A207644 a(n) = 1 + (n-1) + (n-2)*[(n-3)/2] + (n-3)*[(n-4)/2]*[(n-5)/3] + (n-4)*[(n-5)/2]*[(n-6)/3]*[(n-7)/4] +... where [x] = floor(x), with summation extending over the initial [n/2+1] products only.
1, 1, 2, 3, 4, 8, 10, 17, 30, 42, 55, 116, 172, 220, 391, 683, 1024, 1616, 2050, 3675, 6520, 9504, 12505, 22421, 35572, 56918, 85701, 138110, 202765, 326231, 503632, 860497, 1376870, 1927446, 2818531, 4892966, 7784671, 11432772, 17287295, 30423457, 46453786, 71810414
Offset: 0
Examples
a(3) = 1 + 2 = 3; a(4) = 1 + 3 + 2*[1/2] = 4; a(5) = 1 + 4 + 3*[2/2] = 8; a(6) = 1 + 5 + 4*[3/2] + 3*[2/2]*[1/3] = 10; a(7) = 1 + 6 + 5*[4/2] + 4*[3/2]*[2/3] = 17; a(8) = 1 + 7 + 6*[5/2] + 5*[4/2]*[3/3] + 4*[3/2]*[2/3]*[1/4] = 30; a(9) = 1 + 8 + 7*[6/2] + 6*[5/2]*[4/3] + 5*[4/2]*[3/3]*[2/4] = 42; ...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..500
Programs
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Mathematica
a[n_] := 1 + Sum[ Product[ Floor[ (n-k-j+1)/j ], {j, 1, k}], {k, 1, n/2}]; Table[a[n], {n, 0, 41}] (* Jean-François Alcover, Mar 06 2013 *)
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PARI
{a(n)=1+sum(k=1,n\2,prod(j=1,k,floor((n-k-j+1)/j)))} for(n=0,60,print1(a(n),", "))
Formula
a(n) = 1 + Sum_{k=1..[n/2]} Product_{j=1..k} floor( (n-k-j+1) / j ).
Equals the antidiagonal sums of triangle A207645.
Comments