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A360425
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Indices of records in A018804.
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0
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1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 72, 80, 84, 90, 105, 108, 120, 140, 144, 168, 180, 210, 240, 252, 270, 280, 288, 300, 330, 336, 360, 420, 480, 504, 540, 600, 630, 660, 720, 840, 990, 1008, 1080, 1200
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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a(n) seems to be divisible by any positive integer as n approaches infinity. Example: There seem to be 6 terms without a 2 (i.e., 1, 3, 5, 9, 15, 105), 14 terms without a 3 (as large as 280), 22 terms without a factor of 4 (as large as 6930), and 29 terms without a 5 (as large as 3276). True for a(n) < 10^8.
For any term a(n) > 1, it seems that there exists at least one term a(x) such that a(x) * prime(y) = a(n) where prime(y) <= (7/2) * A053669(a(x)). True for a(n) < 10^8.
The ratio of prime(n)/a(n) for indices {1..8, 11, 15, 17} is almost 2 (with an error of at most 1 on the numerator). The exact ratios are 2/1, 3/2, 5/3, 7/4, 11/5, 13/6, 17/8, 19/9, 31/15, 47/24, 59/30. - Matthew Russell Downey, Jul 25 2023
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LINKS
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FORMULA
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EXAMPLE
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A018804(36) = 168 is the largest value among the first 36 terms of A018804, so 36 is a term here; since it is the 18th value that sets a new record, a(18) = 36.
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MATHEMATICA
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f[p_, e_] := (e*(p - 1)/p + 1)*p^e; pil[n_] := Times @@ (f @@@ FactorInteger[n]); seq[nmax_] := Module[{p, pm = 0, s = {}}, Do[If[(p = pil[n]) > pm, pm = p; AppendTo[s, n]], {n, 1, nmax}]; s]; seq[1200] (* Amiram Eldar, Feb 13 2023 *)
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PROG
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(Python)
from sympy import factorint
from math import prod
def A018804(m): return prod(p**(e-1)*((p-1)*e+p) for p, e in factorint(m).items())
record = 0
for m in range(1, 2000):
if value > record:
record = value
print(m, end=", ")
(PARI) f(n) = sumdiv(n, d, n*eulerphi(d)/d); \\ A018804
lista(nn) = my(r=0, list=List()); for (n=1, nn, my(m=f(n)); if (m > r, listput(list, n); r = m); ); Vec(list); \\ Michel Marcus, Feb 26 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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