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A163455
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a(n) = binomial(5*n-1,n).
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3
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1, 4, 36, 364, 3876, 42504, 475020, 5379616, 61523748, 708930508, 8217822536, 95722852680, 1119487075980, 13136858812224, 154603005527328, 1824010149372864, 21566576904406820, 255485622301674660, 3031718514166879020, 36030431772522503316
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OFFSET
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0,2
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COMMENTS
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Also, number of terms in A163142 with n zeros in binary representation.
All terms >= 4 are divisible by 4.
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LINKS
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FORMULA
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a(n) = (5n-1)!/(n!(4n-1)!).
G.f.: A(x)=x*B'(x)/B(x), where B(x)/x is g.f. for A118971. Also a(n) = Sum_{k=0..n} (binomial(n-1,n-k)*binomial(4*n,k)). - Vladimir Kruchinin, Oct 06 2015
a(n) = (-1)^n * binomial(-4*n, n).
a(n) = hypergeom([1 - 4*n, -n], [1], 1).
A(x) satisfies A(x/(1 + x)^5) = 1/(1 - 4*x). (End)
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EXAMPLE
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a(1)=4 because there are 4 terms in A163142 with 1 zero in binary representation {23,27,29,30}_10 ={10111,11011,11101,11110}_2
a(2)=36 because there are 36 terms in A163142 with 2 zeros in binary representation: {639,703,735,751,759,763,765,766,831,863,879,887,891,893,894,927,943,951,955,957,958,975,983,987,989,990,999,1003,1005,1006,1011,1013,1014,1017,1018,1020}_10={1001111111,...,1111111100}_2
a(3)=364 terms in A163142 from 18431 to 32760 with 3 zeros in binary representation 18431_10=100011111111111_2 and 32760_10=111111111111000_2
a(4)=3876 terms in A163142 from 557055 to 1048560 with 4 zeros in binary representation, etc.
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MATHEMATICA
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Table[(5*n-1)!/ n!/(4*n-1)!, {n, 20}]
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PROG
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(Maxima)
B(x):=sum(binomial(5*n-2, n-1)/(n)*x^n, n, 1, 30);
taylor(x*diff(B(x), x, 1)/B(x), x, 0, 10);
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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