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Search: id:A054108
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| A054108 |
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a(n)=(-1)^(n+1)*sum(k=0,n,(-1)^k*binomial(2*k,k)). |
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3
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| 1, 5, 15, 55, 197, 727, 2705, 10165, 38455, 146301, 559131, 2145025, 8255575, 31861025, 123256495, 477823895, 1855782325, 7219352975, 28125910825, 109720617995, 428537256445, 1675561707275, 6557869020325, 25689734662775
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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p divides a((p-3)/2) for p=11,19,29,31,41,59,61,71..=A045468 Primes congruent to {1, 4} mod 5. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
The sequence 1,1,5,15,55,... has general term sum{k=0..n, (-1)^(n-k)*C(2k,k)}. Its Hankel transform is A082761. - Paul Barry (pbarry(AT)wit.ie), Apr 10 2007
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FORMULA
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Also: a(n)=C[2n, n]-a(n-1) with a(0)=1 - Labos E. (labos(AT)ana.hu), Apr 26 2003
C(2n,n) - C(2n-2,n-1) + ... +(-1)^(k+n)*C(2k,k)+ ... + (-1)^(1+n)*C(2,1) + (-1)^n*C(0,0), where C(2k,k)=(2k)!/(k!)^2 - central binomial coefficients A000984[k]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
a(n) = Sum[(-1)^(k+n)*((2k)!/(k!)^2),{k,0,n}]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
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MATHEMATICA
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Table[Sum[(-1)^(k+n)*((2k)!/(k!)^2), {k, 0, n}], {n, 1, 50}] - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 05 2006
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PROGRAM
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(PARI) a(n)=(-1)^(n+1)*sum(k=0, n, (-1)^k*binomial(2*k, k))
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CROSSREFS
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T(2n,n), array T as in A054106.
Cf. A066796, A000984, A054109, A006134, A045468.
Adjacent sequences: A054105 A054106 A054107 this_sequence A054109 A054110 A054111
Sequence in context: A007714 A123011 A006358 this_sequence A114947 A137959 A001552
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KEYWORD
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nonn
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AUTHOR
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Clark Kimberling (ck6(AT)evansville.edu)
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EXTENSIONS
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Formula from Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 29 2002
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