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Revision History for A348717

(Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

a(n) is the least k such that A003961^i(k) = n for some i >= 0 (where A003961^i denotes the i-th iterate of A003961).
(history; published version)

#28 by Michael De Vlieger at Fri May 20 13:00:20 EDT 2022

STATUS

proposed

approved

#27 by Peter Munn at Fri May 20 12:12:50 EDT 2022

STATUS

editing

proposed

#26 by Peter Munn at Fri May 20 12:04:41 EDT 2022

FORMULA

Equivalently, A156552(a(n)) = A000265(A156552(n)).

CROSSREFS

Also see formula section for the relationship to: A000265, A003961, A004277, A005940, A020639, A046523, A055396, A071364, A122111, A156552, A243055, A243074, A297845, A322993.

#25 by Antti Karttunen at Thu May 19 04:34:58 EDT 2022

CROSSREFS

Cf. A246277 (terms halved), A305897 (restricted growth sequence transform), A354185 (Möbius transform), A354186 (Dirichlet inverse), A354187 (sum with it).

Discussion

Fri May 20

12:02

Peter Munn: No worries, Antti, I have been away from OEIS for 2 days, anyway.

#24 by Antti Karttunen at Thu May 19 04:19:09 EDT 2022

FORMULA

a(n) = A005940(1+A322993(n)) = A005940(1+A000265(A156552(n))).

#23 by Peter Munn at Sat May 14 07:40:05 EDT 2022

FORMULA

From Peter Munn, and _Antti Karttunen_, May 12 2022: (Start)

a(n) >= A071364(n).

CROSSREFS

Also see formula section for the relationship to: A003961, A004277, A020639, A046523, A055396, A071364, A122111, A243055, A243074, A297845.

Sequence Sequences with comparable definitiondefinitions: A304776, A316437.

Cf. A020639, A046523, A071364, A243055, A243074, A297845, A316437.

Discussion

Sat May 14

13:15

Antti Karttunen: @07:32 Not ringing any bells now. Encodes how, not bijectively?

Mon May 16

10:46

Peter Munn: I've found it - A156552. I think a(A156552(n)) = oddpart(a(n)).

10:49

Peter Munn: I mean A156552(a(n)) = oddpart(A156552(n)).

10:55

Peter Munn: So a(n) = A005940(1+oddpart(A156552(n)))? Do I have that correct?

Wed May 18

14:21

Antti Karttunen: Yes, a(n) = A005940(1+A000265(A156552(n))). A156552 encodes the prime factorization in the binary, and A000265 just shifts the trailing 0's away, so the effect after Doudna is the same as here, where we shift with A064989, until we get an even number. And sorry for replying so lately, almost forgot this one!

14:22

Antti Karttunen: I even have this https://oeis.org/A322993 separately.

#22 by Peter Munn at Sat May 14 07:25:24 EDT 2022

COMMENTS

Smallest number generated by uniformly decrementing the indices of the prime factors of n. Thus, for n > 1, the smallest m > 1 such that the first differences of the indices of the ordered prime factors (including repetitions) are the same for m and n. As a function, a(.) preserves properties such as prime signature. - Peter Munn, May 12 2022

FORMULA

A071364(a(n)) = A071364(n).

CROSSREFS

Cf. Positions of particular values (see formula section): A000040, A001248, A003961, A004277, A006094, A030078, A030514, A046301, A050997, A055396, A090076, A090090, A122111, A166329, A251720, A304776.

Also see formula section for the relationship to: A003961, A004277, A055396, A122111.

Sequence with comparable definition: A304776.

Discussion

Sat May 14

07:30

Peter Munn: I went straight from my copy of revision #21, clicked edit, and only afterwards saw your pink box, Antti. Like minds, clearly.

07:32

Peter Munn: @Antti. I feel you must know a sequence, Axxx, that encodes a number using the gaps between prime indices. So can we get a formula that relates Axxx(n) to Axxx(a(n))?

#21 by Peter Munn at Thu May 12 12:22:40 EDT 2022

COMMENTS

Smallest number generated by uniformly decrementing the indices of the prime factors of n. Thus, for n > 1, the smallest m > 1 such that the first differences of the indices of the ordered prime factors (including repetitions) are the same for m and n. - Peter Munn, May 12 2022

FORMULA

From Peter Munn, May 12 2022: (Start)

a(1) = 1; a(2n) = 2n; a(A003961(n)) = a(n). [complete definition]

A297845(a(n), A020639(n)) = n.

A046523(a(n)) = A046523(n).

A243055(a(n)) = A243055(n).

(End)

CROSSREFS

Cf. A020639, A046523, A071364, A243055, A243074, A297845, A316437.

STATUS

approved

editing

Discussion

Thu May 12

16:15

Antti Karttunen: One could also add that A071364(a(n)) = A071364(n). Note how A071364 is intermediate "in strictness" between this and A046523. Also a(n) >= A071364(n) for all n.

#20 by Michael De Vlieger at Thu Feb 24 21:06:31 EST 2022

STATUS

proposed

approved

#19 by Antti Karttunen at Thu Feb 24 10:54:05 EST 2022

STATUS

editing

proposed

Discussion

Thu Feb 24

12:45

Peter Munn: Yes, you were welcome to sign with my name, Antti.