cp's OEIS Frontend

A355915 Number of ways to write n as a sum of numbers of the form 2^r * 3^s, where r and s are >= 0, and no summand divides another.

%I A355915 #35 Sep 21 2022 15:18:22
%S A355915 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,2,1,1,1,1,1,1,2,1,2,1,2,1,
%T A355915 2,1,2,2,1,1,2,1,1,2,1,1,2,1,3,1,1,1,2,1,1,1,2,2,1,1,1,2,1,1,3,2,2,1,
%U A355915 1,2,2,1,2,2,1,2,1,1,2,1,1,2,2,1,3,1,2,2,3,1,2,1,2,2,2,1,3,3,2,1
%N A355915 Number of ways to write n as a sum of numbers of the form 2^r * 3^s, where r and s are >= 0, and no summand divides another.
%C A355915 It is a theorem of Erdos [Erdős] that this representation is always possible.
%C A355915 Without the divisibility constraint the answer is A062051.
%C A355915 See A356792 for when k first appears.
%H A355915 Michael S. Branicky, <a href="/A355915/b355915.txt">Table of n, a(n) for n = 1..10000</a>
%H A355915 Michael S. Branicky, <a href="/A355915/a355915_1.py.txt">Python Program</a>
%H A355915 William Lowell Putman Mathematical Competition, <a href="http://math.hawaii.edu/home/pdf/putnam/2005.pdf">Number 66, 2005, Problem A-1</a>.
%e A355915 Illustration of initial terms:
%e A355915 1 = 2^0
%e A355915 2 = 2^1
%e A355915 3 = 3^1
%e A355915 4 = 2^2
%e A355915 5 = 2+3
%e A355915 6 = 2*3
%e A355915 7 = 2^2+3
%e A355915 8 = 2^3
%e A355915 9 = 3^2
%e A355915 10 = 2^2 + 2*3
%e A355915 11 = 2+3^2 = 2^3+3 (this is the first time there are 2 solutions)
%e A355915 12 = 2^2*3
%e A355915 13 = 2^2+3^2
%e A355915 14 = 2^3+2*3
%e A355915 ...
%o A355915 (Python) # see linked program
%Y A355915 Cf. A062051, A356792.
%K A355915 nonn
%O A355915 1,11
%A A355915 _Michael S. Branicky_ and _N. J. A. Sloane_, Sep 21 2022
%E A355915 More than the usual number of terms are shown, to distinguish this from similar sequences.