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#24 by Vaclav Kotesovec at Fri Mar 10 12:19:45 EST 2023
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#23 by Vaclav Kotesovec at Fri Mar 10 12:19:41 EST 2023
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#22 by Vaclav Kotesovec at Fri Jan 06 09:33:11 EST 2023
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#21 by Joerg Arndt at Fri Jan 06 05:42:01 EST 2023
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#20 by Amiram Eldar at Fri Jan 06 03:31:17 EST 2023
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#19 by Amiram Eldar at Fri Jan 06 02:58:01 EST 2023
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| FORMULA
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From Amiram Eldar, Jan 06 2023: (Start)
Multiplicative with a(p^e) = (p-1)^3*p^(3*e-3).
Sum_{k=1..n} a(k) ~ c * n^4, where c = (1/4) * Product_{p prime}(1 - 3/p^2 + 3/p^3 - 1/p^4) = 0.08429696844... .
Sum_{k>=1} 1/a(k) = Product_{p prime} (1 + p^3/((p-1)^3*(p^3-1))) = 2.47619474816... (A335818). (End)
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| MATHEMATICA
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a[n_] := EulerPhi[n]^3; Array[a, 100] (* Amiram Eldar, Jan 06 2023 *)
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| CROSSREFS
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Cf. A000010, A127473, A335818.
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| STATUS
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approved
editing
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#18 by Michael De Vlieger at Fri Dec 02 07:03:33 EST 2022
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#17 by Kevin Ryde at Thu Dec 01 23:32:24 EST 2022
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#16 by Kevin Ryde at Tue Nov 29 01:27:45 EST 2022
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Discussion
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| Tue Nov 29
| 01:43
| Param Mayurkumar Parekh: respected @Jon E. Schoenfield sir, the edits that you have made mean the same as I intended
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| 03:23
| Param Mayurkumar Parekh: sir, this series can also be generalized to the multiplication of m variables(in Z_n) and taking gcd with n. answer for that will be phi(n)^m.
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| 11:56
| Michel Marcus: generalization: I think we can stop at m=3 now
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| Thu Dec 01
| 04:49
| Param Mayurkumar Parekh: respected sir, any updates regarding the approval of A358714?
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#15 by Kevin Ryde at Tue Nov 29 01:27:21 EST 2022
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| KEYWORD
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nonn,easy,mult,changed
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Discussion
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| Tue Nov 29
| 01:27
| Kevin Ryde: A000010 rates "easy" so might as well here too.
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