjicostas, Table of n, a(n) for n = 0..300
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's Interesting Series, arXiv:1009.4274 [math-ph], 2010-2011.
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's interesting series, Amer. Math. Monthly, 120 (2013), 116-130.
D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92(7) (1985), 449-457.

Cf. A098830 (s_n), A181334 (r_n), A181374, A180875, A014307.

# The function LehmerSer is defined in A181334.
a := n -> ilog[3](denom(LehmerSer(n))):
seq(a(n), n=0..57); # Peter Luschny, May 15 2020

f[n_] := Sum[j^n/Binomial[2*j, j], {j, 1, Infinity}];
a[n_] := 1 + Log[3, Denominator[Expand[FunctionExpand[f[n]]][[2, 1]]]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 60}] (* Jean-François Alcover, Nov 24 2017 *)

a(n) = logint(denominator(2*sum(m=1, n+1, sum(p=0, m-1, (-1)^p*(m!/((p+1)*3^(m+2)))*stirling(n+1,m,2)*binomial(2*p,p)*binomial(m-1,p)))), 3) \\ Petros Hadjicostas, May 14 2020

a(n) = ilog[3](denominator(2*Sum_{m=1..n+1} Sum_{p=0..m-1} (-1)^p * (m!/((p+1)*3^(m+2))) * Stirling2(n+1,m) * binomial(2*p,p) * binomial(m-1,p))), where ilog[3](3^k) = k. [It follows from Theorem 1 in Dyson et al. (2013).] - Petros Hadjicostas, May 14 2020

a(11)-a(57) from Nathaniel Johnston, Apr 07 2011

A181374 Let f(n) = Sum_{j>=1} j^n3^j/binomial(2j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives s_n.

Original entry on oeis.org

3, 18, 156, 1890, 29496, 563094, 12709956, 331109658, 9777612432, 322738005150, 11775245575836, 470571509329506, 20441566147934568, 959052902557542246, 48330130399621041396, 2603558645653906065834, 149306059777139762896704, 9081311859252750219451182, 583927964165576868953730636

Offset: 0

N. J. A. Sloane, Feb 09 2011, following a suggestion from Herb Conn

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..200
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's Interesting Series, arXiv:1009.4274 [math-ph], 2010-2011.
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's interesting series, Amer. Math. Monthly, 120 (2013), 116-130.
D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92(7) (1985), 449-457.

Cf. A185672 (r_n), A180875 and A014307 (2^j rather than 3^j).

f[n_] := Sum[j^n*3^j/Binomial[2*j, j], {j, 1, Infinity}];
a[n_] := FindIntegerNullVector[{Pi/Sqrt[3], 1, N[-f[n], 20]}][[2]];
Table[s = a[n]; Print[s]; s, {n, 0, 8}] (* Jean-François Alcover, Sep 05 2018 *)
Table[Expand[FunctionExpand[FullSimplify[Sum[j^n*3^j/Binomial[2*j, j], {j, 1, Infinity}]]]][[1]], {n, 0, 20}] (* Vaclav Kotesovec, May 14 2020 *)
S[k_, z_] := Sum[n!*(z/(4 - z))^n* StirlingS2[k + 1, n]*(1/n + Sum[(-1)^p*Pochhammer[1/2, p]/(p + 1)!* Binomial[n - 1, p]*(4/z)^(p + 1)*(Sqrt[z/(4 - z)]*ArcSin[Sqrt[z]/2] - 1/2*Sum[Gamma[l]/Pochhammer[1/2, l]*(z/4)^l, {l, 1, p}]), {p, 0, n - 1}]), {n, 1, k + 2}]; Table[Expand[Simplify[S[j, 3]]][[1]], {j, 0, 20}] (* Vaclav Kotesovec, May 15 2020 *)

a(n) ~ sqrt(2) * Pi * n^(n+1) / (3 * exp(n) * (log(4/3))^(n + 3/2)). - Vaclav Kotesovec, May 15 2020

More terms from Vaclav Kotesovec, May 14 2020

A181334 Let f(n) = Sum_{j>=1} j^n/binomial(2j,j) = r_nPi*sqrt(3)/3^{t_n} + s_n/3; sequence gives r_n.

Original entry on oeis.org

2, 2, 10, 74, 238, 938, 13130, 23594, 1298462, 26637166, 201403930, 5005052234, 135226271914, 1315508114654, 13747435592810, 153590068548062, 202980764290906, 69141791857625242, 2766595825017102650, 38897014541363246798, 1724835471991750464238, 80219728936311383557694

Offset: 0

N. J. A. Sloane, Feb 09 2011, following a suggestion from Herb Conn

nonn

Petros Hadjicostas, Table of n, a(n) for n = 0..101
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's Interesting Series, arXiv:1009.4274 [math-ph], 2010-2011.
F. J. Dyson, N. E. Frankel and M. L. Glasser, Lehmer's interesting series, Amer. Math. Monthly, 120 (2013), 116-130.
D. H. Lehmer, Interesting series involving the central binomial coefficient, Amer. Math. Monthly, 92(7) (1985), 449-457.

Cf. A098830 (s_n), A185585 (t_n), A181374, A180875, A014307.

LehmerSer := n -> 2*add(add((-1)^p*(m!/((p+1)*3^(m+2)))*Stirling2(n+1,m)
*binomial(2*p, p)*binomial(m-1, p), p=0..m-1), m=1..n+1):
a := n -> numer(LehmerSer(n)): seq(a(n), n=0..21);
# (after Petros Hadjicostas) Peter Luschny, May 15 2020

f[n_] := Sum[j^n/Binomial[2*j, j], {j, 1, Infinity}];
a[n_] := Expand[ FunctionExpand[ f[n] ] ][[2, 1]] // Numerator;
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 24 2017 *)

a(n)=numerator(2*sum(m=1, n+1, sum(p=0, m-1, (-1)^p*(m!/((p+1)*3^(m+2)))*stirling(n+1,m,2)*binomial(2*p,p)*binomial(m-1,p)))) \\ Petros Hadjicostas, May 15 2020

a(n) = numerator(2*Sum_{m=1..n+1} Sum_{p=0..m-1} (-1)^p * (m!/((p+1)*3^(m+2))) * Stirling2(n+1,m) * binomial(2*p,p) * binomial(m-1,p)). [It follows from Theorem 1 in Dyson et al. (2010-2011, 2013).] - Petros Hadjicostas, May 15 2020

a(11)-a(21) from Nathaniel Johnston, Apr 07 2011

A093568 Let M = the 2 X 2 matrix [0 1 / -1 2+sqrt(8)]. Perform the operation M^n * [1 1] = [x y]; then a(n) = floor(x), a(n+1) = floor(y).

Original entry on oeis.org

1, 3, 17, 80, 371, 1714, 7904, 36451, 168098, 775200, 3574898, 16485939, 76026256, 350601298, 1616826563, 7456127936, 34384543809, 158567136067, 731245317072, 3372197587729, 15551164985523, 71715469249777, 330721752004000

Offset: 1

Gary W. Adamson, Mar 31 2004, suggested by Herb Conn

nonn

Lim_{n->infinity} a(n)/a(n-1) = 4.61158178931... = e^(arccosh(1+sqrt(2))). This constant is found in the Descartes circle formula given ("curvature" = 1/r), which states that given 4 mutually tangent circles with curvatures a,b,c,d, (a^2 + b^2 + c^2 + d^2) = (1/2)*(a + b + c + d)^2. Let a=b=1. Find the radius of circle c such that the radius of the innermost circle d = 1/(r of c). The answer is r of c = 4.611581789... and r of d = 1/4.611581... = 0.216845335437. Let C = 4.611581... C + 1/C = 2 + sqrt(8), where C may be found through an iterative method:
1. Begin with the formula for the radius r of the innermost tangential circle surrounded by 3 mutually tangent circles with radii a, b, c. Given a = b = 1, with c variable, then r = f(c) = (2*sqrt(1 + 2/c) - 2 - 1/c) / (4/c - 1/c^2).
2. Using an iterative operation, take any real number 1 or > 1 (we can use 1). Find f(c), then repeatedly use 1/f(c) as the new c. After about a dozen of such operations, the fixed constant becomes 4.61158178931... This constant C has the property that C + 1/C = 4.828427124... = 2 + sqrt(8).

			M^5 * [1 1] = [371.676... 1714.013...]. a(5) = 371, a(6) = 1714. a(10)/a(9) = 775200/168098 = 4.611595...

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Table[ Floor[ MatrixPower[{{0, 1}, {-1, 2 + Sqrt[8]}}, n].{1, 1}][[1]], {n, 24}] (* Robert G. Wilson v, Apr 09 2004 *)

A082649 Triangle of coefficients in expansion of sinh^2(n*x) in powers of sinh(x).

Original entry on oeis.org

1, 4, 4, 16, 24, 9, 64, 128, 80, 16, 256, 640, 560, 200, 25, 1024, 3072, 3456, 1792, 420, 36, 4096, 14336, 19712, 13440, 4704, 784, 49, 16384, 65536, 106496, 90112, 42240, 10752, 1344, 64, 65536, 294912, 552960, 559104, 329472, 114048, 22176, 2160, 81, 262144, 1310720, 2785280, 3276800, 2329600

Offset: 1

Gary W. Adamson, May 16 2003, suggested by Herb Conn

Using arcsin(x) = Pi/2 - arccos(x), valid for -1 < = x <= 1, we find sin^2(k*arcsin(x)) = sin^2(k*arccos(x)) for k odd, while sin^2(k*arcsin(x)) = 1 - sin^2(k*arccos(x)) for k even. Thus the expansion of sin^2(n*x) in powers of cos(x) will produce a similar table of coefficients. See the example section below. - Peter Bala, Feb 02 2017

			sinh^2 x = sinh^2 x
sinh^2 2x = 4 sinh^4 x + 4 sinh^2 x
sinh^2 3x = 16 sinh^6 x + 24 sinh^4 x + 9 sinh^2 x
sinh^2 4x = 64 sinh^8 x + 128 sinh^6 x + 80 sinh^4 x + 16 sinh^2 x
sinh^2 5x = 256 sinh^10 x + 640 sinh^8 x + 560 sinh^6 x + 200 sinh^4 x + 25 sinh^2 x
From _Peter Bala_, Feb 02 2016: (Start)
sin^2(x) = 1 - cos^2(x);
sin^2(2*x) = -4*cos^4(x) + 4*cos^2(x);
sin^2(3*x) = 1 - (16*cos^6(x) - 24*cos^4(x) + 9*cos^2(x));
sin^2(4*x) = -64*cos^8(x) + 128*cos^6(x) - 80*cos^4(x) + 16*cos^2(x);
sin^2(5*x) = 1 - (256*cos^10(x) - 640*cos^8(x) + 560*cos^6(x) - 200*cos^4(x) + 25*cos^2(x)). (End)

Robert Israel, Table of n, a(n) for n = 1..10011 (rows 0 to 140, flattened)

A001108 gives row sums.

Closely related to A123583 and A123588.

g:= (1+x*y)/((1-x*y)*(1-(4+2*y)*x+x^2*y^2)):
S:= series(g,x,15):
seq(seq(coeff(coeff(S,x,n),y,k),k=0..n),n=0..14); # Robert Israel, Dec 20 2017

Table[Reverse[CoefficientList[1/x TrigExpand[Sinh[n ArcSinh[Sqrt[x]]]^2], x]], {n, 7}] // Flatten (* Eric W. Weisstein, Apr 05 2017 *)
Abs[Table[CoefficientList[x^n Piecewise[{{1 - ChebyshevT[n, 1/Sqrt[x]]^2, Mod[n, 2] == 0}, {ChebyshevT[n, 1/Sqrt[x]]^2, Mod[n, 2] == 1}}], x], {n, 10}]] // Flatten (* Eric W. Weisstein, Apr 05 2017 *)

Coefficients are: 4^(n-1), (2n)4^(n-2), (2n)(2n-3)4^(n-3)/2!, (2n)(2n-4)(2n-5)4^(n-4)/3!, (2n)(2n-5)(2n-6)(2n-7)4^(n-5)/4!, (2n)(2n-6)(2n-7)(2n-8)(2n-9)4^(n-6)/5!...

G.f. as triangle: (1+x*y)/((1-x*y)*(1-(4+2*y)*x+x^2*y^2)). - Robert Israel, Dec 20 2017

More terms from Robert Israel, Dec 20 2017

cp's OEIS Frontend

User: Herb Conn

A185672 Let f(n) = Sum_{j>=1} j^n*3^j/binomial(2*j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives r_n.

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A185585 Let f(n) = Sum_{j>=1} j^n/binomial(2*j,j) = r_n*Pi*sqrt(3)/3^{t_n} + s_n/3; sequence gives t_n.

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A181374 Let f(n) = Sum_{j>=1} j^n*3^j/binomial(2*j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives s_n.

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A181334 Let f(n) = Sum_{j>=1} j^n/binomial(2*j,j) = r_n*Pi*sqrt(3)/3^{t_n} + s_n/3; sequence gives r_n.

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A093568 Let M = the 2 X 2 matrix [0 1 / -1 2+sqrt(8)]. Perform the operation M^n * [1 1] = [x y]; then a(n) = floor(x), a(n+1) = floor(y).

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A082649 Triangle of coefficients in expansion of sinh^2(n*x) in powers of sinh(x).

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A185672 Let f(n) = Sum_{j>=1} j^n3^j/binomial(2j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives r_n.

A185585 Let f(n) = Sum_{j>=1} j^n/binomial(2j,j) = r_nPi*sqrt(3)/3^{t_n} + s_n/3; sequence gives t_n.

A181374 Let f(n) = Sum_{j>=1} j^n3^j/binomial(2j,j) = r_n*Pi/sqrt(3) + s_n; sequence gives s_n.

A181334 Let f(n) = Sum_{j>=1} j^n/binomial(2j,j) = r_nPi*sqrt(3)/3^{t_n} + s_n/3; sequence gives r_n.