Original entry on oeis.org
1, 4, 26, 318, 8362, 518136, 79762944, 31347878526, 31875040276682, 84415672202652576, 584440009839756835096, 10604211348847550536329728, 505234836942870203845022012928, 63320076924994011498092218964764672, 20908605974267058444111310750652481533952
Offset: 0
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A055612:= n -> mul(binomial(n,m)+1,m=1..n):
seq(A055612(n+1)-A055612(n),n=0..20); # Robert Israel, Oct 30 2017
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Differences@ Array[Product[1 + Binomial[#, m], {m, #}] &, 16, 0] (* Michael De Vlieger, Oct 30 2017 *)
Original entry on oeis.org
1, 3, 9, 41, 391, 9103, 535951, 80825743, 31508994061, 31937977439061, 84479516648536637, 584608936935076469309, 10605380482241904040731709, 505256047119225750718027007037, 63321087426482869467351750978047037, 20908732615936655362731019703436410620989
Offset: 0
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ListTools:-PartialSums([seq(mul(binomial(n,m)+1,m=1..n),n=0..20)]); # Robert Israel, Oct 30 2017
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Accumulate@ Array[Product[1 + Binomial[#, m], {m, #}] &, 16, 0] (* Michael De Vlieger, Oct 30 2017 *)
Original entry on oeis.org
2, 4, 12, 64, 700, 17424, 1053696, 160579584, 62856336636, 63812936890000, 168895157342195152, 1169048914836855865344, 21209591746609937928524800, 1010490883477487017627972550656, 126641164340871500483202065902080000
Offset: 0
Row 2 is 1 2 1 and we have 000, 001, 010, 011, 020, 021, 100, 101, 110, 111, 120 and 121 so a(2)=12.
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Table[Product[Binomial[n, k] + 1, {k, 0, n}], {n, 0, 15}] (* T. D. Noe, Mar 21 2013 *)
A129824
a(n) = Product_{k=0..n} (1 + binomial(n,k)).
Original entry on oeis.org
2, 4, 12, 64, 700, 17424, 1053696, 160579584, 62856336636, 63812936890000, 168895157342195152, 1169048914836855865344, 21209591746609937928524800, 1010490883477487017627972550656, 126641164340871500483202065902080000, 41817338589698457759723104703370865147904
Offset: 0
a(4) = (1+1)(1+4)(1+6)(1+4)(1+1) = 2*5*7*5*2 = 700.
- H. W. Gould, A product analog of the binomial expansion, unpublished manuscript, Jun 03 2007.
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A129824:= func< n | (&*[1 + Binomial(n,k): k in [0..n]]) >;
[A129824(n): n in [0..20]]; // G. C. Greubel, Apr 26 2024
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Table[Product[1 + Binomial[n,k], {k,0,n}], {n,0,15}] (* Vaclav Kotesovec, Oct 27 2017 *)
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{ a(n) = prod(k=0,n, 1 + binomial(n,k))}
for(n=0,15,print1(a(n),", ")) \\ Paul D. Hanna, Oct 27 2017
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def A129824(n): return product(1 + binomial(n,k) for k in range(n+1))
[A129824(n) for n in range(21)] # G. C. Greubel, Apr 26 2024
A338772
The number of different probabilities p for which a coin that lands heads with probability p can, using n flips, perfectly model one flip of a fair coin.
Original entry on oeis.org
1, 3, 19, 271, 8635, 623533
Offset: 1
For n = 2 the a(2) = 3 different values of p are, in increasing order:
1 - sqrt(1/2), which can model a fair flip with the partition (HH, HT, TH), (TT);
1/2, which can model a fair flip with the partition (HH, HT), (TH, TT) (i.e., by ignoring the second flip); and
sqrt(1/2), which can model a fair flip with the partition (HH), (HT, TH, TT).
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P. = QQ[]
def polystream(nn, pol=P(0), kk=0):
if kk >= nn:
yield pol - 1
else:
for ii in sxrange(binomial(nn, kk) + 1):
for xx in polystream(nn, pol + 2 * ii * p^kk * (1-p)^(nn-kk), kk + 1):
yield xx
def calculate(nn):
solutions = Set()
for pol in polystream(nn):
rootlist = [xx[0] for xx in pol.roots(ring=QQbar)]
for root in rootlist:
if root.real() == root and 0 <= root <= 1:
solutions += Set([root])
return len(solutions)
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