A324974 - OEIS

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

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OFFSET

1,1

COMMENTS

While two polygonal numbers of different ranks can be equal (e.g., P(6,n) = P(3,2n-1)), that cannot occur for special polygonal numbers, since for fixed p the value of P(r,p) is strictly increasing with r. Thus the rank of a special polygonal number is well-defined.

The Carmichael numbers A002997 and primary Carmichael numbers A324316 are special polygonal numbers (see Kellner and Sondow 2019). Their ranks form the subsequences A324975 and A324976.

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.

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.)

EXAMPLE

If m = A324973(4) = 70 = 2*5*7, then p = 7, so a(4) = 2+2*((70/7)-1)/(7-1) = 5.

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[#]] &];