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A122057
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a(n) = (n+1)! * (H(n+1) - H(2)), where H(n) are the harmonic numbers.
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1
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0, 2, 14, 94, 684, 5508, 49104, 482256, 5185440, 60668640, 767940480, 10462227840, 152698210560, 2377651449600, 39350097561600, 689874448435200, 12773427499929600, 249097496204390400, 5103595024496640000, 109608397522606080000
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OFFSET
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1,2
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COMMENTS
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Former title (corrected): A Legendre-based recurrence sequence. Let b(n) = ((4*n+2)*x -(2*n+1) )/(n+1)*b(n-1) - (n/(n+1))*b(n-2), where x=1, then a(n) = (n+1)!*b(n)/6. - G. C. Greubel, Oct 03 2019
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REFERENCES
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Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964, 9th Printing (1970), pp. 782
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LINKS
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M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
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FORMULA
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Let b(n) = ((-2*n-1) +(4*n+2)*x)/(n+1)*b(n-1) - (n/(n+1))*b(n-2) with x=1, then a(n) = b(n)*(n+1)!/6.
a(n) = (n+1)! * Sum_{k=3..n+1} 1/k. - Gary Detlefs, Jul 15 2010
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MAPLE
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a:=n-> (n+1)!*add(1/k, k=3..n+1): seq(a(n), n=1..30); # Gary Detlefs, Jul 15 2010
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MATHEMATICA
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x=1; b[1]:=0; b[2]:=2; b[n_]:= b[n]= ((-2*n-1) +(4*n+2)*x)/(n+1)*b[n-1] - (n/(n+1))*b[n-2]; Table[b[n]*(n+1)!/6, {n, 30}]
Table[(n+1)!*(HarmonicNumber[n+1] - 3/2), {n, 30}] (* G. C. Greubel, Oct 03 2019 *)
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PROG
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(PARI) vector(30, n, (n+1)!*(sum(k=1, n+1, 1/k) - 3/2) ) \\ G. C. Greubel, Oct 03 2019
(Magma) [Factorial(n+1)*(HarmonicNumber(n+1) - 3/2): n in [1..30]]; // G. C. Greubel, Oct 03 2019
(Sage) [factorial(n+1)*(harmonic_number(n+1) - 3/2) for n in (1..30)] # G. C. Greubel, Oct 03 2019
(GAP) List([1..30], n-> Factorial(n+1)*(Sum([1..n+1], k-> 1/k) - 3/2) ); # G. C. Greubel, Oct 03 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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If all terms are really negative, sequence should probably be negated. - N. J. A. Sloane, Oct 01 2006
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STATUS
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approved
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