consecutive prime gaps and explicit bound

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I am aware of the theorem that $p_n+1- p_n leq n^0.525$ which is true for all sufficiently large numbers due to Baker, but if i want to make the implicit "for all sufficiently large numbers" explicit, is it known that $p_n+1-p_n leq c n^alpha$ for all $n geq 1$ and for small $c$, lets say $c leq 2$ and $alpha leq 0.55$ ?



Any ref that can give me the explicit numbers or a way to construct them would be great.



Thank you, also i posted the question yesterday on MSE






nt.number-theory reference-request analytic-number-theory prime-numbers prime-gaps







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    up vote
    2
    down vote

    favorite
    1












    I am aware of the theorem that $p_n+1- p_n leq n^0.525$ which is true for all sufficiently large numbers due to Baker, but if i want to make the implicit "for all sufficiently large numbers" explicit, is it known that $p_n+1-p_n leq c n^alpha$ for all $n geq 1$ and for small $c$, lets say $c leq 2$ and $alpha leq 0.55$ ?



    Any ref that can give me the explicit numbers or a way to construct them would be great.



    Thank you, also i posted the question yesterday on MSE






    nt.number-theory reference-request analytic-number-theory prime-numbers prime-gaps







    share|cite|improve this question

























      up vote
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      down vote

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      I am aware of the theorem that $p_n+1- p_n leq n^0.525$ which is true for all sufficiently large numbers due to Baker, but if i want to make the implicit "for all sufficiently large numbers" explicit, is it known that $p_n+1-p_n leq c n^alpha$ for all $n geq 1$ and for small $c$, lets say $c leq 2$ and $alpha leq 0.55$ ?



      Any ref that can give me the explicit numbers or a way to construct them would be great.



      Thank you, also i posted the question yesterday on MSE






      nt.number-theory reference-request analytic-number-theory prime-numbers prime-gaps







      share|cite|improve this question















      I am aware of the theorem that $p_n+1- p_n leq n^0.525$ which is true for all sufficiently large numbers due to Baker, but if i want to make the implicit "for all sufficiently large numbers" explicit, is it known that $p_n+1-p_n leq c n^alpha$ for all $n geq 1$ and for small $c$, lets say $c leq 2$ and $alpha leq 0.55$ ?



      Any ref that can give me the explicit numbers or a way to construct them would be great.



      Thank you, also i posted the question yesterday on MSE






      nt.number-theory reference-request analytic-number-theory prime-numbers prime-gaps




      nt.number-theory reference-request analytic-number-theory prime-numbers prime-gaps






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      edited 3 hours ago









      GH from MO

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      asked 4 hours ago









      Ahmad

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          The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.






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          • Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
            – Ahmad
            1 hour ago











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          accepted










          The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.






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          • Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
            – Ahmad
            1 hour ago















          up vote
          4
          down vote



          accepted










          The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.






          share|cite|improve this answer






















          • Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
            – Ahmad
            1 hour ago













          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.






          share|cite|improve this answer














          The result you quote is due to Baker-Harman-Pintz (2000). I am not aware of any concrete effective version of this result, but if you increase the exponent $0.525$ to $2/3$, then such a variant is available by the work of Dudek. See also my response to this MO question.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          GH from MO

          56.4k5138214




          56.4k5138214











          • Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
            – Ahmad
            1 hour ago

















          • Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
            – Ahmad
            1 hour ago
















          Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
          – Ahmad
          1 hour ago





          Thx, so is there $alpha <1$ such that $cleq 2$ and $n_0$ or $x_0$ is small say less than $10^10$ or $10^12 $ ?
          – Ahmad
          1 hour ago


















           

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