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A328969
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Irregular table T(n,k), n >= 2, k=1..pi(n). arising in expressing the sequence A006022 as the coefficients depending on the maximal k-th prime factor pk of the formula for A006022(n) of its unique prime factor equation.
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1
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1, 0, 1, 3, 0, 0, 0, 1, 3, 1, 0, 0, 0, 0, 1, 7, 0, 0, 0, 0, 4, 0, 0, 5, 0, 1, 0, 0, 0, 0, 0, 1, 9, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 7, 0, 0, 1, 0, 0, 0, 5, 1, 0, 0, 0, 15, 0, 0, 0, 0, 0
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OFFSET
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2,4
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COMMENTS
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The length of the n-th row is pi(n) (A000720), i.e., 1,2,2,3,... for n>2.
The sum of the rows equals the sequence A006022.
When n is prime the entire row is 0 except at p=n where T(p,p)=1.
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LINKS
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FORMULA
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Let p_k be the k-th prime, where k is the column index, p_k <= n, and n >= 2, and m_k is the multiplicity of p_k occurring in n:
T(n,p_k) = n * 1/(p_1^m_1*p_2^m_2*...*p_k^m_k) * (p_k^m_k-1)/(p_k-1), if p_k divides n;
T(n,p_k) = 0; if p_k does not divide n.
T(2*n,2) = A129527(n); T(2*n+1,2) = 0.
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EXAMPLE
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First few rows are:
1;
0, 1;
3, 0;
0, 0, 1;
3, 1, 0;
0, 0, 0, 1;
7, 0, 0, 0;
0, 4, 0, 0;
5, 0, 1, 0;
0, 0, 0, 0, 1;
...
Examples (see the p_k formulas)
T(2^3,1) = (2^3-1) / (2-1) = 7
T(3^2,1) = (3^2-1) / (3-1) = 4
T(3*2,2) = (6/(2*3)) * (3^2-1) / (3-1) = 4
T(12,1) = (12/(2^2)) * (2^2-1) / (2-1) = 9
T(12,2) = (12/(2^2*3)) * (3-1) / (3-1) = 1
T(15,2) = (15/3) * (3-1) / (3-1) = 5
T(15,3) = (15/(2^2*3)) * (3-1) / (3-1) = 1
T(2*3*5^2*7,3) = (2*3*5^2*7/(2*3*5^2)) * (5^2-1) / (5-1) = 42
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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