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A176346
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A dual remainder symmetrical triangle sequence T(n, m) = 1 + 2*(n+1) - (m+1)*floor((n+1)/(m+1)) - (n-m+1)*floor((n+1)/( n-m+1)), read by rows.
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2
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1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 3, 5, 3, 1, 1, 2, 3, 3, 2, 1, 1, 3, 4, 7, 4, 3, 1, 1, 2, 5, 4, 4, 5, 2, 1, 1, 3, 3, 5, 9, 5, 3, 3, 1, 1, 2, 4, 6, 5, 5, 6, 4, 2, 1, 1, 3, 5, 7, 6, 11, 6, 7, 5, 3, 1, 1, 2, 3, 4, 7, 6, 6, 7, 4, 3, 2, 1, 1, 3, 4, 5, 8, 7, 13, 7, 8, 5, 4, 3, 1
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OFFSET
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0,5
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COMMENTS
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This sequence comes from computability functions.
Row sums are : {1, 2, 5, 6, 13, 12, 23, 24, 33, 36, 55,...}.
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REFERENCES
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N. J. Cutland, "Computability, An introduction to recursive function theory", Cambridge University Press, London, 1980, page 37.
Martin Davis, "Computability and Unsolvability", Dover Press, New York, page 43.
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LINKS
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FORMULA
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T(n, m) = 1 + 2*(n+1) - (m+1)*floor((n+1)/(m+1)) - (n-m+1)*floor((n+1)/( n-m+1)).
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EXAMPLE
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Triangle begins as:
1;
1, 1;
1, 3, 1;
1, 2, 2, 1;
1, 3, 5, 3, 1;
1, 2, 3, 3, 2, 1;
1, 3, 4, 7, 4, 3, 1;
1, 2, 5, 4, 4, 5, 2, 1;
1, 3, 3, 5, 9, 5, 3, 3, 1;
1, 2, 4, 6, 5, 5, 6, 4, 2, 1;
1, 3, 5, 7, 6, 11, 6, 7, 5, 3, 1;
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MATHEMATICA
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T[n_, m_]:= 3 +2*n -(m+1)*Floor[(n+1)/(m+1)] -(n-m+1)*Floor[(n+1)/(n-m+1 )]; Table[T[n, m], {n, 0, 12}, {m, 0, n}]//Flatten (* modified by G. C. Greubel, Apr 26 2019 *)
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PROG
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(PARI) {T(n, m) = 1 + 2*(n+1) - (m+1)*floor((n+1)/(m+1)) - (n-m+1)* floor((n+1)/(n-m+1))}; \\ G. C. Greubel, Apr 26 2019
(Magma) [[1 + 2*(n+1) - (m+1)*Floor((n+1)/(m+1)) - (n-m+1)*Floor((n+1)/( n-m+1)): m in [0..n]]: n in [0..12]]; // G. C. Greubel, Apr 26 2019
(Sage) [[1 + 2*(n+1) - (m+1)*floor((n+1)/(m+1)) - (n-m+1)*floor((n+1)/( n-m+1)) for m in (0..n)] for n in (0..12)] # G. C. Greubel, Apr 26 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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