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a(n) = binomial coefficient C(n,24).
4

%I #35 Dec 15 2023 10:34:38

%S 1,25,325,2925,20475,118755,593775,2629575,10518300,38567100,

%T 131128140,417225900,1251677700,3562467300,9669554100,25140840660,

%U 62852101650,151584480450,353697121050,800472431850,1761039350070,3773655750150,7890371113950,16123801841550

%N a(n) = binomial coefficient C(n,24).

%H T. D. Noe, <a href="/A010977/b010977.txt">Table of n, a(n) for n = 24..1000</a>

%H <a href="/index/Rec#order_25">Index entries for linear recurrences with constant coefficients</a>, signature (25, -300, 2300, -12650, 53130, -177100, 480700, -1081575, 2042975, -3268760, 4457400, -5200300, 5200300, -4457400, 3268760, -2042975, 1081575, -480700, 177100, -53130, 12650, -2300, 300, -25, 1).

%F G.f.: x^24/(1-x)^25. - _Zerinvary Lajos_, Aug 04 2008 [Corrected by _Georg Fischer_, May 19 2019]

%F a(n) = n/(n-24) * a(n-1), n > 24. - _Vincenzo Librandi_, Mar 26 2011

%F From _Amiram Eldar_, Dec 11 2020: (Start)

%F Sum_{n>=24} 1/a(n) = 24/23.

%F Sum_{n>=24} (-1)^n/a(n) = A001787(24)*log(2) - A242091(24)/23! = 201326592*log(2) - 15566188845789952/111546435 = 0.9627768409... (End)

%p seq(binomial(n,24),n=24..41); # _Zerinvary Lajos_, Aug 04 2008

%t Table[Binomial[n,24],{n,24,50}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 26 2011 *)

%o (Magma) [ Binomial(n,24): n in [24..90]]; // _Vincenzo Librandi_, Mar 26 2011

%o (PARI) x='x+O('x^50); Vec(x^24/(1-x)^25) \\ _G. C. Greubel_, Nov 23 2017

%Y Pascal's triangle A007318 diagonal. - _Zerinvary Lajos_, Aug 04 2008

%Y Cf. A010970, A010971, A010972, A001787, A242091.

%K nonn

%O 24,2

%A _N. J. A. Sloane_