A324974 Rank of the n-th special polygonal number A324973(n).
3, 3, 3, 5, 3, 3, 6, 3, 6, 3, 11, 5, 3, 3, 8, 10, 5, 6, 12, 3, 15, 9, 3, 5, 3, 8, 3, 8, 19, 14, 5, 7, 3, 6, 6, 36, 21, 66, 22, 3, 10, 5, 6, 3, 3, 50, 10, 20, 5, 14, 11, 51, 3, 10, 21, 6, 13, 5, 16, 25, 3, 3, 6, 6, 12, 14, 10, 68, 5, 28, 3, 11, 29, 3, 56, 6, 19
Offset: 1
Keywords
Examples
If m = A324973(4) = 70 = 2*5*7, then p = 7, so a(4) = 2+2*((70/7)-1)/(7-1) = 5.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Bernd C. Kellner and Jonathan Sondow, On Carmichael and polygonal numbers, Bernoulli polynomials, and sums of base-p digits, Integers 21 (2021), #A52, 21 pp.; arXiv preprint, arXiv:1902.10672 [math.NT], 2019-2021.
- Bernd C. Kellner, On primary Carmichael numbers, Integers 22 (2022), #A38, 39 pp.; arXiv preprint, arXiv:1902.11283 [math.NT], 2019-2022.
- Wikipedia, Polygonal number.
Programs
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Mathematica
GPF[n_] := Last[Select[Divisors[n], PrimeQ]]; T = Select[Flatten[Table[{p, (p^2*(r - 2) - p*(r - 4))/2}, {p, 3, 150}, {r, 3, 100}], 1], SquareFreeQ[Last[#]] && First[#] == GPF[Last[#]] &]; TT = Take[Union[Table[Last[T[[i]]], {i, Length[T]}]], 47]; Table[2 + 2*(t/GPF[t] - 1)/(GPF[t] - 1), {t, TT}]
Formula
a(n) = 2 + 2*((m/p)-1)/(p-1), where m = A324973(n) and p is its greatest prime factor. (Proof: solve m = P(r,p) = (p^2*(r-2) - p*(r-4))/2 for r.)
Extensions
Several missing terms inserted by and more terms from Jinyuan Wang, Feb 18 2021
Comments