A210843
Level of the n-th plateau of the column k of the square array A195825, when k -> infinity.
Original entry on oeis.org
1, 4, 13, 35, 86, 194, 415, 844, 1654, 3133, 5773, 10372, 18240, 31449, 53292, 88873, 146095, 236977, 379746, 601656, 943305, 1464501, 2252961, 3436182, 5198644, 7805248, 11634685, 17224795, 25336141, 37038139, 53828275, 77792869
Offset: 1
Column 1 of A195825 is A000041 which starts: [1, 1], 2, 3, 5, 7, 11... The column contains only one plateau: [1, 1] which has level 1 and length 2. So a(1) = 1.
Column 3 of A195825 is A036820 which starts: [1, 1, 1, 1], 2, 3, [4, 4], 5, 7, 10... The column contains only two plateaus: [1, 1, 1, 1], [4, 4], which have levels 1, 4 and lengths 4, 2. So a(1)= 1 and a(2) = 2.
Column 6 of A195825 is A195850 which starts: [1, 1, 1, 1, 1, 1, 1], 2, 3, [4, 4, 4, 4, 4], 5, 7, 10, 12, [13, 13, 13], 14, 16, 21... The column contains three plateaus: [1, 1, 1, 1, 1, 1, 1], [4, 4, 4, 4, 4], [13, 13, 13], which have levels 1, 4, 13 and lengths 7, 5, 3. So a(1) = 1, a(2) = 4 and a(3) = 13.
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CoefficientList[Series[1/(1-x)*Product[1/(1-x^k)^3,{k,1,50}],{x,0,50}],x] (* Vaclav Kotesovec, Aug 16 2015 *)
A303070
a(n) = [x^n] (1/(1 - x))*Product_{k>=1} 1/(1 - x^k)^n.
Original entry on oeis.org
1, 2, 8, 35, 164, 787, 3857, 19147, 96004, 485009, 2465013, 12589315, 64555985, 332158127, 1714001409, 8866730665, 45968787524, 238778897128, 1242417984179, 6474394344503, 33784931507529, 176515163156311, 923265560495737, 4834081924982522, 25334170138318345, 132883719945537587
Offset: 0
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Table[SeriesCoefficient[1/(1 - x) Product[1/(1 - x^k)^n, {k, 1, n}], {x, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[1/(1 - x) Exp[n Sum[x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 25}]
A146023
Triangle read by rows, square of A027293.
Original entry on oeis.org
1, 2, 1, 5, 2, 1, 10, 5, 2, 1, 20, 10, 5, 2, 1, 36, 20, 10, 5, 2, 1, 65, 36, 20, 10, 5, 2, 1, 110, 65, 36, 20, 10, 5, 2, 1, 185, 110, 65, 36, 20, 10, 5, 2, 1, 300, 185, 110, 65, 36, 20, 10, 5, 2, 1
Offset: 0
First few rows of the triangle =
1;
2, 1;
5, 2, 1;
10, 5, 2, 1;
20, 10, 5, 2, 1;
36, 20, 10, 5, 2, 1;
65, 36, 20, 10, 5, 2, 1;
110, 65, 36, 20, 10, 5, 2, 1;
...
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lim = 10;
A000712 = Table [Length@IntegerPartitions[n, All, Range@n~Join~Range@n], {n, 0, lim - 1}]
Table[Reverse[Take[A000712, n]], {n, lim}] // Flatten (* Robert Price, Jun 15 2020 *)
A210764
Square array T(n,k), n>=0, k>=0, read by antidiagonals in which column k gives the partial sums of column k of A144064.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 8, 4, 1, 1, 12, 18, 13, 5, 1, 1, 19, 38, 35, 19, 6, 1, 1, 30, 74, 86, 59, 26, 7, 1, 1, 45, 139, 194, 164, 91, 34, 8, 1, 1, 67, 249, 415, 416, 281, 132, 43, 9, 1, 1, 97, 434, 844, 990, 787, 447, 183, 53, 10, 1
Offset: 0
Array begins:
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
1, 4, 8, 13, 19, 26, 34, 43, 53,
1, 7, 18, 35, 59, 91, 132, 183,
1, 12, 38, 86, 164, 281, 447,
1, 19, 74, 194, 416, 787,
1, 30, 139, 415, 990,
1, 45, 249, 844,
1, 67, 434,
1, 97,
1,
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with(numtheory):
etr:= proc(p) local b;
b:= proc(n) option remember; `if`(n=0, 1,
add(add(d*p(d), d=divisors(j))*b(n-j), j=1..n)/n)
end
end:
A:= (n, k)-> etr(j-> k +`if`(j=1, 1, 0))(n):
seq(seq(A(d-k, k), k=0..d), d=0..14); # Alois P. Heinz, May 20 2013
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etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[ j]}]*b[n-j], {j, 1, n}]/n]; b]; A[n_, k_] := etr[Function[{j}, k + If[j == 1, 1, 0]]][n]; Table[Table[A[d-k, k], {k, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Mar 05 2015, after Alois P. Heinz *)
A093010
Triangle, read by rows, such that the convolution of the n-th row with the natural numbers forms the n-th diagonal, for n>=0, where each row begins with 1.
Original entry on oeis.org
1, 1, 2, 1, 4, 3, 1, 6, 7, 4, 1, 8, 14, 10, 5, 1, 10, 22, 22, 13, 6, 1, 12, 33, 40, 30, 16, 7, 1, 14, 45, 66, 58, 38, 19, 8, 1, 16, 60, 100, 104, 76, 46, 22, 9, 1, 18, 76, 146, 168, 142, 94, 54, 25, 10, 1, 20, 95, 202, 262, 242, 180, 112, 62, 28, 11, 1, 22, 115, 272, 386, 394, 316
Offset: 0
T(7,3) = 66 = 1*4+8*3+14*2+10*1 = T(4,0)*4+T(4,1)*3+T(4,2)*2+T(4,3)*1; this is also the third term of the 4th-diagonal.
The 6th antidiagonal is {1,10,14,4}, which has a sum of 29 = A000990(6) = number of 2-line partitions of 6.
Rows begin:
{1},
{1,2},
{1,4,3},
{1,6,7,4},
{1,8,14,10,5},
{1,10,22,22,13,6},
{1,12,33,40,30,16,7},
{1,14,45,66,58,38,19,8},
{1,16,60,100,104,76,46,22,9},
{1,18,76,146,168,142,94,54,25,10},
{1,20,95,202,262,242,180,112,62,28,11},
{1,22,115,272,386,394,316,218,130,70,31,12},...
Showing 1-5 of 5 results.
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