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 1-3 of 3 results.

A174502 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086903(n)) ), where A086903(n) = (4+sqrt(15))^n + (4-sqrt(15))^n.

Original entry on oeis.org

1, 6, 1, 60, 1, 486, 1, 3840, 1, 30246, 1, 238140, 1, 1874886, 1, 14760960, 1, 116212806, 1, 914941500, 1, 7203319206, 1, 56711612160, 1, 446489578086, 1, 3515205012540, 1, 27675150522246, 1, 217885999165440, 1, 1715412842801286, 1
Offset: 0

Views

Author

Paul D. Hanna, Mar 21 2010

Keywords

Examples

			Let L = Sum_{n>=1} 1/(n*A086903(n)) or, more explicitly,
L = 1/8 + 1/(2*62) + 1/(3*488) + 1/(4*3842) + 1/(5*30248) +...
so that L = 0.1338200441271648228100625767561479963630539052200...
then exp(L) = 1.1431870779045667085973926071888878662387686835715...
equals the continued fraction given by this sequence:
exp(L) = [1;6,1,60,1,486,1,3840,1,30246,1,238140,1,...]; i.e.,
exp(L) = 1 + 1/(6 + 1/(1 + 1/(60 + 1/(1 + 1/(486 + 1/(1 +...)))))).
Compare these partial quotients to A086903(n), n=1,2,3,...:
[8,62,488,3842,30248,238142,1874888,14760962,116212808,914941502,...].

Links

Crossrefs

Cf. A086903, A174500, A174501, A174503, A174508.

Programs

  • Mathematica
    LinearRecurrence[{0,9,0,-9,0,1},{1,6,1,60,1,486},50] (* Harvey P. Dale, Jun 09 2013 *)
  • PARI
    {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((4+sqrt(15))^m+(4-sqrt(15))^m))));contfrac(exp(L))[n]}

Formula

a(2n-2) = 1, a(2n-1) = A086903(n) - 2, for n>=1 [conjecture].
The above conjectures are correct. See the Bala link for details. - Peter Bala, Jan 08 2013
a(n) = 9*a(n-2)-9*a(n-4)+a(n-6). G.f.: -(x^4+6*x^3-8*x^2+6*x+1) / ((x-1)*(x+1)*(x^4-8*x^2+1)). [Colin Barker, Jan 20 2013]

A174509 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A086927(n)) ), where A086927(n) = (5+sqrt(26))^n + (5-sqrt(26))^n.

Original entry on oeis.org

1, 9, 101, 1, 1029, 10401, 1, 105049, 1060901, 1, 10714069, 108201601, 1, 1092730089, 11035502501, 1, 111447755109, 1125513053601, 1, 11366578291129, 114791295964901, 1, 1159279537940149, 11707586675366401, 1
Offset: 0

Views

Author

Paul D. Hanna, Mar 21 2010

Keywords

Examples

			Let L = Sum_{n>=1} 1/(n*A086927(n)) or, more explicitly,
L = 1/10 + 1/(2*102) + 1/(3*1030) + 1/(4*10402) + 1/(5*105050) +...
so that L = 0.1052516947742519131304505213983109248819463097531...
then exp(L) = 1.1109902055968924364755807035083159869000358017128...
equals the continued fraction given by this sequence:
exp(L) = [1;9,101,1,1029,10401,1,105049,1060901,1,...]; i.e.,
exp(L) = 1 + 1/(9 + 1/(101 + 1/(1 + 1/(1029 + 1/(10401 +1/(1+...)))))).
Compare these partial quotients to A086927(n), n=1,2,3,...:
[10,102,1030,10402,105050,1060902,10714070,108201602,...].

Links

Crossrefs

Cf. A086927, A174503, A174508, A174510.

Programs

  • Mathematica
    LinearRecurrence[{0,0,103,0,0,-103,0,0,1},{1,9,101,1,1029,10401,1,105049,1060901},30] (* Harvey P. Dale, Dec 24 2014 *)
  • PARI
    {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((5+sqrt(26))^m+(5-sqrt(26))^m))));contfrac(exp(L))[n]}

Formula

a(3n-3) = 1, a(3n-2) = A086927(2n-1) - 1, a(3n-1) = A086927(2n) - 1, for n>=1 [conjecture].
a(n) = 103*a(n-3)-103*a(n-6)+a(n-9). G.f.: -(x^2-x+1)*(x^6-10*x^5-10*x^4-2*x^3+110*x^2+10*x+1) / ((x-1)*(x^2+x+1)*(x^6-102*x^3+1)). [Colin Barker, Jan 20 2013]
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A086927(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(26) - 5. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 10. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(26) - 5}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(26) - 5}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
(End)

A174507 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A085447(n)) ), where A085447(n) = (3+sqrt(10))^n + (3-sqrt(10))^n.

Original entry on oeis.org

1, 5, 37, 1, 233, 1441, 1, 8885, 54757, 1, 337433, 2079361, 1, 12813605, 78960997, 1, 486579593, 2998438561, 1, 18477210965, 113861704357, 1, 701647437113, 4323746327041, 1, 26644125399365, 164188498723237, 1, 1011775117738793
Offset: 0

Views

Author

Paul D. Hanna, Mar 21 2010

Keywords

Examples

			Let L = Sum_{n>=1} 1/(n*A085447(n)) or, more explicitly,
L = 1/6 + 1/(2*38) + 1/(3*234) + 1/(4*1442) + 1/(5*8886) +...
so that L = 0.1814484777922995750614847484088330644558009487798...
then exp(L) = 1.1989527624251050123398509513177598419795554140316...
equals the continued fraction given by this sequence:
exp(L) = [1;5,37,1,233,1441,1,8885,54757,1,337433,...]; i.e.,
exp(L) = 1 + 1/(5 + 1/(37 + 1/(1 + 1/(233 + 1/(1441 + 1/(1 +...)))))).
Compare these partial quotients to A085447(n), n=1,2,3,...:
[6,38,234,1442,8886,54758,337434,2079362,12813606,78960998,...].

Links

Crossrefs

Cf. A085447, A174501, A174506, A174508.

Programs

  • Mathematica
    LinearRecurrence[{0,0,39,0,0,-39,0,0,1},{1,5,37,1,233,1441,1,8885,54757},30] (* Harvey P. Dale, Aug 07 2016 *)
  • PARI
    {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(10))^m+(3-sqrt(10))^m))));contfrac(exp(L))[n]}

Formula

a(3n-3) = 1, a(3n-2) = A085447(2n-1) - 1, a(3n-1) = A085447(2n) - 1, for n>=1 [conjecture].
a(n) = 39*a(n-3)-39*a(n-6)+a(n-9). G.f.: -(x^2-x+1) * (x^6 -6*x^5 -6*x^4 -2*x^3 +42*x^2 +6*x +1) / ((x-1)*(x^2+x+1)*(x^6-38*x^3+1)). [Colin Barker, Jan 20 2013]
From Peter Bala, Jan 25 2013: (Start)
The above conjectures are correct. The real number exp( Sum {n>=1} 1/(n*A085447(n)) ) is equal to the infinite product F(x) := product {n >= 0} (1 + x^(4*n+3))/(1 - x^(4*n+1)) evaluated at x = sqrt(10) - 3. Ramanujan has given a continued fraction expansion for the product F(x). Using this we can find the simple continued fraction expansion of the numbers F(1/2*(sqrt(N^2 + 4) - N)), N a positive integer. The present case is when N = 6. See the Bala link for details.
The theory also provides the simple continued fraction expansion of the numbers F({sqrt(10) - 3}^(2*k+1)), k = 1, 2, 3, ...: if [1; c(1), c(2), 1, c(3), c(4), 1, ...] denotes the present sequence then the simple continued fraction expansion of F({sqrt(10) - 3}^(2*k+1)) is given by [1; c(2*k+1), c(2*(2*k+1)), 1, c(3*(2*k+1)), c(4*(2*k+1)), 1, ...].
(End)
Showing 1-3 of 3 results.

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