A292419
a(n) = [x^n] Product_{k>=1} (1 + n*x^k) / (1 - n*x^k).
Original entry on oeis.org
1, 2, 12, 96, 872, 9960, 138180, 2298016, 44686224, 995739498, 24993249820, 697309946784, 21396151468536, 715827315312200, 25926440773118340, 1010478298772398080, 42162515927954808352, 1875027040759682964144, 88527520717734462201756, 4422273966757678408594560
Offset: 0
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nmax = 25; Table[SeriesCoefficient[Product[(1+n*x^k)/(1-n*x^k), {k, 1, n}], {x, 0, n}], {n, 0, nmax}]
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{a(n)= polcoef(prod(k=1, n, ((1+n*x^k)/(1-n*x^k) +x*O(x^n))), n)};
for(n=0,20, print1(a(n), ", ")) \\ G. C. Greubel, Feb 02 2019
A124578
Define p(alpha,2) to be the number of H-conjugacy classes where H is an infant subgroup ( similar to Young subgroups of S_n) of type alpha of the hyperoctahedral group B_n. Then a(n) = sum p(alpha,2) where |alpha| = n and alpha has at most n parts.
Original entry on oeis.org
2, 16, 150, 1784, 25460
Offset: 1
Richard Bayley (r.t.bayley(AT)qmul.ac.uk), Nov 12 2006
E.g p((2,1),2) = # H-conjugacy classes of B_3 where H = Inft((2,1)) isom B_2 times B_1 . Then a(3) = p((3),2) + p((2,1),2) + p((2,0,1),2) + p((1,2),2) + p((1,1,1),2)+ p((1,0,2),2)+ p((0,3),2) + p((0,2,1),2) + p((0,1,2),2) + p((0,0,3),2) =10 + 16 + 16 + 16 + 24 + 16 + 10 + 16 + 16 +10 = 150
- Richard Bayley, Relative Character Theory and the Hyperoctahedral Group, Ph.D. thesis, Queen Mary College, University of London, to be published 2007.
- Steve Donkin, Invariant functions on Matrices, Math. Proc. Camb. Phil. Soc. 113 (1993) 23-43.
A303914
a(n) = [x^n] (1/(1 - x))*Product_{k>=1} 1/(1 - n*x^k).
Original entry on oeis.org
1, 2, 9, 55, 465, 5051, 69265, 1147287, 22307905, 497211049, 12484203601, 348391613615, 10691846920081, 357749800027465, 12958472141161457, 505088781523073326, 21076091000708067585, 937322034938743608556, 44256147057318887809993, 2210813717869831566759857, 116492226446226314836976401
Offset: 0
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b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
b(n, i-1, k) +`if`(i>n, 0, k*b(n-i, i, k))))
end:
a:= n-> add(b(j$2, n), j=0..n):
seq(a(n), n=0..20); # Alois P. Heinz, May 02 2018
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Table[SeriesCoefficient[1/(1 - x) Product[1/(1 - n x^k), {k, 1, n}], {x, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[1/(1 - x) Exp[Sum[n^k x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 20}]
A302836
a(n) = [x^n] Product_{k>=1} 1/(1 - n*x^(k^2)).
Original entry on oeis.org
1, 1, 4, 27, 260, 3150, 46872, 825944, 16810048, 387952668, 10010010100, 285526191874, 8921263237056, 303013028232642, 11116057874586840, 438023675344410000, 18451248777413066768, 827408674110381669305, 39353155876513869320412, 1978708139249503877752798
Offset: 0
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Table[SeriesCoefficient[Product[1/(1 - n x^k^2), {k, 1, n}], {x, 0, n}], {n, 0, 19}]
A302837
a(n) = [x^n] Product_{k>=1} 1/(1 - n*x^prime(k)).
Original entry on oeis.org
1, 0, 2, 3, 16, 30, 252, 399, 4672, 8100, 111200, 193006, 3279888, 5631093, 114137072, 196030350, 4600308224, 7873377297, 210745840824, 359490499457, 10808589776800, 18389567969010, 613522956193532, 1041400845949218, 38179982045681856, 64675343605547500, 2584515863960831060
Offset: 0
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Table[SeriesCoefficient[Product[1/(1 - n x^Prime[k]), {k, 1, n}], {x, 0, n}], {n, 0, 26}]
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