cp's OEIS Frontend

A064686 a(n) = number of n-digit base-3 biquams.

%I A064686 #45 Aug 10 2021 10:43:44
%S A064686 0,2,7,23,73,227,697,2123,6433,19427,58537,176123,529393,1590227,
%T A064686 4774777,14332523,43013953,129074627,387289417,1161999323,3486260113,
%U A064686 10459304627,31378962457,94138984523,282421147873,847271832227
%N A064686 a(n) = number of n-digit base-3 biquams.
%C A064686 A biquam or biquanimous number (A064544) is a number whose digits can be split into two groups with equal sum.
%C A064686 This is the same as A083313 (apart from the initial term). Proof: Let sum(w) denote the sum of the digits of w. There are 2*3^(n-1) n-digit base-3 numbers: w = (w_1,w_2,...,w_n) with w_i in {0,1,2} for all i and w_1 != 0. Partition them into 4 classes: (i) sum(w) is odd, (ii) sum(w) is even, w contains no 1's and has an odd number of 2s, (iii) sum(w) is even, w contains no 1's and has an even number of 2s and (iv) sum(w) is even and w contains some 1's. Clearly, no biquams occur in cases (i) and (ii), case (iii) consists entirely of biquams and, we claim, so does case (iv). For case (iv) forces an even number, say 2k, of 1's. An even number of 2s clearly gives a biquam and an odd number 2m+1 of 2s does too because {m 2s, (k+1) 1's} and {(m+1) 2s, (k-1) 1's} is a biquam split. There are 3^(n-1) w's in case (i) and 2^(n-2) w's in case (ii) and hence 2*3^(n-1) - (3^(n-1) + 2^(n-2)) = 3^(n-1) - 2^(n-2) (A083313) biquams among n-digit base-3 numbers. - _David Callan_, Sep 15 2004
%C A064686 a(n) % 100 = 23  for n = 4*k-1, k>=1; a(n) % 100 = 27  for n = 4*k+1, k>=1. - _Alex Ratushnyak_, Jul 03 2012
%C A064686 The fraction of biquams for any base approaches 1/2 as the number of digits grows but only if you count leading zeros. Without counting leading zeros, the fraction appears to converge to (b-1)/2b where b is the base used. For base 3 this is 1/3 which fits the data in this sequence (see paper cited below for proofs and the OEIS data collated as fractions). - _Timothy Varghese_, Aug 08 2021
%H A064686 Timothy Varghese and George Varghese, <a href="https://web.cs.ucla.edu/~varghese/partitionableintegersforOEIS.pdf">The Stash-Repair Method applied to Partitionable Numbers</a>,
%H A064686 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-6).
%F A064686 a(1) = 0, a(n) = 3^(n-1)-2^(n-2) for n>=2. - _Alex Ratushnyak_, Jul 02 2012
%F A064686 a(n) = 5*a(n-1)-6*a(n-2) for n>3. G.f.: -x^2*(3*x-2) / ((2*x-1)*(3*x-1)). - _Colin Barker_, May 27 2013
%o A064686 (Python)
%o A064686 print([0]+[3**n - 2**(n-1) for n in range(1,29)])
%o A064686 # _Alex Ratushnyak_, Jul 02 2012
%Y A064686 Essentially the same as A083313.
%Y A064686 Cf. A053152 (partial sums).
%K A064686 base,easy,nonn
%O A064686 1,2
%A A064686 _David W. Wilson_, Oct 10 2001