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A102468
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a(n)! is the smallest factorial divisible by the numerator of Sum_{k=0...n} 1/k!, with a(0) = 1.
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1
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1, 2, 5, 4, 13, 163, 103, 137, 863, 98641, 10687, 31469, 1540901, 522787, 5441, 226871807, 13619, 1276861, 414026539, 2124467, 12670743557, 838025081381, 44659157, 323895443, 337310723185584470837549, 54352957
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OFFSET
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0,2
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COMMENTS
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It appears that a(n) = A102469(n) (largest prime factor of the same numerator) except when n = 3. The smallest factorial divisible by the corresponding denominator is n!. Omitting the 0th term in the sum, it appears that the Kempner number (A002034) and the largest prime factor, of the numerator of Sum_{k=1...n} 1/k! are both equal to A096058(n).
The Mathematica program given below was used to generate the sequence. If the numerator of Sum_{k=0...n}(1/k!) is squarefree, the program prints the value of the numerator's largest prime factor, which must equal a(n). Otherwise, the program prints the complete factorization of the numerator so a(n) can be determined by inspection. - Ryan Propper, Jul 31 2005
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LINKS
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A. J. Kempner, Miscellanea, Amer. Math. Monthly, 25 (1918), 201-210 [ See Section II, "Concerning the smallest integer m! divisible by a given integer n". ]
J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e? (and a link to the primes 2, 5, 13, 37, 463), II, Gems in Experimental Mathematics (T. Amdeberhan, L. A. Medina, and V. H. Moll, eds.), Contemporary Mathematics, vol. 517, Amer. Math. Soc., Providence, RI, 2010.
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FORMULA
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EXAMPLE
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Sum_{k=0...3} 1/k! = 8/3 and 4! is the smallest factorial divisible by 8, so a(3) = 4.
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MATHEMATICA
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Do[l = FactorInteger[Numerator[Sum[1/k!, {k, 0, n}]]]; If[Length[l] == Plus @@ Last /@ l, Print[Max[First /@ l]], Print[l]], {n, 1, 30}] (* Ryan Propper, Jul 31 2005 *)
nmax = 30; Clear[a]; Do[f = FactorInteger[ Numerator[ Sum[1/k!, {k, 0, n}] ] ]; a[n] = If[Length[f] == Total[f[[All, 2]] ], Max[f[[All, 1]] ], f[[-1, 1]] ], {n, 0, nmax}]; a[3] = 4; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Sep 16 2015, adapted from Ryan Propper's script *)
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PROG
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(PARI) a(n) = {j = 1; s = numerator(sum(k=0, n, 1/k!)); while (j! % s, j++); j; } \\ Michel Marcus, Sep 16 2015
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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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STATUS
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approved
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