Need help finding a formula for this sequence:
$ A(n) = 52, 103, 155, 206, 258, 309, 368, 412, 463, 515, 566, 618, 669, 747, 772, 823, 875, 926, 978, 1029, 1103, 1132, 1183, 1235, 1286, 1338, 1389, 1494, 1492, 1543, 1595, 1646, 1698, 1749, 1838, 1852, 1903, 1955, 2006, 2058, 2109, 2291, ...$
Is there a single formula that can give this sequence, ideally as a non-recursive polynomial? If so, what is it?
Details of the sequence
The above sequence is $A(n) = B(60n)$ for $n > 0$ where $B(x) = x - floor(\frac{x}{y})$ and where $y$ is the least non-divisor of $x$. Note that $B(x)$ is the OEIS sequence A213634; you may refer to this lookup table for mapping $x$ up to $10,000$.
Where I'm getting stuck is that I'm not sure how to represent $y$ (the least non-divisor of $x$) with a standard mathematical equation. On a computer, we can find the least non-divisor of $x$ using a looping function such as the one given in the OEIS page for the sequence of least non-divisors, A007978:
def a(n):
k = 2
while not n%k: k += 1
return k
print([a(n) for n in range(1, 101)])
Interestingly the sequence of least non-divisors exhibits striking fractal symmetry based on $a(n) = lcm(1,2,...,x)$ where $x$ is the $n$-th prime power (OEIS A000961), though I'm not sure why.
Properties & patterns in the sequence
I've found some bizarre properties and patterns in this sequence $A(n)$.
The subsequence $Ae(n) = A(2n) = 103, 206, 309, ...$ initially seems to be given by $103n$. This holds for n < 7, but but:
- for $Ae(n)$ from 1 thru 6, we have $103n = ((n-1)+\frac{103}{103})103$
- for $Ae(7)$, we have $747 = 103n+26 = ((n-1)+\frac{129}{103})103$
- for $Ae(n)$ from 8 thru 13, we have $103n-1 = ((n-1)+\frac{102}{103})103$
- for $Ae(14)$, we have $1494 = (103n-1)+53 = ((n-1)+\frac{155}{103})103$
- for $Ae(n)$ from 15 thru 20, we have $103n-2 = ((n-1)+\frac{101}{103})103$
- for $Ae(21)$, we have $2291 = (103n-2)+130 = ((n-1)+\frac{231}{103})103$
- for $Ae(n)$ from 22 thru 27, we have $103n-3 = ((n-1)+\frac{100}{103})103$
- for $Ae(28)$, we have $2987 = (103n-3)+106 = ((n-1)+\frac{206}{103})103$
- for $Ae(n)$ from 29 thru 34, we have $103n-4 = ((n-1)+\frac{99}{103})103$
- for $Ae(35)$, we have $3734 = (103n-4)+133 = ((n-1)+\frac{232}{103})103)$
Clearly therefore we might have some term like $-floor(\frac{n}{14}-1)$ in the equation, but I'm not sure how to account for the additional $26, 52, 130, 106, 133, ...$ that gets tacked onto every 14th member of the sequence.
Any help would be appreciated, I feel at this point I'm seeing the trees and not the forest. Thanks.
Update, 3/9/23
I discovered that this sequence appears to be based on the sequence $a(n)=\{52,103,155,206,258,309,361,412,463,515,566,618,669,721,772,...\}$ defined by the recurrence relation $a(n+8)=-a(n)+a(n+1)+a(n+7)$, which can be defined non-recursively as:
- $a(n)=51\lfloor\frac{n}{7}\rfloor +51\lfloor\frac{n+1}{7}\rfloor+52\lfloor\frac{n+2}{7}\rfloor+51\lfloor\frac{n+3}{7}\rfloor+52\lfloor\frac{n+4}{7}\rfloor+51\lfloor\frac{n+5}{7}\rfloor+52\lfloor\frac{n}{7}\rfloor$
The original sequence $A(n)$ differs from this $a(n)$ base sequence only where $n=7m,m\in ℕ_+$, according to the following bizarre series of differences: $\{7,26,23,53,37,130,52,106,...\}$ for which I have yet to determine a discernible pattern (although searching for the first six with any wildcards in between them does yield these hits in OEIS).
Well, I cannot see how to describe your sequence in a single form easily.
However, an excellent linear form is found when one excludes multiples of 420 from the basic form $A(n)=B(60n)$. Looking at the OEIS sequence bfile through n=10000, this separates your data into 143 main points with 23 to be treated separately. What results is $A(n)$ ~ $round(K*n)$, where K~ $0.857218087051$, for $n (mod 420) !=0$. The result is never more than 1 away from the true value. The remaining 23 points can also be fit with a line (R^2>0.999) but the maximum error is far less satisfactory. The second coefficient I found for the remnant was ~ 0.883926919519 for $n (mod 420) ==0$.