A186375 -id:A186375 - cp's OEIS frontend

A318245 Scaled g.f. T(v) = Sum_{n>=0} a(n)(3v/64)^n satisfies 9(5v-4)T + d/dv(16v(v-1)(3v-4)T') = 0, and a(0)=1; sequence gives a(n).

1, 12, 180, 2928, 49860, 875952, 15754704, 288722880, 5373771876, 101334517680, 1932405892560, 37208369165760, 722497419680400, 14132680228175040, 278236490874120000, 5508974545258860288, 109624581377872629156, 2191185332414847848880, 43971545517545956240464

Offset: 0

Bradley Klee, Aug 22 2018

nonn

The linked document "Proof Certificate" explains that period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_4 symmetry (also cf. Examples and A186375).

			Period function T_{O}(w): Take T_{C3}(u) and T_{C4}(v) from A186375 and A318245 respectively. Set (u,v)=(w-2/3,2-w), with u in [0,1/3], v in [0,1], and w in [2/3,2]. Define piecewise function T_{O}(w) = T_{C3}(w-2/3) if w in [2/3,1] or T_{O}(w) = T_{C4}(2-w) if w in [1,2].
Geometric Singular Points: Construct a family of algebraic sphere curves by intersecting a sphere 1=X^2+Y^2+Z^2 with the octahedral surface w=2*(X^4+Y^4+Z^4). Four cube vertex axes--(x+y+z, -x+y+z, x-y+z, x+y-z)--intersect the sphere in eight circular points with w=2/3. Three octahedron vertex axes--(x, y, z)--intersect the sphere in six circular points with w=2. Six cuboctahedron vertex axes--(x+y, x-y, y+z, y-z, z+x, z-x)--intersect the sphere in twelve hyperbolic points with w=1.

W. G. Harter and C. W. Patterson, Rotational energy surfaces and high-J eigenvalue structure of polyatomic molecules, The Journal of Chemical Physics, 80 (1984), 4252.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Bradley Klee, Proof Certificate.
Bradley Klee, Checking Weierstrass data, 2023.
Eric Weisstein's World of Mathematics, Goursat's Surface.

Periods: A186375, A318417.

CoefficientList[Expand[Normal@Series[Divide[Sqrt[S],Sqrt[1-4*S*x]*Sqrt[S-8 + 8*Sqrt[1-4*S*x]]], {x, 0, 13}]/.{S->12+4*Q^2}]/.{Q^n_:>(1/2)^n*Binomial[n, n/2]} /.{x->1/3*x}, x]
RecurrenceTable[{3*n^2*a[n] - 4*(28*n^2-28*n+9)*a[n-1] + 64*(4*n-5)*(4*n-3)*a[n-2] == 0, a[0]==1, a[1]==12}, a, {n,0,1000}]

3*n^2*a(n) - 4*(28*n^2-28*n+9)*a(n-1) + 64*(4*n-5)*(4*n-3)*a(n-2) = 0.
For n > 0, a(n) mod 3 = 0 (conjecture, tested up to n=3*10^6).
From Bradley Klee, May 30 2023: (Start)
The defining ODE can be derived from the following Weierstrass data:
g2 = (3/16)*(256 - 576*x + 405*x^2 - 81*x^3);
g3 = (1/64)*(4096 - 13824*x + 17496*x^2 - 9963*x^3 + 2187*x^4);
which determine an elliptic surface with four singular fibers. (End)

A186378 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^2 + x^p)^n found at sufficiently large p for some fixed k>0.

Original entry on oeis.org

1, 7, 95, 1609, 30271, 606057, 12636689, 271026455, 5934011839, 131978406553, 2971793928145, 67586972435495, 1549805136840625, 35783848365934663, 831089570101489423, 19400246240227360809, 454864027237803296703

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: (1 + 2*x + x^2 + x^p)^n for all p>2(n+1).
...
More generally, let B(x) equal the g.f. for the least sum of the squares of the coefficients in (F(x^k) + x^p)^n where F(x) is a finite polynomial in x with degree d and p>(n+1)dk for some fixed k>0,
then B(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} c(n)/n!^2] where c(n) equals the sum of the squares of the coefficients in the polynomial F(x)^n.

			G.f.: A(x) = 1 + 7*x + 95*x^2/2!^2 + 1609*x^3/3!^2 + 30271*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = C(x) * BesselI(0, 2*sqrt(x)) where
C(x)= 1 + 6*x + 70*x^2/2!^2 + 924*x^3/3!^2 + 12870*x^4/4!^2 +...+ (4n)!/(2n)!^2*x^n/n!^2

Cf. A186375, A186376, A186377, A186391, A186392.

Table[Sum[Binomial[n,k]^2 * Binomial[4*k,2*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

{a(n)=local(V=Vec((1+2*x+x^2+x^(2*n+3))^n));V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*(4*k)!/(2*k)!^2)}

{a(n)=n!^2*polcoeff(sum(m=0,n,(4*m)!/(2*m)!^2*x^m/m!^2)*sum(m=0,n,x^m/m!^2+x*O(x^n)),n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(4k,2k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} (4n)!/(2n)!^2 *x^n/n!^2] *[Sum_{n>=0} x^n/n!^2].
Recurrence: n^2*(2*n-1)^2*(1152*n^4 - 8160*n^3 + 21040*n^2 - 23376*n + 9467)*a(n) = 3*(55296*n^8 - 503808*n^7 + 1891456*n^6 - 3812256*n^5 + 4504864*n^4 - 3193428*n^3 + 1326995*n^2 - 296732*n + 27900)*a(n-1) - 3*(451584*n^8 - 4607232*n^7 + 19744768*n^6 - 46227488*n^5 + 64243016*n^4 - 53731348*n^3 + 26049967*n^2 - 6596672*n + 675000)*a(n-2) + (n-2)^2*(2230272*n^6 - 16991232*n^5 + 49582720*n^4 - 69169056*n^3 + 46825856*n^2 - 13847412*n + 1451547)*a(n-3) - 900*(n-3)^2*(n-2)^2*(1152*n^4 - 3552*n^3 + 3472*n^2 - 1168*n + 123)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 5^(2*n+3/2) / (2^(7/2) * Pi * n). - Vaclav Kotesovec, Feb 12 2015

A186376 a(n) is the sum of the squares of the coefficients of (x + 2y + 3z)^n.

Original entry on oeis.org

1, 14, 294, 7292, 198310, 5717124, 171485916, 5290989816, 166688596998, 5335746337460, 172951272017524, 5662840724506056, 186960502253087836, 6215612039963043368, 207865952390729881080, 6987002286567227550192

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + 3*x^p)^n found at sufficiently large p for some fixed k>0.
Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:
. (1 + 3*x^k + 2*x^p)^n, or
. (2 + x^k + 3*x^p)^n, or
. (3 + 2*x^k + x^p)^n, etc.,
for all p>(n+1)k and fixed k>0.
a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n. - Michael Somos, Aug 25 2018

			G.f.: A(x) = 1 + 14*x + 294*x^2/2!^2 + 7292*x^3/3!^2 +...
The g.f. may be expressed as:
[Sum_{n>=0}x^n/n!^2]*[Sum_{n>=0}(4x)^n/n!^2]*[Sum_{n>=0}(9x)^n/n!^2].

Cf. A186375, A186377, A186378, A186391.

Table[Sum[Binomial[n,k]^2 * Sum[Binomial[k,j]^2 * 4^(k-j) * 9^j, {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[(x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 2, z -> 3}]]; (* Michael Somos, Aug 25 2018 *)

{a(n)=local(V=Vec((1+2*x+3*x^(n+2))^n));V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*sum(j=0,k,binomial(k,j)^2*4^(k-j)*9^j))}

{a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)*sum(m=0,n,(2^2*x)^m/m!^2)*sum(m=0,n,(3^2*x)^m/m!^2),n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2 *Sum_{j=0..k} C(k,j)^2*4^(k-j)*9^j.
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x) * B(2^2*x) * B(3^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*(4*n - 7)*a(n) = 14*(16*n^3 - 44*n^2 + 34*n - 9)*a(n-1) - 196*(2*n - 3)^2*(4*n - 3)*a(n-2) + 144*(4*n - 9)*(4*n - 7)*(4*n - 3)*a(n-3). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 2^(2*n-1) * 3^(2*n+1) / (Pi*n). - Vaclav Kotesovec, Feb 12 2015

Name changed to match the definition given by Michael Somos. - Paul D. Hanna, Sep 05 2018

A186377 a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + x^p + x^q)^n found at sufficiently large p and q>(n+1)p for some fixed k>0.

Original entry on oeis.org

1, 7, 79, 1129, 18559, 333577, 6365089, 126652183, 2598628543, 54577439833, 1167481074529, 25346459683783, 557042221952881, 12368307313680871, 277027947337574911, 6251808554314780009, 142015508983550880703

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:
. (2 + x^k + x^p + x^q)^n, or
. (1 + x^k + 2*x^p + x^q)^n, or
. (1 + x^k + x^p + 2*x^q)^n,
for all p>(n+1)k and q>(n+1)p and fixed k>0.

			G.f.: A(x) = 1 + 7*x + 79*x^2/2!^2 + 1129*x^3/3!^2 + 18559*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2]^3 *[Sum_{n>=0} (4x)^n/n!^2] where
[Sum_{n>=0} x^n/n!^2]^3 = 1 + 3*x + 15*x^2/2!^2 + 93*x^3/3!^2 + 639*x^4/4!^2 + 4653*x^5/5!^2 +...+ A002893(n)*x^n/n!^2 +...

Cf. A186375, A186376, A186378.

Table[Sum[Binomial[n,k]^2 * 4^(n-k) *Sum[Binomial[k,j]^2 * Binomial[2j,j], {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

{a(n)=local(V=Vec((1+2*x+x^(n+2)+x^(n^2+2*n+3))^n));V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*4^(n-k)*sum(j=0,k,binomial(k,j)^2*binomial(2*j,j)))}

{a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^3*sum(m=0,n,(2^2*x)^m/m!^2),n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2 *4^(n-k) *Sum_{j=0..k} C(k,j)^2*C(2j,j).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x)^3 * B(2^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: (n-1)*n^3*(3*n - 5)*a(n) = 2*(n-1)*(54*n^4 - 174*n^3 + 192*n^2 - 99*n + 20)*a(n-1) - 2*(441*n^5 - 2604*n^4 + 6102*n^3 - 7107*n^2 + 4111*n - 940)*a(n-2) + 2*(n-2)^2*(726*n^3 - 3076*n^2 + 4188*n - 1655)*a(n-3) - 225*(n-3)^2*(n-2)^2*(3*n - 2)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 5^(2*n+2) / (2^(7/2) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Feb 12 2015

A318495 Scaled g.f. T(u) = Sum_{n>=0} a(n)(u/16)^n satisfies 5(21u-16)T + d/du( 4u(u-1)(27u-32)*T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 10, 120, 1540, 20500, 279480, 3876600, 54496200, 774468900, 11107261000, 160553895040, 2336799457200, 34219387524400, 503846306168800, 7455357525594000, 110811908027490960, 1653792126235140900, 24774309852363829800, 372404448149589213600

Offset: 0

Bradley Klee, Aug 27 2018

nonn

The linked document "Proof Certificate" gives data for verifying that period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_5 symmetry (also cf. Examples and A318496).

			Period function T_{I}(w): Take T_{C5}(u) and T_{C3}(v) from A318495 and A318496 respectively. Set (u,v)=(1-w,w+5/27), with u in [0,1], v in [0,5/27], and w in [-5/27,1]. Define piecewise function T_{I}(w) = T_{C5}(1-w) if w in [0,1] or T_{I}(w) = T_{C3}(w+5/27) if w in [-5/27,0].
Geometric Singular Points: Construct a family of algebraic sphere curves by intersecting a sphere 1=X^2+Y^2+Z^2 with the icosahedral surface w=Z^6 - 5*(X^2+Y^2)*Z^4 + 5*(X^2+Y^2)^2*Z^2 - 2*(X^4-10*X^2*Y^2+5*Y^4)*X*Z. Six icosahedron vertex axes intersect the sphere in twelve circular points with w=1. Ten dodecahedron vertex axes intersect the sphere in twenty circular points with w=-5/27. Fifteen icosidodecahedron vertex axes intersect the sphere in thirty hyperbolic points with w=0.

É. Goursat, Étude des surfaces qui admettent tous les plans de symétrie d'un polyèdre régulier, Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 4 (1887), 166-170.
W. G. Harter and D. E. Weeks, Rotation-vibration spectra of icosahedral molecules. I. Icosahedral symmetry analysis and fine structure, Journal of Chemical Physics, 90 (1989), 4370.
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Bradley Klee, Proof Certificate.
Bradley Klee, Checking Weierstrass data, 2023.
O. Laporte, Polyhedral Harmonics, Zeitschrift für Naturforschung A, 8-11 (1948), 450.

Cf. A318496. Periods: A186375, A318245, A318417.

a:=[1,10];; for n in [3..20] do a[n]:=(1/(2*(n-1)^2))*(( (59*(n^2-3*n+2)+20)*a[n-1]-(12*(6*n-13)*(6*n-11))*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018

RecurrenceTable[{2 n^2 a[n] - (59 n^2 - 59 n + 20) a[n - 1] + 12 (6 n - 7) (6 n - 5) a[n - 2] == 0, a[0] == 1, a[1] == 10}, a, {n, 0, 1000}]

2*n^2*a(n) - (59*n^2-59*n+20)*a(n-1) + 12*(6*n-7)*(6*n-5)*a(n-2) = 0.
For n > 0, a(n) mod 10 = 0 (conjecture, tested up to n=10^6).
From Bradley Klee, May 30 2023: (Start)
The defining ODE can be derived from the following Weierstrass data:
g2 = (243/256)*(256-640*x+520*x^2-135*x^3);
g3 = (729/8192)*(8192-30720*x+44160*x^2-29680*x^3+8775*x^4-729*x^5);
which determine an elliptic surface with four singular fibers. (End)
G.f.: hypergeom([1/12, 5/12],[1],1728*x^5*(27*x-2)^3*(16*x-1)^2/(2160*x^3-520*x^2+40*x-1)^3)/(1-40*x+520*x^2-2160*x^3)^(1/4). - Mark van Hoeij, Dec 13 2024

A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)(v/16)^n satisfies 15(189v-80)T + d/dv(4v(27v-5)(27v-32)T') = 0, and a(0)=1; sequence gives a(n).

Original entry on oeis.org

1, 30, 1440, 85260, 5606100, 391231080, 28360117800, 2110794125400, 160187289344100, 12339496371120600, 961855480344860640, 75700880007230883600, 6005580964527420946800, 479651805879329497831200, 38529018420812424368031600, 3110295017383730347887664560

Offset: 0

Bradley Klee, Aug 27 2018

nonn

Period function T(v) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318495.

É. Goursat, Étude des surfaces qui admettent tous les plans de symétrie d'un polyèdre régulier, Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 4 (1887), 166-170.

Cf. A318495. Periods: A186375, A318245, A318417.

a:=[1,30];; for n in [3..20] do a[n]:=(1/(10*(n-1)^2))*(3*(333*(n^2-3*n+2)+100)*a[n-1]-(324*(6*n-13)*(6*n-11)*a[n-2])); od; a; # Muniru A Asiru, Sep 24 2018

RecurrenceTable[{10 n^2 a[n] - 3 (333 n^2 - 333 n + 100) a[n-1] + 324 (6*n - 7) (6 n - 5) a[n-2] == 0, a[0] == 1, a[1] == 30}, a, {n, 0, 15}]

10*n^2*a(n) - 3*(333*n^2-333*n+100)*a(n-1) + 324*(6*n-7)*(6*n-5)*a(n-2) = 0.
For n > 0, a(n) mod 30 = 0 (conjecture, tested up to n=10^6).
G.f.: hypergeom([1/12, 5/12],[1],-1728*(432*x-5)^2*x^3*(27*x-2)^5/(1-120*x+3240*x^2+174960*x^3)^3)/(1-120*x+3240*x^2+174960*x^3)^(1/4). - Mark van Hoeij, Dec 13 2024

cp's OEIS Frontend

A318245 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(3*v/64)^n satisfies 9*(5*v-4)*T + d/dv(16*v*(v-1)*(3*v-4)*T') = 0, and a(0)=1; sequence gives a(n).

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A186378 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^2 + x^p)^n found at sufficiently large p for some fixed k>0.

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A186376 a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n.

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A186377 a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + x^p + x^q)^n found at sufficiently large p and q>(n+1)p for some fixed k>0.

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A318495 Scaled g.f. T(u) = Sum_{n>=0} a(n)*(u/16)^n satisfies 5*(21*u-16)*T + d/du( 4*u*(u-1)*(27*u-32)*T') = 0, and a(0)=1; sequence gives a(n).

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A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)*(v/16)^n satisfies 15*(189*v-80)*T + d/dv(4*v*(27*v-5)*(27*v-32)*T') = 0, and a(0)=1; sequence gives a(n).

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A318245 Scaled g.f. T(v) = Sum_{n>=0} a(n)(3v/64)^n satisfies 9(5v-4)T + d/dv(16v(v-1)(3v-4)T') = 0, and a(0)=1; sequence gives a(n).

A186376 a(n) is the sum of the squares of the coefficients of (x + 2y + 3z)^n.

A318495 Scaled g.f. T(u) = Sum_{n>=0} a(n)(u/16)^n satisfies 5(21u-16)T + d/du( 4u(u-1)(27u-32)*T') = 0, and a(0)=1; sequence gives a(n).

A318496 Scaled g.f. T(v) = Sum_{n>=0} a(n)(v/16)^n satisfies 15(189v-80)T + d/dv(4v(27v-5)(27v-32)T') = 0, and a(0)=1; sequence gives a(n).