Is $198585576189$ a member of OEIS sequence A228059?

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I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me.



So here it goes:




Is the Descartes spoof
$$mathscrD = 3^2cdot7^2cdot11^2cdot13^2cdot22021 = 198585576189$$
a member of OEIS sequence A228059?




There is an existing Mathematica code in the OEIS hyperlink to test this. Thanks!




number-theory elementary-number-theory mathematica divisor-sum perfect-numbers




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  • Note that $22021 = 19^2cdot61$,
    – Jose Arnaldo Bebita Dris
    Aug 3 at 6:56






  • 1




    Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
    – Claude Leibovici
    Aug 3 at 7:57










  • Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
    – Jose Arnaldo Bebita Dris
    Aug 3 at 8:14






  • 1




    As you say, who knows ?
    – Claude Leibovici
    Aug 3 at 8:38










  • I have just also posted a closely related question here.
    – Jose Arnaldo Bebita Dris
    Aug 3 at 9:05














up vote
0
down vote

favorite












I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me.



So here it goes:




Is the Descartes spoof
$$mathscrD = 3^2cdot7^2cdot11^2cdot13^2cdot22021 = 198585576189$$
a member of OEIS sequence A228059?




There is an existing Mathematica code in the OEIS hyperlink to test this. Thanks!




number-theory elementary-number-theory mathematica divisor-sum perfect-numbers




share|cite|improve this question



















  • Note that $22021 = 19^2cdot61$,
    – Jose Arnaldo Bebita Dris
    Aug 3 at 6:56






  • 1




    Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
    – Claude Leibovici
    Aug 3 at 7:57










  • Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
    – Jose Arnaldo Bebita Dris
    Aug 3 at 8:14






  • 1




    As you say, who knows ?
    – Claude Leibovici
    Aug 3 at 8:38










  • I have just also posted a closely related question here.
    – Jose Arnaldo Bebita Dris
    Aug 3 at 9:05












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me.



So here it goes:




Is the Descartes spoof
$$mathscrD = 3^2cdot7^2cdot11^2cdot13^2cdot22021 = 198585576189$$
a member of OEIS sequence A228059?




There is an existing Mathematica code in the OEIS hyperlink to test this. Thanks!




number-theory elementary-number-theory mathematica divisor-sum perfect-numbers




share|cite|improve this question











I currently do not have enough computing power, so please pardon me for my question, which occurred just recently to me.



So here it goes:




Is the Descartes spoof
$$mathscrD = 3^2cdot7^2cdot11^2cdot13^2cdot22021 = 198585576189$$
a member of OEIS sequence A228059?




There is an existing Mathematica code in the OEIS hyperlink to test this. Thanks!





number-theory elementary-number-theory mathematica divisor-sum perfect-numbers





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 3 at 6:08









Jose Arnaldo Bebita Dris

5,26631940




5,26631940











  • Note that $22021 = 19^2cdot61$,
    – Jose Arnaldo Bebita Dris
    Aug 3 at 6:56






  • 1




    Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
    – Claude Leibovici
    Aug 3 at 7:57










  • Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
    – Jose Arnaldo Bebita Dris
    Aug 3 at 8:14






  • 1




    As you say, who knows ?
    – Claude Leibovici
    Aug 3 at 8:38










  • I have just also posted a closely related question here.
    – Jose Arnaldo Bebita Dris
    Aug 3 at 9:05
















  • Note that $22021 = 19^2cdot61$,
    – Jose Arnaldo Bebita Dris
    Aug 3 at 6:56






  • 1




    Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
    – Claude Leibovici
    Aug 3 at 7:57










  • Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
    – Jose Arnaldo Bebita Dris
    Aug 3 at 8:14






  • 1




    As you say, who knows ?
    – Claude Leibovici
    Aug 3 at 8:38










  • I have just also posted a closely related question here.
    – Jose Arnaldo Bebita Dris
    Aug 3 at 9:05















Note that $22021 = 19^2cdot61$,
– Jose Arnaldo Bebita Dris
Aug 3 at 6:56




Note that $22021 = 19^2cdot61$,
– Jose Arnaldo Bebita Dris
Aug 3 at 6:56




1




1




Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
– Claude Leibovici
Aug 3 at 7:57




Looking at the Mathematica code, I suppose that it will take a long time to generate the 10th, 11th, ... terms.
– Claude Leibovici
Aug 3 at 7:57












Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
– Jose Arnaldo Bebita Dris
Aug 3 at 8:14




Yes, essentially that is the problem, @ClaudeLeibovici. Note that the Descartes spoof might be the 10th or 11th term. Who knows? =)
– Jose Arnaldo Bebita Dris
Aug 3 at 8:14




1




1




As you say, who knows ?
– Claude Leibovici
Aug 3 at 8:38




As you say, who knows ?
– Claude Leibovici
Aug 3 at 8:38












I have just also posted a closely related question here.
– Jose Arnaldo Bebita Dris
Aug 3 at 9:05




I have just also posted a closely related question here.
– Jose Arnaldo Bebita Dris
Aug 3 at 9:05










1 Answer
1






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accepted










The answer to my question is NO, since the abundancy index $I(x):=sigma(x)/x$ (where $sigma(x)$ is the sum of the divisors of $x in mathbbN$) of the first $9$ terms of OEIS sequence A228059 are:



$$I(45) = frac2615 approx 1.73333$$
WolframAlpha computation here
$$I(405) = frac242135 approx 1.79259$$
WolframAlpha computation here
$$I(2205) = frac494245 approx 2.01633$$
WolframAlpha computation here
$$I(26325) = frac5251426325 approx 1.99483$$
WolframAlpha computation here
$$I(236925) = frac474362236925 approx 2.00216$$
WolframAlpha computation here
$$I(1380825) = frac307086153425 approx 2.00154$$
WolframAlpha computation here
$$I(1660725) = frac33231381660725 approx 2.00102$$
WolframAlpha computation here
$$I(35698725) = frac7139653435698725 approx 1.99997$$
WolframAlpha computation here
$$I(3138290325) = frac7748803438744325 approx 1.99998$$
WolframAlpha computation here



Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.



Therefore, since
$$I(198585576189) = frac2362211011 approx 2.14531,$$
it follows that the Descartes spoof
$$mathscrD = 198585576189$$
is not a member of OEIS sequence A228059.






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    up vote
    0
    down vote



    accepted










    The answer to my question is NO, since the abundancy index $I(x):=sigma(x)/x$ (where $sigma(x)$ is the sum of the divisors of $x in mathbbN$) of the first $9$ terms of OEIS sequence A228059 are:



    $$I(45) = frac2615 approx 1.73333$$
    WolframAlpha computation here
    $$I(405) = frac242135 approx 1.79259$$
    WolframAlpha computation here
    $$I(2205) = frac494245 approx 2.01633$$
    WolframAlpha computation here
    $$I(26325) = frac5251426325 approx 1.99483$$
    WolframAlpha computation here
    $$I(236925) = frac474362236925 approx 2.00216$$
    WolframAlpha computation here
    $$I(1380825) = frac307086153425 approx 2.00154$$
    WolframAlpha computation here
    $$I(1660725) = frac33231381660725 approx 2.00102$$
    WolframAlpha computation here
    $$I(35698725) = frac7139653435698725 approx 1.99997$$
    WolframAlpha computation here
    $$I(3138290325) = frac7748803438744325 approx 1.99998$$
    WolframAlpha computation here



    Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.



    Therefore, since
    $$I(198585576189) = frac2362211011 approx 2.14531,$$
    it follows that the Descartes spoof
    $$mathscrD = 198585576189$$
    is not a member of OEIS sequence A228059.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      The answer to my question is NO, since the abundancy index $I(x):=sigma(x)/x$ (where $sigma(x)$ is the sum of the divisors of $x in mathbbN$) of the first $9$ terms of OEIS sequence A228059 are:



      $$I(45) = frac2615 approx 1.73333$$
      WolframAlpha computation here
      $$I(405) = frac242135 approx 1.79259$$
      WolframAlpha computation here
      $$I(2205) = frac494245 approx 2.01633$$
      WolframAlpha computation here
      $$I(26325) = frac5251426325 approx 1.99483$$
      WolframAlpha computation here
      $$I(236925) = frac474362236925 approx 2.00216$$
      WolframAlpha computation here
      $$I(1380825) = frac307086153425 approx 2.00154$$
      WolframAlpha computation here
      $$I(1660725) = frac33231381660725 approx 2.00102$$
      WolframAlpha computation here
      $$I(35698725) = frac7139653435698725 approx 1.99997$$
      WolframAlpha computation here
      $$I(3138290325) = frac7748803438744325 approx 1.99998$$
      WolframAlpha computation here



      Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.



      Therefore, since
      $$I(198585576189) = frac2362211011 approx 2.14531,$$
      it follows that the Descartes spoof
      $$mathscrD = 198585576189$$
      is not a member of OEIS sequence A228059.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The answer to my question is NO, since the abundancy index $I(x):=sigma(x)/x$ (where $sigma(x)$ is the sum of the divisors of $x in mathbbN$) of the first $9$ terms of OEIS sequence A228059 are:



        $$I(45) = frac2615 approx 1.73333$$
        WolframAlpha computation here
        $$I(405) = frac242135 approx 1.79259$$
        WolframAlpha computation here
        $$I(2205) = frac494245 approx 2.01633$$
        WolframAlpha computation here
        $$I(26325) = frac5251426325 approx 1.99483$$
        WolframAlpha computation here
        $$I(236925) = frac474362236925 approx 2.00216$$
        WolframAlpha computation here
        $$I(1380825) = frac307086153425 approx 2.00154$$
        WolframAlpha computation here
        $$I(1660725) = frac33231381660725 approx 2.00102$$
        WolframAlpha computation here
        $$I(35698725) = frac7139653435698725 approx 1.99997$$
        WolframAlpha computation here
        $$I(3138290325) = frac7748803438744325 approx 1.99998$$
        WolframAlpha computation here



        Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.



        Therefore, since
        $$I(198585576189) = frac2362211011 approx 2.14531,$$
        it follows that the Descartes spoof
        $$mathscrD = 198585576189$$
        is not a member of OEIS sequence A228059.






        share|cite|improve this answer













        The answer to my question is NO, since the abundancy index $I(x):=sigma(x)/x$ (where $sigma(x)$ is the sum of the divisors of $x in mathbbN$) of the first $9$ terms of OEIS sequence A228059 are:



        $$I(45) = frac2615 approx 1.73333$$
        WolframAlpha computation here
        $$I(405) = frac242135 approx 1.79259$$
        WolframAlpha computation here
        $$I(2205) = frac494245 approx 2.01633$$
        WolframAlpha computation here
        $$I(26325) = frac5251426325 approx 1.99483$$
        WolframAlpha computation here
        $$I(236925) = frac474362236925 approx 2.00216$$
        WolframAlpha computation here
        $$I(1380825) = frac307086153425 approx 2.00154$$
        WolframAlpha computation here
        $$I(1660725) = frac33231381660725 approx 2.00102$$
        WolframAlpha computation here
        $$I(35698725) = frac7139653435698725 approx 1.99997$$
        WolframAlpha computation here
        $$I(3138290325) = frac7748803438744325 approx 1.99998$$
        WolframAlpha computation here



        Notice that, by the definition of OEIS sequence A228059, $|I(x_i)-2|$ must be a (strictly?) decreasing sequence.



        Therefore, since
        $$I(198585576189) = frac2362211011 approx 2.14531,$$
        it follows that the Descartes spoof
        $$mathscrD = 198585576189$$
        is not a member of OEIS sequence A228059.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 3 at 12:06









        Jose Arnaldo Bebita Dris

        5,26631940




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