1,3

Also called the Smarandache Crescendo Pyramidal sequence.

The following sequences all have the same parity: A004737, A006590, A027052, A071028, A071797, A078358, A078446. - Jeremy Gardiner (jeremy.gardiner(AT)btinternet.com), Mar 16 2003

a(A002061(n)) = n; a(A000290(n)) = a(A002522(n)) = 1. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 10 2006

The ordinal transform of a sequence b_0, b_1, b_2, ... is the sequence a_0, a_1, a_2, ... where a_n is the number of times b_n has occurred in {b_0 ... b_n}.

Jerry Brown et al., Problem 4619, School Science and Mathematics, USA, Vol. 97 (4), 1997, pp. 221-222.

F. Smarandache, "Numerical Sequences", University of Craiova, 1975.

F. Smarandache, "Collected Papers", Vol. II, Tempus Publ. Hse., Bucharest, 1996.

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

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

F. Smarandache, Collected Papers, Vol. II

F. Smarandache, Collected Papers, Vol. II.

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

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

a(n) = if n<3 then 1 else (if a(n-1)=1 then 1 + 0^(a(n-2)-1) else a(n-1) - 0^X + (a(n-1)-a(n-2))*(1 - 0^X)), where X = A003059(n-1)-a(n-1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 10 2006

Let b(n) = floor(sqrt(n-1)). Then a(n) = min(n - b(n)^2, (b(n)+1)^2 - n + 1). - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Jun 09 2006

Ordinal transform of A004741. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Aug 28 2006

If the sequence is read as a triangular array, beginning [1]; [1,2,1]; [1,2,3,2,1]; ..., then the o.g.f. is (1+qx)/((1-x)(1-qx)(1-q^2x)) = 1 + x(1 + 2q + q^2) + x^2(1 + 2q + 3q^2 + 2q^3 +q^4) + .... The row polynomials for this triangle are (1 + q + ... + q^n)^2 =[n,2]_q + q[n-1,2]_q, where [n,2]_q are Gaussian polynomials (see A008967). - Peter Bala (pbala(AT)toucansurf.com), Sep 23 2007

a(n+1)=a(n)+{floor[sqrt(n)]-floor[sqrt(n-1)]-1}*(-1)^{floor[(n-1)/floor[(1/2)*(1+sqrt(4*n-3))]]-floor[(1/2)*(1+sqrt(4*n-3))]+1}, with a(1)=1 [From Paolo P. Lava (ppl(AT)spl.at), Nov 13 2008]

P:=proc(i) local a, b, c, n; a:=1; print(a); for n from 1 by 1 to i do a:=a+(floor(sqrt(n))-floor(sqrt(n-1))-1)*(-1)^(floor((n-1)/floor(1/2*(1+sqrt(4*n\ -3))))-floor(1/2*(1+sqrt(4*n-3)))+1); print(a); od; end: P(100); [From Paolo P. Lava (ppl(AT)spl.at), Nov 13 2008]

Cf. A004738, A008967.

Sequence in context: A072782 A122563 A080345 this_sequence A014600 A165475 A098280

Adjacent sequences: A004734 A004735 A004736 this_sequence A004738 A004739 A004740

nonn,easy

R. Muller

More terms from Patrick De Geest (pdg(AT)worldofnumbers.com), Jun 15 1998.