The distribution of zeroes zeros is the same as in A276204.
The distribution of zeroes zeros is the same as in A276204.
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a(0) = a(1) = 0 and for . For n>1 a(n) is the smallest nonnegative integer such that there is no arithmetic progression m_1,m_2,...,m_i,n (2<=i<n) such that a(m_1)+a(m_2)+...+a(m_i) = a(n).
0, 0, 1, 0, 0, 2, 4, 1, 1, 0, 0, 5, 0, 0, 3, 7, 2, 3, 2, 4, 1, 1, 12, 1, 5, 1, 1, 0, 0, 13, 0, 0, 10, 9, 6, 7, 0, 0, 18, 0, 0, 15, 4, 14, 7, 6, 8, 2, 6, 3, 16, 3, 3, 2, 3, 7, 1, 10, 25, 8, 5, 1, 1, 1, 4, 14, 27, 4, 1, 1, 10, 2, 2, 6, 1, 26, 8, 1, 19, 1, 1, 0, 0, 13, 0, 0, 7, 24, 2, 19, 0, 0, 34, 0, 0, 29, 32, 32, 5, 15, 21, 14, 15, 6, 6, 24, 13, 39, 0, 0, 24, 0, 0, 14, 12, 12, 34, 0, 0, 23, 0, 0, 7, 5, 20, 3, 8, 3, 22, 12, 9, 12, 4, 5, 11, 4, 27, 11, 9, 11, 11, 18, 24, 5, 6, 10, 48, 13, 28, 11, 6, 36, 1, 16, 41, 6, 17, 7, 3, 6, 2, 2, 6, 5, 1, 4, 4, 5, 4, 1, 1, 9, 58, 23, 9, 61, 1, 45, 2, 50, 2, 3, 1, 62, 2, 1, 31, 12, 1, 1, 67, 8, 2, 33, 1, 26, 20, 2, 86, 47, 1, 47, 8, 35, 20, 1, 1
One can easily notice the The distribution of zeroes is the same as in A276204.
for For n = 5 we have that:
there is no such arithmetic progression m_1,m_2,...,m_i,5 that a(m_1)+a(m_2)+...+a(m_i)=2, so a(5) = 2.
nonn,unkn,uned,changed
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allocated a(0)=a(1)=0 and for Michal Urbanskin>1 a(n) is the smallest nonnegative integer such that there is no arithmetic progression m_1,m_2,...,m_i,n (2<=i<n) such that a(m_1)+a(m_2)+...+a(m_i)=a(n
0, 0, 1, 0, 0, 2, 4, 1, 1, 0, 0, 5, 0, 0, 3, 7, 2, 3, 2, 4, 1, 1, 12, 1, 5, 1, 1, 0, 0, 13, 0, 0, 10, 9, 6, 7, 0, 0, 18, 0, 0, 15, 4, 14, 7, 6, 8, 2, 6, 3, 16, 3, 3, 2, 3, 7, 1, 10, 25, 8, 5, 1, 1, 1, 4, 14, 27, 4, 1, 1, 10, 2, 2, 6, 1, 26, 8, 1, 19, 1, 1, 0, 0, 13, 0, 0, 7, 24, 2, 19, 0, 0, 34, 0, 0, 29, 32, 32, 5, 15, 21, 14, 15, 6, 6, 24, 13, 39, 0, 0, 24, 0, 0, 14, 12, 12, 34, 0, 0, 23, 0, 0, 7, 5, 20, 3, 8, 3, 22, 12, 9, 12, 4, 5, 11, 4, 27, 11, 9, 11, 11, 18, 24, 5, 6, 10, 48, 13, 28, 11, 6, 36, 1, 16, 41, 6, 17, 7, 3, 6, 2, 2, 6, 5, 1, 4, 4, 5, 4, 1, 1, 9, 58, 23, 9, 61, 1, 45, 2, 50, 2, 3, 1, 62, 2, 1, 31, 12, 1, 1, 67, 8, 2, 33, 1, 26, 20, 2, 86, 47, 1, 47, 8, 35, 20, 1, 1
0,6
One can easily notice the distribution of zeroes is the same as in A276204
for n=5
a(5)>0, because a(3)+a(4)=0 and 3,4,5 is an arithmetic progression
a(5)>1, because a(2)+a(3)+a(4)=1 and 2,3,4,5 is an arithmetic progression
there is no such arithmetic progression m_1,m_2,...,m_i,5 that a(m_1)+a(m_2)+...+a(m_i)=2, so a(5)=2
allocated
nonn,unkn,uned
Michal Urbanski, Aug 24 2016
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allocated for Michal Urbanski
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