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Mar 1, 2023 at 18:42 comment added Sylvain JULIEN I may have mixed up admissible $k$-tuples and prime constellations. I wrote my comment in my train after work and somewhat lacked focus.
Mar 1, 2023 at 17:46 comment added Steven Clark @SylvainJULIEN Both of the lists I posted in the comments above contain numbers that are not in OEIS entry A008407 (see oeis.org/A008407). For example, both lists contain the number $4$. Or did I misinterpret your comment?
Mar 1, 2023 at 17:09 comment added Sylvain JULIEN @StevenClark the values you listed look like diameters of prime constellations. Is it merely coincidental?
Feb 28, 2023 at 14:03 answer added Pavel Gubkin timeline score: 1
Feb 24, 2023 at 7:37 answer added Pavel Gubkin timeline score: 7
Feb 24, 2023 at 5:33 comment added Joachim König I suggest to additionally allow $a, b, c, d$ to be $0$, and show that with this convention, $f(n) =\varphi(n)$ for all $n>1$ (equivalently, when disallowing $0$, $f(n) =\varphi(n)-2$ for all $n>2$ as observed by Peter Taylor) . Indeed, there seems to be a 1-1 correspondence between such $(a, b, c, d)$ and coprime residues mod $n$ via $(a, b, c, d) \mapsto a+b$. Surely this can be seen via some simple trick?
Feb 23, 2023 at 21:55 comment added Peter Taylor @StevenClark, for $n > 2$ that simplifies to $f(n) \stackrel{?}{=} \varphi(n) - 2$, which holds for $3 \le n \le 4500$.
Feb 23, 2023 at 17:28 comment added Steven Clark @PeterTaylor Thanks, with this correction for the first 30 values I get $\{0,0,0,0,2,0,4,2,4,2,8,2,10,4,6,6,14,4,16,6,10,8,20,6,18,10,16,10,26,6\}$ which seems to correspond to $f(n)=\text{A181830}(n)+\text{A070824}(n-1)$ except for $n=1$ (see oeis.org/A070824 ).
Feb 23, 2023 at 17:27 comment added Pavel Gubkin @FFCH, the question would be much nicer if you could add the first values of $f(n)$ into the question
Feb 23, 2023 at 17:06 comment added Peter Taylor @StevenClark, no. All of the < should be <=.
Feb 23, 2023 at 16:35 comment added Martin Rubey isn't it simply en.wikipedia.org/wiki/B%C3%A9zout%27s_identity?
Feb 23, 2023 at 16:31 comment added Steven Clark Here's my Mathematica code for verification: $$f(\text{n$\_$})\text{:=}\text{Block}[\{a=1,b,c,d,\text{cnt}=0\},\text{While}[a<n-3,b=1;\text{While}[a+b<n-2,c=1;\text{While}[a+b+c<n-1,d=n-(a+b+c);\text{If}[a d-b c=1,\text{cnt}\text{++}];c\text{++}];b\text{++}];a\text{++}];\text{cnt}]$$ Does it correctly implement the function $f(n)$?
Feb 23, 2023 at 16:22 comment added Steven Clark @MartinRubey For the first 30 values I get $\{0,0,0,0,1,0,2,2,2,1,6,2,6,4,4,4,11,4,12,6,6,6,18,6,12,9,14,8,22,6\}$ which seems to correspond to oeis.org/A181830 except for $n=1$.
Feb 23, 2023 at 16:12 comment added Martin Rubey oeis.org/A055684
Feb 23, 2023 at 15:41 comment added Puzzled Did you consider the condition $ad-bc=1$?
Feb 23, 2023 at 15:35 comment added Peter Taylor If my calculations are correct and the sequence starts [2, 0, 4, 2, 4, 2, 8, 2, 10, 4, 6, 6, 14, 4, 16, 6, 10, 8, 20, 6, 18, 10, 16, 10, 26, 6, 28, 14, 18, 14, 22, 10, 34, 16, 22, 14, 38, 10, 40, 18, 22, 20, 44, 14, 40, 18, 30, 22, 50, 16, 38, 22, 34, 26, 56] with offset 5 then it's not in OEIS.
Feb 23, 2023 at 15:34 review Close votes
Feb 28, 2023 at 3:07
Feb 23, 2023 at 15:19 comment added Puzzled I want to take the order into account. So the partitions of $10$ you wrote are different. I added a condition on $a,b,c,d$ to my question.
Feb 23, 2023 at 15:18 history edited Puzzled CC BY-SA 4.0
added 22 characters in body
Feb 23, 2023 at 15:11 comment added Stefan Kohl Are you counting ordered or unordered partitions, i.e. do you count 10 = 1+2+3+4 and 10 = 4+3+2+1 as the same or as two distinct decompositions?
Feb 23, 2023 at 15:05 history asked Puzzled CC BY-SA 4.0