A099237 a(n) = Sum_{k=0..n} binomial(n*(n-k), k).
1, 1, 3, 10, 45, 251, 1624, 11908, 97545, 880660, 8664546, 92096731, 1050304775, 12778138842, 165033693175, 2253204163256, 32401745953105, 489207829112931, 7733130368443057, 127664099576228184, 2196149923000824756
Offset: 0
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..490
Programs
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Magma
A099237:= func< n | (&+[Binomial(n*j, n-j): j in [0..n]]) >; [A099237(n): n in [0..30]]; // G. C. Greubel, Mar 09 2021
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Maple
A099237:= n-> add( binomial(n*j, n-j), j=0..n ); seq(A099237(n), n=0..30); # G. C. Greubel, Mar 09 2021
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Mathematica
Table[Sum[Binomial[n*(n - k), k], {k, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Feb 19 2018 *) -
Sage
def A099237(n): return sum( binomial(n*j, n-j) for j in (0..n)) [A099237(n) for n in (0..30)] # G. C. Greubel, Mar 09 2021
Formula
From Vaclav Kotesovec, Feb 19 2018: (Start)
a(n)^(1/n) ~ n^(n/w) * (n+1-w)^(1 - (n+1)/w) * (w-1)^(1/w - 1), where w = LambertW(exp(1)*n),
a(n)^(1/n) ~ n/log(n), but the convergence is too slow. (End)
From Peter Bala, Jan 19 2023: (Start)
Conjectures: a(2^k) == 1 (mod 2^k) and a(3^k) == 1 (mod 3^(k+1)); a(p^k) == 1 (mod p^(k+1)) for all primes p >= 5.
Let m be a positive integer. Similar recurrences may hold for the sequence whose n-th term is given by Sum_{k = 0..n} binomial(m*n*k, n-k). Cf. A359842. (End)
Comments