A276659
Accumulation of the upper left triangle used in binomial transform of nonnegative integers.
Original entry on oeis.org
0, 2, 11, 39, 114, 300, 741, 1757, 4052, 9162, 20415, 44979, 98214, 212888, 458633, 982905, 2097000, 4456278, 9436995, 19922735, 41942810, 88080132, 184549101, 385875669, 805306044, 1677721250, 3489660551, 7247756907, 15032385102, 31138512432, 64424508945
Offset: 0
Starting from the triangle:
0, 1, 2, 3, 4, 5, ...
1, 3, 5, 7, 9, ...
4, 8, 12, 16, ...
12, 20, 28, ...
32, 48, ...
80, ...
...
the first terms are:
a(0) = 0;
a(1) = a(0) + 1 + 1 = 2;
a(2) = a(1) + 4 + 3 + 2 = 11;
a(3) = a(2) + 12 + 8 + 5 + 3 = 39, etc.
First column is A001787: n*2^(n-1).
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[(2^(n+2)-n-3)*n/2: n in [0..40]]; // Vincenzo Librandi, Sep 13 2016
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A276659:=n->n*(2^(n+2) - n - 3)/2: seq(A276659(n), n=0..50); # Wesley Ivan Hurt, Sep 16 2017
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t[0, k_] := k; t[n_, k_] := t[n, k] = t[n - 1, k] + t[n - 1, k + 1]; a[n_] := Sum[t[m, k], {m, 0, n}, {k, 0, n - m}]; Table[a[n], {n, 0, 30}]
Table[(2^(n + 2) - n - 3) n / 2, {n, 0, 30}] (* Vincenzo Librandi, Sep 13 2016 *)
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x='x+O('x^99); concat(0, Vec(x*(2-3*x)/((1-x)^3*(1-2*x)^2))) \\ Altug Alkan, Sep 14 2017
A327916
Triangle T(k, n) read by rows: Array A(k, n) = 2^k*(k + 1 + 2*n), k >= 0, n >= 0, read by antidiagonals upwards.
Original entry on oeis.org
1, 4, 3, 12, 8, 5, 32, 20, 12, 7, 80, 48, 28, 16, 9, 192, 112, 64, 36, 20, 11, 448, 256, 144, 80, 44, 24, 13, 1024, 576, 320, 176, 96, 52, 28, 15, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 11264, 6144, 3328, 1792, 960, 512, 272, 144, 76, 40, 21
Offset: 0
The triangle T(k, n) begins:
k\n 0 1 2 3 4 5 6 7 8 9 10 ...
-----------------------------------------------------
0: 1
1: 4 3
2: 12 8 5
3: 32 20 12 7
4: 80 48 28 16 9
5: 192 112 64 36 20 11
6: 448 256 144 80 44 24 13
7: 1024 576 320 176 96 52 28 15
8: 2304 1280 704 384 208 112 60 32 17
9: 5120 2816 1536 832 448 240 128 68 36 19
10: 11264 6144 3328 1792 960 512 272 144 76 40 21
...
The sequence of (sub)diagonal k, for k >= 0, is the row k sequence of array A: {(k + 2*n + 1)*2^k}_{k >= 0}.
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Table[2^#*(# + 1 + 2 n) &[k - n], {k, 0, 10}, {n, 0, k}] // Flatten (* Michael De Vlieger, Oct 03 2019 *)
A374419
Triangle read by rows: T(n,k) = number of permutations in symmetric group S_n with an even number of non-fixed point cycles, without k<=n particular fixed points.
Original entry on oeis.org
1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 4, 3, 3, 3, 3, 36, 32, 29, 26, 23, 20, 296, 260, 228, 199, 173, 150, 130, 2360, 2064, 1804, 1576, 1377, 1204, 1054, 924, 19776, 17416, 15352, 13548, 11972, 10595, 9391, 8337, 7413, 180544, 160768, 143352, 128000, 114452, 102480, 91885, 82494, 74157, 66744
Offset: 0
Triangle array T(n,k) begins:
n: {k<=n}
0: {1}
1: {1, 0}
2: {1, 0, 0}
3: {1, 0, 0, 0}
4: {4, 3, 3, 3, 3}
5: {36, 32, 29, 26, 23, 20}
6: {296, 260, 228, 199, 173, 150, 130}
7: {2360, 2064, 1804, 1576, 1377, 1204, 1054, 924}
T(n,0) = A373339(n) = the number of permutations in S_n without k=0 particular fixed points (i.e., not filtered, so all permutations) with an even number of cycles.
T(n,n) = A216778(n) = the number of permutations in S_n without k=n particular fixed points (i.e., filtered down to just the derangements) with an even number of cycles.
T(4,1<=k<=4) = 3 because S_4 contains 3 permutations with an even number of non-fixed point cycles without k=1,2,3 or 4 particular fixed points, namely the 3 (2,2)-cycles: (12)(34), (13)(24), (14)(23).
T(4,0) = 4 is one more than the above because it includes the permutation without k=0 particular fixed points, i.e., the identity permutation of 4 fixed points.
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Table[Table[1/2*(Sum[(-1)^j*Binomial[k, j]*(n - j)!, {j, 0, k}] + 2^(n - k - 1)*(2 - n - k)), {k, 0, n}], {n, 0, 10}]
A374420
Triangle T(n, k) for the number of permutations of symmetric group S_n with an odd number of non-fixed point cycles, without k <= n particular fixed points.
Original entry on oeis.org
0, 0, 0, 1, 1, 1, 5, 4, 3, 2, 20, 15, 11, 8, 6, 84, 64, 49, 38, 30, 24, 424, 340, 276, 227, 189, 159, 135, 2680, 2256, 1916, 1640, 1413, 1224, 1065, 930, 20544, 17864, 15608, 13692, 12052, 10639, 9415, 8350, 7420, 182336, 161792, 143928, 128320, 114628, 102576, 91937, 82522, 74172, 66752
Offset: 0
Triangle array T(n,k)
n: {k<=n}
0: {0}
1: {0, 0}
2: {1, 1, 1}
3: {5, 4, 3, 2}
4: {20, 15, 11, 8, 6}
5: {84, 64, 49, 38, 30, 24}
6: {424, 340, 276, 227, 189, 159, 135}
7: {2680, 2256, 1916, 1640, 1413, 1224, 1065, 930}
T(n,0) = A373340(n) = the number of permutations in S_n without k=0 particular fixed points (i.e., not filtered, so all permutations) with an odd number of cycles.
T(n,n) = A216779(n) = the number of permutations in S_n without k=n particular fixed points (i.e., filtered down to just the derangements) with an odd number of cycles.
T(2,k) = 1 because S_2 contains 1 permutation with an odd number of non-fixed point cycles without k=0,1 or 2 particular fixed points, namely the derangement (12).
T(3,2) = 3 because S_3 contains 3 permutations with an odd number of non-fixed point cycles without k=2 particular fixed points: say, without fixed points (1) and (2), namely (12)(3), (123), (132).
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Table[Table[1/2*(Sum[(-1)^j*Binomial[k, j]*(n - j)!, {j, 0, k}] - 2^(n - k - 1)*(2 - n - k)), {k, 0, n}], {n, 0, 10}]
Original entry on oeis.org
0, 1, -1, 4, -1, -2, 12, 0, -3, -3, 32, 4, -4, -5, -4, 80, 16, -4, -8, -7, -5, 192, 48, 0, -12, -12, -9, -6, 448, 128, 16, -16, -20, -16, -11, -7, 1024, 320, 64, -16, -32, -28, -20, -13, -8, 2304, 768, 192, 0, -48, -48, -36, -24, -15, -9
Offset: 0
Triangle begins : 0 ; 1,-1 ; 4,-1,-2 ; 12,0,-3,-3 ; 32,4,-4,-5,-4 ; ...
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