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A235353
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Numbers m such that phi(m) and tau(m) divide m, where phi = A000010 and tau = A000005.
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6
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1, 2, 8, 12, 18, 24, 36, 72, 96, 108, 128, 288, 384, 864, 972, 1152, 1944, 3456, 6144, 6912, 7776, 13122, 18432, 26244, 31104, 32768, 52488, 55296, 62208, 69984, 98304, 209952, 279936, 294912, 497664, 559872, 708588, 839808, 884736, 1679616, 3538944, 4478976
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OFFSET
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1,2
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COMMENTS
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m is always of the form (2^i)(3^j) where i>0, j>=0.
If j=0, then m is a deficient number, and sigma(m)=2m-1. The deficiency is always 1.
If j>0, then m is an abundant number. (End)
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LINKS
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MATHEMATICA
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Select[Range@ 1000000, And[Mod[#, EulerPhi@ #] == 0, Mod[#, DivisorSigma[0, #]] == 0] &] (* Michael De Vlieger, May 05 2015 *)
Select[Range[55*10^5], Mod[#, EulerPhi[#]]==Mod[#, DivisorSigma[0, #]]==0&] (* Harvey P. Dale, Feb 22 2023 *)
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PROG
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(Haskell)
a235353 n = a235353_list !! (n-1)
a235353_list = filter (\x -> mod x (a000005 x) == 0) a007694_list
(PARI) for(n=1, 10^6, if(!(n%numdiv(n)+n%eulerphi(n)), print1(n, ", "))) \\ Derek Orr, Apr 30 2015
(PARI) sm3(n)=if(n<1, 0, n>>=valuation(n, 2); 3^valuation(n, 3)==n)
list(lim)=my(v=List([1]), t); for(i=1, log(lim)\log(2), if(!sm3(i+1), next); for(j=0, log(lim>>i)\log(3), t=2^i*3^j; if(t%((i+1)*(j+1))==0, listput(v, t)))); Set(v) \\ Charles R Greathouse IV, May 05 2015
(Python)
from itertools import count, islice
from math import prod
from sympy import factorint
def A235353_gen(startvalue=1): # generator of terms >= startvalue
for k in count(max(startvalue, 1)):
f = factorint(k)
t = prod(p**(e-1)*(p-1) for p, e in f.items())
s = prod(e+1 for e in f.values())
if not (k%s or k%t):
yield k
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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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