A335818 - OEIS

A335818

Decimal expansion of Sum_{k>=1} 1/phi(k)^3, where phi is the Euler totient function.

2, 4, 7, 6, 1, 9, 4, 7, 4, 8, 1, 6, 5, 0, 2, 5, 7, 9, 4, 3, 2, 6, 8, 5, 5, 4, 4, 4, 1, 2, 5, 1, 4, 5, 1, 6, 0, 0, 4, 5, 4, 5, 6, 8, 5, 6, 3, 5, 5, 2, 8, 4, 3, 8, 4, 3, 4, 5, 7, 0, 7, 8, 7, 9, 1, 5, 0, 9, 4, 9, 0, 3, 0, 1, 1, 7, 5, 1, 2, 4, 5, 8, 1, 7, 6, 2, 8, 0, 1, 3, 4, 6, 1, 5, 2, 6, 7, 3, 8, 9, 3, 3, 2, 8, 9

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OFFSET

1,1

COMMENTS

Sum_{k>=1} 1/phi(k)^m is convergent iff m > 1 (reference Monier). - Bernard Schott, Jan 14 2021

REFERENCES

Jean-Marie Monier, Analyse, Exercices corrigés, 2ème année MP, Dunod, 1997, Exercice 3.2.21, pp. 281 and 294.

LINKS

Table of n, a(n) for n=1..105.

Eric Weisstein's World of Mathematics, Totient Function.

Eric Weisstein's World of Mathematics, Totient Summatory Function.

Wikipedia, Euler's totient function.

FORMULA

Equals Product_{primes p} (1 + 1/((1 - 1/p)^3 * (p^3 - 1))).

EXAMPLE

2.476194748165025794326855444125145160045456856355284384345707879150949...

MATHEMATICA

$MaxExtraPrecision = 1000; f[p_] := (1 + 1/((1 - 1/p)^s * (p^s - 1))) /. s -> 3; Do[cc = Rest[CoefficientList[Series[Log[f[1/x]], {x, 0, m}], x]]; Print[f[2] * Exp[N[Sum[Indexed[cc, n]*(PrimeZetaP[n] - 1/2^n), {n, 2, m}], 120]]], {m, 100, 1000, 100}]

PROG

(PARI) prodeulerrat(1 + 1/((1 - 1/p)^3 * (p^3 - 1))) \\ Amiram Eldar, Mar 15 2021

CROSSREFS

Cf. A000010, A065484, A109695, A335319.

Sequence in context: A261076 A302991 A015791 * A129980 A198137 A126786

Adjacent sequences: A335815 A335816 A335817 * A335819 A335820 A335821

KEYWORD

nonn,cons

AUTHOR

Vaclav Kotesovec, Jun 25 2020

STATUS

approved