A207643
a(n) = 1 + (n-1) + (n-1)*[n/2-1] + (n-1)*[n/2-1]*[n/3-1] + (n-1)*[n/2-1]*[n/3-1]*[n/4-1] +... for n>0 with a(0)=1, where [x] = floor(x).
Original entry on oeis.org
1, 1, 2, 3, 7, 9, 26, 31, 71, 129, 262, 291, 1222, 1333, 2198, 5139, 11881, 12673, 39594, 41923, 117326, 251841, 354292, 371163, 1870453, 2598577, 3456926, 7103955, 16665859, 17283113, 72923314, 75437911, 165990152, 335534913, 422310802, 695765699, 3589651696
Offset: 0
a(2) = 1 + 1 = 2; a(3) = 1 + 2 = 3;
a(4) = 1 + 3 + 3*[4/2-1] = 7;
a(5) = 1 + 4 + 4*[5/2-1] = 9;
a(6) = 1 + 5 + 5*[6/2-1] + 5*[6/2-1]*[6/3-1] = 26;
a(7) = 1 + 6 + 6*[7/2-1] + 6*[7/2-1]*[7/3-1] = 31;
a(8) = 1 + 7 + 7*[8/2-1] + 7*[8/2-1]*[8/3-1] + 7*[8/2-1]*[8/3-1]*[8/4-1] = 71; ...
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a[n_] := 1 + Sum[ Product[ Floor[(n-j)/j], {j, 1, k}], {k, 1, n/2}]; Table[a[n], {n, 0, 36}] (* Jean-François Alcover, Mar 06 2013 *)
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{a(n)=1+sum(k=1,n,prod(j=1,k,floor(n/j-1)))}
for(n=0,50,print1(a(n),", "))
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a(n)=my(t=1);1+sum(k=1,n,t*=n\k-1) \\ Charles R Greathouse IV, Feb 20 2012
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.
Original entry on oeis.org
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
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; ...
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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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{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),", "))
A207645
Triangle where T(n,k) = Product_{j=1..k} floor(n/j - 1), as read by rows n>=0, columns k=0..[n/2].
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 1, 3, 3, 1, 4, 4, 1, 5, 10, 10, 1, 6, 12, 12, 1, 7, 21, 21, 21, 1, 8, 24, 48, 48, 1, 9, 36, 72, 72, 72, 1, 10, 40, 80, 80, 80, 1, 11, 55, 165, 330, 330, 330, 1, 12, 60, 180, 360, 360, 360, 1, 13, 78, 234, 468, 468, 468, 468, 1, 14, 84, 336, 672, 1344, 1344, 1344
Offset: 0
Triangle begins with row n=0 as:
1;
1;
1, 1;
1, 2;
1, 3, 3;
1, 4, 4;
1, 5, 10, 10;
1, 6, 12, 12;
1, 7, 21, 21, 21;
1, 8, 24, 48, 48;
1, 9, 36, 72, 72, 72;
1, 10, 40, 80, 80, 80;
1, 11, 55, 165, 330, 330, 330;
1, 12, 60, 180, 360, 360, 360;
1, 13, 78, 234, 468, 468, 468, 468;
1, 14, 84, 336, 672, 1344, 1344, 1344;
1, 15, 105, 420, 1260, 2520, 2520, 2520, 2520;
1, 16, 112, 448, 1344, 2688, 2688, 2688, 2688; ...
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t[n_, k_] := Product[Floor[n/j - 1], {j, 1, k}]; Flatten[Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* Jean-François Alcover, Jun 12 2012 *)
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{T(n,k)=if(k==0,1,prod(j=1,k,floor(n/j-1)))}
for(n=0,12,for(k=0,n\2,print1(T(n,k),", "));print(""))
A207646
Product_{k=1..n} floor(2*n/k - 1).
Original entry on oeis.org
1, 1, 3, 10, 21, 72, 330, 468, 2520, 8160, 20520, 60480, 318780, 504000, 2426112, 10523520, 21092400, 53222400, 452390400, 506373120, 3226728960, 11604902400, 21299241600, 76640256000, 431500608000, 844958822400, 3197988864000, 10492449177600, 38109367296000
Offset: 0
Illustration of the initial terms:
a(1) = [2/1-1] = 1;
a(2) = [4/1-1]*[4/2-1] = 3;
a(3) = [6/1-1]*[6/2-1]*[6/3-1] = 10;
a(4) = [8/1-1]*[8/2-1]*[8/3-1]*[8/4-1] = 21;
a(5) = [10/1-1]*[10/2-1]*[10/3-1]*[10/4-1]*[10/5-1] = 72; ...
where [x] = floor(x).
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Table[Product[Floor[(2n)/k-1],{k,n}],{n,0,30}] (* Harvey P. Dale, Aug 27 2017 *)
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{a(n)=prod(k=1,n,floor(2*n/k-1))}
for(n=0,50,print1(a(n),", "))
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a(n)=n*=2;prod(k=1,n/3,n\k-1) \\ Charles R Greathouse IV, Feb 20 2012
Showing 1-4 of 4 results.
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