1,2

"Smarandache Decrescendo Subsequences".

The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals upwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23, 2002

C. Kimberling, "Numeration systems and fractal sequences," Acta Arithmetica 73 (1995) 103-117.

F. Smarandache, "Collected Papers", Vol. II, Tempus Publ. Hse., Bucharest, 1996; F. Smarandache, "Numerical Sequences", University of Craiova, 1975; [ See Arizona State University, Special Collection, Tempe, AZ, USA ].

F. Smarandache, Sequences of Numbers Involved in Unsolved Problems, Hexis, Phoenix, 2006.

H. S. M. Coxeter, Regular Polytopes, 3rd ed., Dover, NY, 1973, pp 159-162 [From Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 25 2009]

C. Kimberling, Fractal sequences

M. L. Perez et al., eds., Smarandache Notions Journal

F. Smarandache, Collected Papers, Vol. II

F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.

M. Somos, Sequences used for indexing triangular or square arrays

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

a(n) = (2 - 2 n + round(SQRT(2 n)) + round(SQRT(2 n))^2)/2. E.g. a(47) = 9. - Brian Tenneson (phoenix(AT)alephnulldimension.net), Oct 11 2003

G.f.: 1 / [(1-x)^2 * (1-xy) ]. - R. Stephan, Jan 23 2005

Contribution from Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 25 2009: (Start)

Recursion:

e(n,k) = (e(n - 1, k)*e(n, k - 1) + 1)/e(n - 1, k - 1); (End)

1; 2,1; 3,2,1; 4,3,2,1; 5,4,3,2,1; ...

Contribution from Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 25 2009: (Start)

{1},

{2, 1},

{3, 2, 1},

{4, 3, 2, 1},

{5, 4, 3, 2, 1},

{6, 5, 4, 3, 2, 1},

{7, 6, 5, 4, 3, 2, 1},

{8, 7, 6, 5, 4, 3, 2, 1},

{9, 8, 7, 6, 5, 4, 3, 2, 1},

{10, 9, 8, 7, 6, 5, 4, 3, 2, 1} (End)

(Excel cell formula) =if(row()>=column(); row()-column()+1; "") [From Mats Granvik (mats.granvik(AT)abo.fi), Jan 19 2009]

Flatten[ Table[ Reverse[ Range[n]], {n, 12}]] (from Robert G. Wilson v Apr 27 2004)

Contribution from Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 25 2009: (Start)

Clear[e, n, k];

e[n_, 0] := n;

e[n_, k_] := 0 /; k >= n;

e[n_, k_] := (e[n - 1, k]*e[n, k - 1] + 1)/e[n - 1, k - 1];

Table[Table[e[n, k], {k, 0, n - 1}], {n, 1, 10}];

Flatten[%] (End)

(PARI) a(n)=1+binomial(1+floor(1/2+sqrt(2*n)), 2)-n

(PARI) t1(n)=binomial(floor(3/2+sqrt(2*n)), 2)-n+1 /* A004736 */

(PARI) t2(n)=n-binomial(floor(1/2+sqrt(2*n)), 2) /* A002260 */

a(n+1)=1+A025581(n). Cf. A002262, A025581, A003056.

Ordinal transform of A002260.

Sequence in context: A141671 A088643 A102482 this_sequence A167288 A167289 A023122

Adjacent sequences: A004733 A004734 A004735 this_sequence A004737 A004738 A004739

nonn,easy,tabl,nice

R. Muller