A025562
a(n) = n!/LCM{1, C(n-1,1), C(n-2,2), ..., C(n-[ n/2 ],[ n/2 ])}.
Original entry on oeis.org
1, 1, 2, 3, 8, 10, 24, 84, 192, 432, 2880, 15840, 34560, 224640, 483840, 3628800, 116121600, 493516800, 1045094400, 9928396800, 20901888000, 219469824000, 9656672256000, 111051730944000, 231760134144000, 2897001676800000, 30128817438720000, 406739035422720000
Offset: 0
A073618
Consider Pascal's triangle A007318; a(n) = LCM of terms at +45 degree slope with the horizontal.
Original entry on oeis.org
1, 1, 1, 2, 3, 12, 30, 60, 210, 840, 1260, 2520, 13860, 27720, 180180, 360360, 180180, 720720, 6126120, 12252240, 116396280, 232792560, 116396280, 232792560, 2677114440, 5354228880, 13385572200, 26771144400, 40156716600, 80313433200, 1164544781400, 2329089562800
Offset: 0
The ninth diagonal is 1,7,15,10,1 and the LCM of the terms = 210 hence a(8) = 30.
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a:= n-> ilcm(seq(binomial(n-i, i), i=0..floor(n/2))):
seq(a(n), n=0..35); # Alois P. Heinz, Nov 27 2023
More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 22 2003
A247651
Maximum number of binary strings of length 2n obtained from a partition of n.
Original entry on oeis.org
1, 2, 3, 12, 30, 60, 210, 840, 2520, 7560, 27720, 83160, 240240, 840840, 2702700, 10810800, 36756720, 122522400, 465585120, 1551950400, 4888643760, 19554575040, 74959204320, 257002986240, 936990054000, 3480248772000, 11745839605500, 40477970332800, 146732642456400, 524045151630000
Offset: 0
n=0 gives the empty string.
n=1 and the only possible partition generate 01 and 10.
For n=2, both possible partitions generate 3 strings (0011,0110 and 1100, and respectively 0101, 1001 and 1010, based on runs of 1's).
For n=3, the optimal partition is {1,2}, generating 12 strings (based on runs of 1's: 001011, 001101, 010011, 010110, 011001, 011010, 100011, 100110, 101100, 110001, 110010, 110100).
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nseq[p_]:=FactorialPower[Total[p]+1,Length[p]]/Apply[Times,Map[Factorial[Count[p,#1]]&,Range[Max[Length[p]]]]];
a[n_]:=Max[Map[nseq,IntegerPartitions[n]]]
Table[a[n],{n,0,20}] (* after A130670 *)
A374440
Triangle read by rows: T(n, k) = T(n - 1, k) + T(n - 2, k - 2), with boundary conditions: if k = 0 or k = 2 then T = 1; if k = 1 then T = n - 1.
Original entry on oeis.org
1, 1, 0, 1, 1, 1, 1, 2, 1, 0, 1, 3, 1, 1, 1, 1, 4, 1, 3, 2, 0, 1, 5, 1, 6, 3, 1, 1, 1, 6, 1, 10, 4, 4, 3, 0, 1, 7, 1, 15, 5, 10, 6, 1, 1, 1, 8, 1, 21, 6, 20, 10, 5, 4, 0, 1, 9, 1, 28, 7, 35, 15, 15, 10, 1, 1, 1, 10, 1, 36, 8, 56, 21, 35, 20, 6, 5, 0
Offset: 0
Triangle starts:
[ 0] 1;
[ 1] 1, 0;
[ 2] 1, 1, 1;
[ 3] 1, 2, 1, 0;
[ 4] 1, 3, 1, 1, 1;
[ 5] 1, 4, 1, 3, 2, 0;
[ 6] 1, 5, 1, 6, 3, 1, 1;
[ 7] 1, 6, 1, 10, 4, 4, 3, 0;
[ 8] 1, 7, 1, 15, 5, 10, 6, 1, 1;
[ 9] 1, 8, 1, 21, 6, 20, 10, 5, 4, 0;
[10] 1, 9, 1, 28, 7, 35, 15, 15, 10, 1, 1;
Cf.
A000032 (Lucas),
A001611 (even sums, Fibonacci + 1),
A000071 (odd sums, Fibonacci - 1),
A001911 (alternating sums, Fibonacci(n+3) - 2),
A025560 (row lcm),
A073028 (row max),
A117671 &
A025174 (central terms),
A057979 (subdiagonal),
A000217 (column 3).
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T := proc(n, k) option remember; if k = 0 or k = 2 then 1 elif k > n then 0
elif k = 1 then n - 1 else T(n - 1, k) + T(n - 2, k - 2) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..9);
T := (n, k) -> ifelse(k = 0, 1, binomial(n - floor(k/2), ceil(k/2)) -
binomial(n - ceil((k + irem(k + 1, 2))/2), floor(k/2))):
A025561
a(n) = sum of the exponents in the prime factorization of LCM{1, n-1, ..., C(n-[ n/2 ],[ n/2 ])}.
Original entry on oeis.org
0, 0, 1, 1, 3, 3, 4, 4, 6, 6, 7, 7, 8, 8, 9, 8, 10, 10, 11, 11, 12, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 18, 17, 18, 17, 18, 18, 19, 18, 19, 19, 20, 20, 21, 20, 21, 21, 22, 22, 23, 22, 23, 23, 24, 23, 24, 23, 24, 24, 25, 25, 26, 25, 27, 26, 27, 27, 28, 27, 28, 28, 29, 29, 30
Offset: 1
Showing 1-5 of 5 results.
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