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

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

Offset

1,1

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)

Links

Robert Israel, Table of n, a(n) for n = 1..448

Peter Bala, A note on A262957 and A263295

Example

(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.

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

Prog

(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

Crossrefs

Same principle as A244279 and A244280 - except here we begin with subtraction rather than addition.

Cf. A263295 (denominators), A346590.

Sequence in context: A128968 A153409 A143893 * A072263 A009178 A141508

Adjacent sequences: A262954 A262955 A262956 * A262958 A262959 A262960

Keyword

nonn,frac

Author

Mohamed Sabba, Nov 19 2015

Extensions

More terms from Mohamed Sabba, Dec 22 2015

Status

approved