1,2

More precisely, number of ways to pick a subinterval [i,i+1,...,j] of [1..n] and to map it to a nondecreasing sequence of the same length with symbols from [1..n]. If s is the length of the subinterval (1 <= s <= n), there are n+1-s ways to choose the subinterval and binomial(n+s-1,s) ways to choose the sequence, for a total of Sum_{s=1..n} (n+1-s)*binomial(n+s-1,s) = binomial(2*n+1, n+1)-(n+1) ways. - N. J. A. Sloane (njas(AT)research.att.com), Feb 02 2009

Arises in the classification of endomorphisms of certain finite-dimensional operator algebras.

Equals binomial transform of A163765: (1, 6, 18, 48, 131, 363, 1017,...). [From Gary W. adamson (qntmpkt(AT)yahoo.com), Aug 03 2009]

David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)

a(n) = binomial(2*n+1, n+1)-(n+1).

a(n) = (1/2)*Sum[Sum[(i+j)!/(i!*j!),{i,1,n}],{j,1,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 04 2006. Corrected by N. J. A. Sloane (njas(AT)research.att.com), Jan 30 2009.

G.f.: (1/(2*x))*(1/sqrt(1-4*x)-1) - 1/(1-x)^2. - N. J. A. Sloane (njas(AT)research.att.com), Feb 02 2009

a(2) = 7 because there are two maps with domain {1}, two with domain {2} and three maps with domain {1,2}.

Table[Sum[Sum[(i+j)!/i!/j!/2, {i, 1, n}], {j, 1, n}], {n, 1, 20}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 04 2006. Corrected by N. J. A. Sloane (njas(AT)research.att.com), Jan 30 2009.

Cf. A001700, A144657.

A163765 [From Gary W. adamson (qntmpkt(AT)yahoo.com), Aug 03 2009]

Sequence in context: A032197 A114289 A147597 this_sequence A125193 A002184 A002588

Adjacent sequences: A048772 A048773 A048774 this_sequence A048776 A048777 A048778

easy,nonn,nice

Stephen C. Power (s.power(AT)lancaster.ac.uk)

More terms from N. J. A. Sloane (njas(AT)research.att.com), Dec 15 2008