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A336835
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Number of iterations of x -> A003961(x) needed before the result is deficient (sigma(x) < 2x), when starting from x=n.
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15
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0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 2
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OFFSET
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1,120
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COMMENTS
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It holds that a(n) <= A336836(n) for all n, because sigma(n) <= A003961(n) for all n (see A286385 for a proof).
The first 3 occurs at n = 19399380, the first 4 at n = 195534950863140268380. See A336389.
If x and y are relatively prime (i.e., gcd(x,y) = 1), then a(x*y) >= max(a(x),a(y)). Compare to a similar comment in A336915.
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LINKS
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FORMULA
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For all n >= 1,
(End)
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EXAMPLE
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For n = 120, sigma(120) = 360 >= 2*120, thus 120 is not deficient, and we get the next number by applying the prime shift, A003961(120) = 945, and sigma(945) = 1920 >= 945*2, so neither 945 is deficient, so we prime shift once again, and A003961(945) = 9625, which is deficient, as sigma(9625) = 14976 < 2*9625. Thus after two iteration steps we encounter a deficient number, and therefore a(120) = 2.
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MATHEMATICA
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Array[-1 + Length@ NestWhileList[If[# == 1, 1, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}]] &, #, DivisorSigma[1, #] >= 2 # &] &, 120] (* Michael De Vlieger, Aug 27 2020 *)
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PROG
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(PARI)
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961
A336835(n) = { my(i=0); while(sigma(n) >= (n+n), i++; n = A003961(n)); (i); };
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CROSSREFS
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Cf. A003961, A005100, A005940, A023196, A046523, A047802, A071364, A108951, A286385, A294934, A336834, A337474, A337475.
Cf. A336389 (position of the first occurrence of a term >= n).
Differs from A294936 for the first time at n=120.
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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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