A163455 a(n) = binomial(5*n-1,n).
1, 4, 36, 364, 3876, 42504, 475020, 5379616, 61523748, 708930508, 8217822536, 95722852680, 1119487075980, 13136858812224, 154603005527328, 1824010149372864, 21566576904406820, 255485622301674660, 3031718514166879020, 36030431772522503316
Offset: 0
Examples
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.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.
Programs
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Magma
[Binomial(5*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
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Mathematica
Table[(5*n-1)!/ n!/(4*n-1)!,{n,20}] Table[Binomial[5 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *) -
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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PARI
a(n) = binomial(5*n-1,n); \\ Michel Marcus, Oct 06 2015
Formula
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
From Peter Bala, Feb 14 2024: (Start)
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)
From Peter Bala, Jun 05 2024: (Start)
Right-hand side of the identity Sum_{k = 0..n} binomial(n+k-1, k)*binomial(4*n-k-1, n-k) = binomial(5*n-1, n).
a(n) = (3/4)*binomial(4*n, 3*n)*hypergeom([n, -n], [1 - 4*n], 1) for n >= 1. (End)
From Karol A. Penson Jan 20 2025: (Start)
G.f.: 4*z*Hypergeometric5F4([1, 6/5, 7/5, 8/5, 9/5], [5/4, 3/2, 7/4, 2], (3125*z)/256) + 1.
G.f. A(z) satisfies: z*(1250*A^3 - 250*A^2 + 25*A - 1) + (-3125*z + 256)*A^4 + (3125*z - 256)*A^5 = 0. (End)
G.f.: 1/(5-4*g) where g = 1+x*g^5 is the g.f. of A002294. - Seiichi Manyama, Aug 17 2025
Extensions
Entry revised by N. J. A. Sloane, Dec 07 2015
Comments