A324370 Product of all primes p not dividing n such that the sum of the base-p digits of n is at least p, or 1 if no such prime exists.

1, 1, 2, 1, 6, 1, 6, 3, 10, 1, 6, 1, 210, 15, 2, 3, 30, 5, 210, 21, 110, 15, 30, 5, 546, 21, 14, 1, 30, 1, 462, 231, 1190, 105, 6, 1, 51870, 1365, 70, 21, 2310, 55, 2310, 105, 322, 105, 210, 35, 6630, 663, 286, 33, 330, 55, 798, 57, 290, 15, 30, 1, 930930, 15015, 1430, 2145, 1122, 85, 82110, 2415, 70, 3, 330, 55, 21111090, 285285

Offset: 1

Author

Bernd C. Kellner and Jonathan Sondow, Feb 24 2019

Keywords

• nonn
• base

Comments

The product is finite, as the sum of the base-p digits of n is n if p > n.

a(198) = 2465 is the only term below 10^6 that is a Carmichael number (A002997).

It appears that a(n)=1 if and only if n is in A094960. - Robert Israel, Mar 30 2020

It turns out that a(n) equals the denominator of the first derivative of the Bernoulli polynomial B(n,x). So a(n)=1 if and only if n is in A094960, also impyling that n+1 is prime. A324370 is also involved in such formulas regarding higher derivatives. See Kellner 2023. - Bernd C. Kellner, Oct 12 2023

Examples

			For p = 2, 3, and 5, the sum of the base p digits of 7 is 1+1+1 = 3 >= 2, 2+1 = 3 >= 3, and 1+2 = 3 < 5, respectively, so a(7) = 2*3 = 6.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..5000
• Bernd C. Kellner, On a product of certain primes, J. Number Theory, 179 (2017), 126-141; arXiv:1705.04303 [math.NT], 2017.
• Bernd C. Kellner, On the finiteness of Bernoulli polynomials whose derivative has only integral coefficients, J. Integer Seq. 27 (2024), Article 24.2.8, 11 pp.; arXiv:2310.01325 [math.NT], 2023.
• Bernd C. Kellner and Jonathan Sondow, Power-Sum Denominators, Amer. Math. Monthly, 124 (2017), 695-709; arXiv:1705.03857 [math.NT], 2017.
• 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:1902.10672 [math.NT], 2019.

Crossrefs

Cf. A002997, A094960, A144845, A195441, A324315, A324316, A324317, A324318, A324319, A324320, A324369, A324371, A324404, A324405, A366168, A366169, A366186, A366187, A366188.

Programs

Maple

N:= 100: # for a(1)..a(N)
V:= Vector(N,1):
p:= 1:
for iter from 1 do
   p:= nextprime(p);
   if p >= N then break fi;
   for n from p+1 to N do
     if n mod p <> 0 and convert(convert(n,base,p),`+`)>= p then
       V[n]:= V[n]*p
     fi
od od:
convert(V,list); # Robert Israel, Mar 30 2020
# Alternatively, note that this formula is suggesting offset 0 and a(0) = 1:
seq(denom(diff(bernoulli(n, x), x)), n = 1..51); # Peter Luschny, Oct 13 2023

Mathematica

SD[n_, p_] := If[n < 1 || p < 2, 0, Plus @@ IntegerDigits[n, p]];
DD2[n_] := Times @@ Select[Prime[Range[PrimePi[(n+1)/(2+Mod[n+1, 2])]]], !Divisible[n, #] && SD[n, #] >= # &];
Table[DD2[n], {n, 1, 100}]
(* From Bernd C. Kellner, Oct 12 2023 (Start) *)
(* Denominator of first derivative of BP *)
k = 1; Table[Denominator[Together[D[BernoulliB[n, x], {x, k}]]], {n, 1, 100}]
(* End *)

Python

from math import prod
from sympy.ntheory import digits
from sympy import primefactors, primerange
def a(n):
    nonpf = set(primerange(1, n+1)) - set(primefactors(n))
    return prod(p for p in nonpf if sum(digits(n, p)[1:]) >= p)
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Jul 03 2022

Formula

a(n) * A324369(n) = A195441(n-1) = denominator(Bernoulli_n(x) - Bernoulli_n).

a(n) * A324369(n) * A324371(n) = A144845(n-1) = denominator(Bernoulli_{n-1}(x)).

a(n+1) = A195441(n)/A324369(n+1) = A144845(n)/A007947(n+1) = A318256(n). Essentially the same as A318256. - Peter Luschny, Mar 05 2019

From Bernd C. Kellner, Oct 12 2023: (Start)

a(n) = denominator(Bernoulli_n(x)').

k-th derivative: let (n)_m be the falling factorial.

For n > k, a(n-k+1)/gcd(a(n-k+1), (n)_{k-1}) = denominator(Bernoulli_n(x)^(k)). Otherwise, the denominator equals 1. (End)