A085765 -id:A085765 - cp's OEIS frontend

A220338 A modified Engel expansion for 8*sqrt(6) - 19.

2, 6, 10, 2, 50, 98, 2, 4802, 9602, 2, 46099202, 92198402, 2, 4250272665676802, 8500545331353602, 2, 36129635465198759610694779187202, 72259270930397519221389558374402, 2, 2610701117696295981568349760414651575095962187244375364404428802

Offset: 1

Peter Bala, Dec 12 2012

For a brief description of the modified Engel expansion of a real number see A220335.

Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(5) = 8*sqrt(6) - 19. Iterating the identity Q(p) = 1/2 + 1/(2*(p+1)) + 1/(2*(p+1)*(2*p)) + 1/(2*(p+1)*(2*p))*Q(2*p^2-1) leads to a representation for Q(p) as an infinite series of unit fractions. The sequence of denominators of these unit fractions can be used to find the modified Engel expansion of Q(p). For further details see the Bala link. The present sequence is the case p = 5. For other cases see A220335 (p = 2), A220336 (p = 3) and A220337 (p = 4).

Peter Bala, A modified Engel expansion for certain quadratic irrationals
Wikipedia, Engel Expansion

Cf. A084765, A220335 (p = 2), A220336 (p = 3), A220337 (p = 4).

Define the harmonic sawtooth map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 8*sqrt(6) - 19. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the map h(x), with the convention h^(0)(x) = x.
a(3*n+2) = 1/2*{2 + (5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} and
a(3*n+3) = {(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} both for n >= 0.
For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*A084765(n)^2 and a(3*n+3) = 4*A085765(n)^2 - 2.
Recurrence equations:
For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
a(3*n+3) = 2*a(3*n+2) - 2.
Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 8*sqrt(6) - 19 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*6) + 1/(2*6*10) + 1/(2*6*10*2) + 1/(2*6*10*2*50) + ....

A086450 a(0) = 1, a(2n+1) = a(n), a(2n) = a(n) + a(n-1) + ... + a(n-m) + ... where a(n<0) = 0.

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 5, 1, 9, 4, 11, 2, 16, 5, 17, 1, 26, 9, 30, 4, 41, 11, 43, 2, 59, 16, 64, 5, 81, 17, 82, 1, 108, 26, 117, 9, 147, 30, 151, 4, 192, 41, 203, 11, 246, 43, 248, 2, 307, 59, 323, 16, 387, 64, 392, 5, 473, 81, 490, 17, 572, 82, 573, 1, 681, 108, 707

Offset: 0

Ralf Stephan, Jul 20 2003

Sequence has itself and its partial sums as bisections.
Setting m=1 gives Stern-Brocot sequence (A002487).
Conjecture: a(n) mod 2 repeats the 7-pattern 1,1,0,1,0,0,1 (A011657).
The conjecture is easily proved by induction: a(0) to a(14) = 1, 1, 2, 1, 4, 2, 5, 1, 9, 4, 11, 2, 16, 5 read mod 2 gives 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1. Assume the conjecture is true up to n = 14k. Then the next 7 odd entries a(14k+1), a(14k+3), ..., a(14k+13) are read from a(7k) to a(7k+6), which follow the correct mod 2 pattern by assumption. For the even entries a(14k), a(14k+10)... a(14k+12), the sum over the first 7k-1 addends is even, simply because of each consecutive 7 addends exactly 4 are odd. So again a(7k) to a(7k+6) determines the outcome and again gives the desired pattern. a(14k) is odd, since a(7k) is odd, a(14k+2) is even, since a(7k) and a(7k+1) are odd and so on ... - Lambert Herrgesell (zero815(AT)googlemail.com), May 08 2007

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A086449.

Partial sums are in A085765.

a:= proc(n) local m; a(n):= `if`(n=0, 1,
      `if`(irem(n, 2, 'm')=1, a(m), s(m)))
    end:
s:= proc(n) s(n):= a(n) +`if`(n=0, 0, s(n-1)) end:
seq(a(n), n=0..100);  # Alois P. Heinz, Sep 26 2013

a[0] = 1; a[n_] := a[n] = If[EvenQ[n], Sum[a[n/2-k], {k, 0, n/2}], a[(n-1)/2]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Jun 16 2015 *)

a(n)=if(n<2,n>=0,if(n%2==0,sum(k=0,n/2,a(n/2-k)),a((n-1)/2)))

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A220338 A modified Engel expansion for 8*sqrt(6) - 19.

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A086450 a(0) = 1, a(2n+1) = a(n), a(2n) = a(n) + a(n-1) + ... + a(n-m) + ... where a(n<0) = 0.

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