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%I #11 Dec 25 2016 01:09:58
%S 1,2,4,5,6,9,9,12,8,13,17,22,13,20,22,29,10,17,25,32,25,38,40,53,17,
%T 28,38,49,32,49,53,70,12,21,33,42,37,56,58,77,33,54,72,93,58,89,97,
%U 128,21,36,54,69,56,85,89,118,42,69,93,120,77,118,128,169
%N Numerator of the rationals >= 1 whose continued fractions consist of only even terms, in ascending order by the sum of the continued fraction terms and descending by lowest order continued fraction terms to highest.
%H T. D. Noe, <a href="/A072726/b072726.txt">Table of n, a(n) for n = 0..1023</a>
%F a(2^k + 2^j + m) = 2(k-j)*a(2^j + m) + a(m) when 2^k > 2^j > m >=0. a(0) = 1, a(2^k) = 2(k+1), a(2^k + 1) = 4*k + 1 (k>0), a(2^k - 1) = the (k+1)-th Pell number.
%e n: a(n)/A072727 has continued fraction:
%e 0: 1/0 = [infinity]
%e 1: 2/1 = [2]
%e 2: 4/1 = [4]
%e 3: 5/2 = [2;2]
%e 4: 6/1 = [6]
%e 5: 9/2 = [4;2]
%e 6: 9/4 = [2;4]
%e 7: 12/5 = [2;2,2]
%e 8: 8/1 = [8]
%e 9: 13/2 = [6;2]
%e 10: 17/4 = [4;4]
%e 11: 22/5 = [4;2,2]
%e 12: 13/6 = [2;6]
%e 13: 20/9 = [2;4,2]
%e 14: 22/9 = [2;2,4]
%e 15: 29/12= [2;2,2,2]
%t a[0] = 1; a[n_] := a[n] = Which[IntegerQ[k = Log[2, n]], 2 (k + 1), IntegerQ[k = Log[2, n - 1]], 4 k + 1, IntegerQ[k = Log[2, n + 1]], Fibonacci[k + 1, 2], True, Clear[k]; Hold[2*(k - j)*a[2^j + m] + a[m]] /. ToRules[Reduce[2^k > 2^j > m >= 0 && n == 2^k + 2^j + m, {k, j, m}, Integers]] // ReleaseHold];
%t Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 63}] (* _Jean-François Alcover_, Jul 13 2016 *)
%Y Cf. A072727, A071585, A071766, A072728, A072729, A000129.
%K easy,frac,nice,nonn
%O 0,2
%A _Paul D. Hanna_, Jul 09 2002
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