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A176346
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.
2
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
OFFSET
0,5
COMMENTS
This sequence comes from computability functions.
Row sums are : {1, 2, 5, 6, 13, 12, 23, 24, 33, 36, 55,...}.
REFERENCES
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.
FORMULA
T(n, m) = 1 + 2*(n+1) - (m+1)*floor((n+1)/(m+1)) - (n-m+1)*floor((n+1)/( n-m+1)).
EXAMPLE
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;
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Cf. A176298.
Sequence in context: A356400 A153066 A126209 * A338878 A073166 A050169
KEYWORD
nonn,tabl
AUTHOR
Roger L. Bagula, Apr 15 2010
EXTENSIONS
Edited by G. C. Greubel, Apr 26 2019
STATUS
approved