A324371 Product of all primes p dividing n such that the sum of the base p digits of n is less than p, or 1 if no such prime.
1, 2, 3, 2, 5, 3, 7, 2, 3, 5, 11, 3, 13, 7, 5, 2, 17, 3, 19, 5, 7, 11, 23, 1, 5, 13, 3, 7, 29, 15, 31, 2, 11, 17, 35, 3, 37, 19, 13, 5, 41, 7, 43, 11, 1, 23, 47, 1, 7, 5, 17, 13, 53, 3, 55, 7, 19, 29, 59, 5, 61, 31, 7, 2, 13, 11, 67, 17, 23, 7, 71, 1, 73, 37, 5, 19, 77, 13, 79, 5, 3, 41, 83, 21
Offset: 1
Examples
For p = 2 and 3, the sum of the base p digits of 6 is 1+1+0 = 2 >= 2 and 2+0 = 2 < 3, respectively, so a(6) = 3.
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- 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 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
Programs
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Maple
f:= n -> convert(select(p -> convert(convert(n,base,p),`+`)
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Mathematica
SD[n_, p_] := If[n < 1 || p < 2, 0, Plus @@ IntegerDigits[n, p]]; LP[n_] := Transpose[FactorInteger[n]][[1]]; DD3[n_] := Times @@ Select[LP[n], SD[n, #] < # &]; Table[DD3[n], {n, 1, 100}]
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Python
from math import prod from sympy.ntheory import digits from sympy import primefactors as pf def a(n): return prod(p for p in pf(n) if sum(digits(n, p)[1:]) < p) print([a(n) for n in range(1, 85)]) # Michael S. Branicky, Jul 03 2022
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