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A067078
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a(1) = 1, a(2) = 2, a(n) = (n-1)*a(n-1) - (n-2)*a(n-2).
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4
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1, 2, 3, 5, 11, 35, 155, 875, 5915, 46235, 409115, 4037915, 43954715, 522956315, 6749977115, 93928268315, 1401602636315, 22324392524315, 378011820620315, 6780385526348315, 128425485935180315, 2561327494111820315
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Successive differences are factorials, or (n+1)-th successive difference divided by (n)-th successive difference = n. i.e. {a(n+2)-a(n+1)}/{a(n+1)-a(n)} = n. - Amarnath Murthy and Meenakshi Srikanth (amarnath_murthy(AT)yahoo.com), Jun 14 2003
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LINKS
| Reinhard Zumkeller, Table of n, a(n) for n = 1..100
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FORMULA
| a(n) =1+sum_{0<=i<=n-2} i! =2*A014288(n-1)+1 =A007489(n-2)+2 (n>1). - Henry Bottomley (se16(AT)btinternet.com), Oct 23 2002 ; Corrected by M. F. Hasler, Dec 16 2007
a(n) = 1+!(n-1) = 1+A003422(n-1) ; a(n+1)=a(n)+(n-1)! - M. F. Hasler, Dec 16 2007
E.g.f. A(x)=x*B(x) satisfies the differential equation B'(x)=B(x)+log(1/(1-x))+1 [From Vladimir Kruchinin (kru(AT)ie.tusur.ru), Jan 19 2011]
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EXAMPLE
| a(6) = 35, a(5)= 11 hence a(7) = 6*35 - 5*11 = 155.
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MATHEMATICA
| a[1] = 1; a[2] = 2; a[n_] := a[n] = (n - 1)*a[n - 1] - (n - 2)*a[n - 2]; Table[ a[n], {n, 1, 25} ]
a=FoldList[Plus, 2, (Range@40)! ]; PrependTo[a, 1] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), May 21 2010]
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PROG
| (PARI) A067078(n)=sum(k=0, n-2, k!, 1) \\ - M. F. Hasler, Dec 16 2007.
(Haskell)
a067078 n = a067078_list !! (n-1)
a067078_list = scanl (+) 1 a000142_list
-- Reinhard Zumkeller, Dec 27 2011
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CROSSREFS
| Cf. A003422, A014288, A007489.
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 16 2009: (Start)
Equals the row sums of A165680.
(End)
Sequence in context: A124627 A064095 A061935 * A124561 A167604 A065510
Adjacent sequences: A067075 A067076 A067077 * A067079 A067080 A067081
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KEYWORD
| nonn
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AUTHOR
| Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jan 05 2002
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EXTENSIONS
| More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Jan 07 2002
Edited by M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Dec 16 2007
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