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.

User: Mohamed Sabba

Mohamed Sabba's wiki page.

Mohamed Sabba has authored 5 sequences.

A362534 Numerators of the ratio of the symmetry-constrained bound to the adiabatic bound on polarization transfer in AXn spin-1/2 systems.

Original entry on oeis.org

1, 1, 6, 6, 15, 15, 140, 140, 315, 315, 1386, 1386, 3003, 3003, 51480, 51480, 109395, 109395, 92378, 92378, 969969, 969969, 2704156, 2704156, 16900975, 16900975, 70204050, 70204050, 145422675, 145422675, 4808643120, 4808643120, 9917826435, 9917826435, 40838108850, 40838108850
Offset: 1

Author

Mohamed Sabba, Apr 24 2023

Keywords

Comments

In spin physics and NMR, these numbers appear as the numerators of the ratio of different classes of upper bounds on the transfer of z-magnetization in AXn spin systems from a group of spin-1/2 nuclei Xn to a single spin-1/2 nucleus A.
The symmetry-constrained upper bounds are given by the function f(n):
(1) for even n, f(n) = (2^(1-n))*n*binomial(n-1, n/2)
(2) for odd n, f(n) = (2^(1-n))*n*binomial(n-1, (n-1)/2)
The adiabatic bounds are given by the function g(n):
(3) for even n, g(n) = 2*(1-(2^(-n))*binomial(n, n/2))
(4) for odd n, g(n) = 2*(1-(2^(-n))*binomial(n, (n-1)/2))
Where we have the relation:
(5) g(n) = 2*(1 - f(n+1)/(n+1))
The sequence a(n) is defined as the numerator of f(n)/g(n):
(6) a(n) = numerator(f(n)/g(n))
(7) for even n, f(n)/g(n) = (n/2)/(2^(n)*binomial(n, n/2)^(-1) - 1)
(8) for odd n, f(n)/g(n) = ((n+1)/2)/(2^(n)*binomial(n, (n+1)/2)^(-1) - 1)
The first few values of the upper symmetry-constrained bounds f(n) are {1, 1, 3/2, 3/2, 15/8, 15/8, 35/16, 35/16, 315/128, 315/128, ...} which appears to be related to A086116 and A001803.
The first few values of the upper adiabatic bounds g(n) are {1, 1, 5/4, 5/4, 11/8, 11/8, 93/64, 93/64, 193/128, 193/128, ...} which appears to be related to A141244 and A120778.
The first few values of f(n)/g(n) are {1, 1, 6/5, 6/5, 15/11, 15/11, 140/93, 140/93, 315/193, 315/193, ...}
Conjecture: the numerator of g(n) is the denominator of f(n)/g(n).

Crossrefs

Cf. A001803, A086116, A120778, A141244 (denominators but shifted).

Programs

  • Mathematica
    Table[Numerator[Ceiling[n/2]  (2^n Binomial[n, Ceiling[n/2]]^-1 - 1 )^-1], {n, 1, 20}]
  • PARI
    a(n) = numerator(ceil(n/2)/(2^(n)*binomial(n,ceil(n/2))^(-1) - 1)); \\ Michel Marcus, Apr 25 2023

Formula

a(n) = numerator(ceiling(n/2)/(2^(n)*binomial(n,ceiling(n/2))^(-1) - 1)).

A262957 Numerators of the n-th iteration of the alternating continued fraction formed from the positive integers, starting with (1 - ...).

Original entry on oeis.org

2, 3, 19, 64, 538, 2833, 29169, 210308, 2572158, 23595915, 334778571, 3732092084, 60305234822, 791741083537, 14359827157009, 217037153818264, 4366918714540522, 74685204276602819, 1651116684587556019, 31524723785455714840, 759659139498065625218, 16017463672140861567617
Offset: 1

Author

Mohamed Sabba, Nov 19 2015

Keywords

Comments

As n->inf, a(n)/A263295(n) converges to 0.57663338973... (A346590); this number has a surprisingly elegant standard continued fraction representation of [0; 1, 1, 2, 1, 3, 4, 1, 5, 6, 1, 7, 8, ...].
From Robert Israel, Dec 22 2015: (Start)
a(n) is the numerator of b(n)/c(n) where
b(1) = 2, b(2) = 3, c(1) = 3, c(2) = 5,
b(n+1) = (((-1)^n*(n-1)+n*(n+2))*b(n) - (1+(-1)^n*(n+1))*b(n-1))/(n-(-1)^n),
c(n+1) = (((-1)^n*(n-1)+n*(n+2))*c(n) - (1+(-1)^n*(n+1))*c(n-1))/(n-(-1)^n).
Conjecture: b(n) and c(n) are coprime for all n, so that a(n) = b(n).
I have verified this for n <= 10000. (End)

Examples

			(1-1/(2+1)) = 2/3, so a(1) = 2;
(1-1/(2+1/(3-1))) = 3/5, so a(2) = 3;
(1-1/(2+1/(3-1/(4+1)))) = 19/33, so a(3) = 19;
(1-1/(2+1/(3-1/(4+1/(5-1))))) = 64/111, so a(4) = 64.
		

Crossrefs

Same principle as A244279 and A244280 - except here we begin with subtraction rather than addition.
Cf. A263295 (denominators), A346590.

Programs

  • Maple
    P[1]:= 2: P[2]:= 3:
    Q[1]:= 3; Q[2]:= 5;
    for i from 2 to 100 do
      P[i+1]:= ((-1)^i*(i-1) + i^2 + 2*i)/(i-(-1)^i)*P[i] + (1 + (i+1)*(-1)^i)/((-1)^i-i)*P[i-1];
      Q[i+1]:= ((-1)^i*(i-1) + i^2 + 2*i)/(i-(-1)^i)*Q[i] + (1 + (i+1)*(-1)^i)/((-1)^i-i)*Q[i-1];
    od:
    seq(numer(P[i]/Q[i]),i=1..100); # Robert Israel, Dec 22 2015
  • PARI
    a(n) = if(n%2==0, s=-1, s=1); t=1; while(n>-1, t=n+1+s/t; n--; s=-s); denominator(t=1/t)
    vector(30, n, a(n)) \\ Mohamed Sabba, Dec 22 2015

Extensions

More terms from Mohamed Sabba, Dec 22 2015

A263295 Denominators of the n-th iteration of the alternating continued fraction formed from the positive integers, starting with (1 - ...).

Original entry on oeis.org

3, 5, 33, 111, 933, 4913, 50585, 364717, 4460647, 40920133, 580574377, 6472209467, 104581586665, 1373040648769, 24902871413201, 376386726269561, 7573128424949291, 129519388933667493, 2863373356440803473, 54670305859684290279, 1317404009250178503245
Offset: 1

Author

Mohamed Sabba, Nov 20 2015

Keywords

Comments

As n->inf, A262957(n)/a(n) converges to 0.57663338973018...; this number has a surprisingly elegant standard continued fraction representation: [0; 1, 1, 2, 1, 3, 4, 1, 5, 6, 1, 7, 8, ...].

Examples

			(1-1/(2+1)) = 2/3, so a(1) = 3;
(1-1/(2+1/(3-1))) = 3/5, so a(2) = 5;
(1-1/(2+1/(3-1/(4+1)))) = 19/33, so a(3) = 33;
(1-1/(2+1/(3-1/(4+1/(5-1))))) = 64/111, so a(4) = 111.
		

Crossrefs

Same principle as A244279 and A244280 - except here we begin with subtraction rather than addition.
Cf. A262957 (numerators).

Programs

  • PARI
    a(n) = if(n%2==0, s=-1, s=1); t=1; while(n>0, t=n+1+s/t; n--; s=-s); denominator(t=1/t)
    vector(30, n, a(n)) \\ corrected by Mohammed Sabba, Dec 22 2015

Extensions

More terms from Mohamed Sabba, Dec 22 2015

A244280 Denominators of the n-th iteration of the alternating continued fraction of the positive integers, initiated with (1 + ...).

Original entry on oeis.org

2, 2, 11, 27, 202, 870, 8129, 50681, 570638, 4673558, 61724211, 627102091, 9514420518, 115483788186, 1980202320561, 27962630844865, 534877446987082, 8615820301234778, 181912525664114699, 3292162161484924619, 76056192127792619858, 1527880958525256735838
Offset: 1

Author

Mohamed Sabba, Jun 24 2014

Keywords

Comments

As n-->inf, a(n) converges to 0.628736607098954801603428...
This is the result of taking the denominator of a continued fraction with alternating signs a(n) = 1/(1+1/(2-1/(3+1/(4-...1/(n +/- 1))))), where addition follows an odd number and subtraction follows an even number.

Examples

			a(1) = 1/(1+1) = 1/2;
a(2) = 1/(1+1/(2-1)) = 1/2;
a(3) = 1/(1+1/(2-1/(3+1))) = 7/11;
a(4) = 1/(1+1/(2-1/(3+1/(4-1)))) = 17/27.
		

Crossrefs

Cf. A244279 (Numerators).

Programs

  • Maple
    seq(denom(numtheory:-cfrac([0, [1,1], seq([(-1)^j,j],j=2..n),[(-1)^(n+1),1]])), n = 1..40); # Robert Israel, Jan 17 2016
  • PARI
    a(n) = if(n%2==0,s=-1,s=1); t=1; while(n>0, t=n+s/t; n--; s=-s); denominator(t=1/t)
    vector(30, n, a(n)) \\ Colin Barker, Jul 20 2014

Extensions

More terms from Colin Barker, Jul 20 2014

A244279 Numerators of the n-th iteration of the alternating continued fraction of the positive integers, initiated with (1 + ...).

Original entry on oeis.org

1, 1, 7, 17, 127, 547, 5111, 31865, 358781, 2938437, 38808271, 394282041, 5982064475, 72608885159, 1245025688399, 17581129642961, 336297031232409, 5417081623572649, 114375064174857015, 2069902867431592833, 47819312187294567447, 960634689914268797707
Offset: 1

Author

Mohamed Sabba, Jun 24 2014

Keywords

Comments

As n-->inf, a(n)/A244280(n) converges to 0.628736607098954801603428... ; this number has a surprisingly elegant standard continued fraction representation of [0; 1, 1, 1, 2, 3, 1, 4, 5, 1, 6, 7, 1, 8, 9...].

Examples

			a(1) = 1/(1+1) = 1/2;
a(2) = 1/(1+1/(2-1)) = 1/2;
a(3) = 1/(1+1/(2-1/(3+1))) = 7/11;
a(4) = 1/(1+1/(2-1/(3+1/(4-1)))) = 17/27.
		

Crossrefs

Cf. A244280 (Denominators).

Programs

  • Maple
    seq(numer(numtheory:-cfrac([0, [1,1], seq([(-1)^j,j],j=2..n),[(-1)^(n+1),1]])), n = 1..40); # Robert Israel, Jan 17 2016
  • PARI
    a(n) = if(n%2==0,s=-1,s=1); t=1; while(n>0, t=n+s/t; n--; s=-s); numerator(t=1/t)
    vector(30, n, a(n)) \\ Colin Barker, Jul 20 2014

Formula

This is the result of taking the numerator of a continued fraction with alternating signs a(n) = 1/(1+1/(2-1/(3+1/(4-...1/(n +/- 1))))), where addition follows an odd number and subtraction follows an even number.

Extensions

More terms from Colin Barker, Jul 20 2014