Displaying 1-10 of 20 results found.
7, 8, 10, 7, 13, 10, 13, 12, 12, 12, 8, 9, 18, 9, 17, 13, 8, 16, 7, 17, 9, 7, 19, 8, 9, 7, 17, 18, 16, 9, 23, 7, 10, 18, 16, 16, 6, 24, 9, 21, 15, 9, 13, 19, 17, 22, 15, 17, 26, 15, 9, 15, 9, 17, 22, 9, 29, 9, 23, 19, 9, 15, 16, 9, 14, 28, 17, 11, 17, 25, 17, 23, 14, 22, 9, 14, 25, 16, 16, 31, 17, 15, 15, 11, 20
PROG
(PARI)
A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
isA337386(n) = { my(x= A003961(n)); (sigma(x)>=2*x); }; A080224(n) = sumdiv(n, d, sigma(d)>2*d)
k=0; for(n=1, 2^12, if(isA337386(n), print1( A080224(n), ", ")));
Abundant numbers (sum of divisors of m exceeds 2m).
(Formerly M4825)
+10
346
12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 198, 200, 204, 208, 210, 216, 220, 222, 224, 228, 234, 240, 246, 252, 258, 260, 264, 270
COMMENTS
A number m is abundant if sigma(m) > 2m (this sequence), perfect if sigma(m) = 2m (cf. A000396), or deficient if sigma(m) < 2m (cf. A005100), where sigma(m) is the sum of the divisors of m ( A000203). While the first even abundant number is 12 = 2^2*3, the first odd abundant is 945 = 3^3*5*7, the 232nd abundant number!
If m is a term so is every positive multiple of m. "Primitive" terms are in A091191. If m=6k (k>=2), then sigma(m) >= 1 + k + 2*k + 3*k + 6*k > 12*k = 2*m. Thus all such m are in the sequence.
According to Deléglise (1998), the abundant numbers have natural density 0.2474 < A(2) < 0.2480. Thus the n-th abundant number is asymptotic to 4.0322*n < n/A(2) < 4.0421*n. - Daniel Forgues, Oct 11 2015 From Bob Selcoe, Mar 28 2017 (prompted by correspondence with Peter Seymour): (Start) Applying similar logic as the proof that all multiples of 6 >= 12 appear in the sequence, for all odd primes p:
i) all numbers of the form j*p*2^k (j >= 1) appear in the sequence when p < 2^(k+1) - 1;
ii) no numbers appear when p > 2^(k+1) - 1 (i.e., are deficient and are in A005100); iii) when p = 2^(k+1) - 1 (i.e., perfect numbers, A000396), j*p*2^k (j >= 2) appear. Note that redundancies are eliminated when evaluating p only in the interval [2^k, 2^(k+1)].
The first few even terms not of the forms i or iii are {70, 350, 490, 550, 572, 650, 770, ...}. (End)
REFERENCES
L. E. Dickson, Theorems and tables on the sum of the divisors of a number, Quart. J. Pure Appl. Math., Vol. 44 (1913), pp. 264-296.
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section B2, pp. 74-84.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 59.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Eric Weisstein's World of Mathematics, Abundance.
FORMULA
a(n) is asymptotic to C*n with C=4.038... (Deléglise, 1998). - Benoit Cloitre, Sep 04 2002
MAPLE
with(numtheory): for n from 1 to 270 do if sigma(n)>2*n then printf(`%d, `, n) fi: od:
isA005101 := proc(n)
simplify(numtheory[sigma](n) > 2*n) ;
option remember ;
local a ;
if n =1 then
12 ;
else
a := procname(n-1)+1 ;
while numtheory[sigma](a) <= 2*a do
a := a+1 ;
end do ;
a ;
end if ;
PROG
(Haskell)
a005101 n = a005101_list !! (n-1)
a005101_list = filter (\x -> a001065 x > x) [1..]
(Python)
from sympy import divisors
def ok(n): return sum(divisors(n)) > 2*n
(Python)
from sympy import divisor_sigma
from itertools import count, islice
def A005101_gen(startvalue=1): return filter(lambda n:divisor_sigma(n) > 2*n, count(max(startvalue, 1))) # generator of terms >= startvalue
CROSSREFS
Cf. A094268 (n consecutive abundant numbers). Cf. A173490 (even abundant numbers).
Primitive abundant numbers: abundant numbers ( A005101) having no abundant proper divisor.
+10
46
12, 18, 20, 30, 42, 56, 66, 70, 78, 88, 102, 104, 114, 138, 174, 186, 196, 222, 246, 258, 272, 282, 304, 308, 318, 354, 364, 366, 368, 402, 426, 438, 464, 474, 476, 498, 532, 534, 550, 572, 582, 606, 618, 642, 644, 650, 654, 678, 748, 762, 786, 812, 822
COMMENTS
This is a supersequence of the primitive abundant number sequence A071395, since many of these numbers will be positive integer multiples of the perfect numbers ( A000396). - Timothy L. Tiffin, Jul 15 2016
EXAMPLE
12 is a term since 1, 2, 3, 4, and 6 (the proper divisors of 12) are either deficient or perfect numbers, and thus not abundant. - Timothy L. Tiffin, Jul 15 2016
MAPLE
isA005101 := proc(n) is(numtheory[sigma](n) > 2*n ); end proc:
isA091191 := proc(n) local d; if isA005101(n) then for d in numtheory[divisors](n) minus {1, n} do if isA005101(d) then return false; end if; end do: return true; else false; end if; end proc:
for n from 1 to 200 do if isA091191(n) then printf("%d\n", n) ; end if; end do: # R. J. Mathar, Mar 28 2011
MATHEMATICA
t = {}; n = 1; While[Length[t] < 100, n++; If[DivisorSigma[1, n] > 2*n && Intersection[t, Divisors[n]] == {}, AppendTo[t, n]]]; t (* T. D. Noe, Mar 28 2011 *) Select[Range@ 840, DivisorSigma[1, #] > 2 # && Times @@ Boole@ Map[DivisorSigma[1, #] <= 2 # &, Most@ Divisors@ #] == 1 &] (* Michael De Vlieger, Jul 16 2016 *)
PROG
(Haskell)
a091191 n = a091191_list !! (n-1)
a091191_list = filter f [1..] where
f x = sum pdivs > x && all (<= 0) (map (\d -> a000203 d - 2 * d) pdivs)
where pdivs = a027751_row x
Sum of the abundant divisors of n.
+10
16
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 18, 0, 20, 0, 0, 0, 36, 0, 0, 0, 0, 0, 30, 0, 0, 0, 0, 0, 66, 0, 0, 0, 60, 0, 42, 0, 0, 0, 0, 0, 84, 0, 0, 0, 0, 0, 72, 0, 56, 0, 0, 0, 122, 0, 0, 0, 0, 0, 66, 0, 0, 0, 70, 0, 162, 0, 0, 0, 0, 0, 78, 0, 140, 0, 0, 0, 138, 0, 0, 0, 88, 0, 138, 0, 0, 0, 0, 0, 180
COMMENTS
Sum of divisors d of n with sigma(d) > 2*d.
EXAMPLE
a(12) = 12 because the divisors of 12 are 1, 2, 3, 4, 6, 12, but of those only 12 is abundant.
a(13) = 0 because the divisors of 13 are 1 and 13, neither of which is abundant.
MAPLE
local a, d;
a :=0 ;
for d in numtheory[divisors](n) do
if numtheory[sigma](d) > 2* d then
a := a+d ;
end if;
end do:
return a;
end proc:
MATHEMATICA
Table[Total@ Select[Divisors@ n, DivisorSigma[1, #] > 2 # &], {n, 96}] (* Michael De Vlieger, Jul 16 2016 *)
PROG
(Python)
from sympy import divisors, divisor_sigma
def A187795(n): return sum(d for d in divisors(n, generator=True) if divisor_sigma(d) > 2*d) # Chai Wah Wu, Sep 22 2021
a(n) is the number of primitive nondeficient numbers ( A006039) dividing n.
+10
11
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, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 2, 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
COMMENTS
As a simple consequence of the definition of a primitive nondeficient number, a(n) is nonzero if and only if n is nondeficient.
FORMULA
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Sum_{n>=1} 1/ A006039(n) = 0.3... (see A006039 for a better estimate of this constant). - Amiram Eldar, Jan 01 2024
EXAMPLE
The least nondeficient number, therefore the least primitive nondeficient number is 6. So a(1) = a(2) = a(3) = a(4) = a(5) = 0 as all primitive nondeficient numbers are larger, and therefore not divisors; and a(6) = 1, as only 1 primitive nondeficient number divides 6, namely 6 itself.
60 has the following 12 divisors: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. Of these, only 6 and 20 are in A006039, thus a(60) = 2.
PROG
(PARI)
A341619(n) = if(sigma(n) < (2*n), 0, fordiv(n, d, if((d<n)&&(sigma(d) >= 2*d), return(0))); (1)); \\ After code in A071395
CROSSREFS
A006039 (or equivalently, its characteristic function, A341619) is used to define this sequence.
See A000203 and A023196 for definitions of deficient and nondeficient. Positions of first appearances are given in A337691. Differs from its derived sequence A341618 for the first time at n=120, where a(120)=2, while A341618(120)=1.
Number of deficient divisors of n.
+10
10
1, 2, 2, 3, 2, 3, 2, 4, 3, 4, 2, 4, 2, 4, 4, 5, 2, 4, 2, 5, 4, 4, 2, 5, 3, 4, 4, 5, 2, 6, 2, 6, 4, 4, 4, 5, 2, 4, 4, 6, 2, 6, 2, 6, 6, 4, 2, 6, 3, 6, 4, 6, 2, 5, 4, 6, 4, 4, 2, 7, 2, 4, 6, 7, 4, 6, 2, 6, 4, 7, 2, 6, 2, 4, 6, 6, 4, 6, 2, 7, 5, 4, 2, 7, 4, 4, 4, 7, 2, 8, 4, 6, 4, 4, 4, 7, 2, 6, 6, 7, 2, 6, 2, 7, 8
COMMENTS
Number of divisors d of n with sigma(d)<2*d (sigma = A000203).
MATHEMATICA
a[n_] := Sum[If[DivisorSigma[1, d] < 2d, 1, 0], {d, Divisors[n]}];
Number of divisors d of n for which A003961(d) > 2*d.
+10
10
0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 3, 0, 1, 1, 3, 0, 3, 0, 3, 1, 0, 0, 5, 0, 0, 2, 3, 0, 4, 0, 4, 0, 0, 1, 6, 0, 0, 1, 5, 0, 4, 0, 2, 3, 0, 0, 7, 1, 2, 0, 2, 0, 5, 0, 5, 1, 0, 0, 8, 0, 0, 3, 5, 0, 2, 0, 2, 1, 4, 0, 9, 0, 0, 2, 2, 0, 3, 0, 7, 3, 0, 0, 8, 0, 0, 0, 4, 0, 8, 1, 2, 0, 0, 0, 9, 0, 3, 2, 5, 0, 2, 0, 4, 4
COMMENTS
Number of terms of A246282 that divide n. Number of divisors d of n for which A048673(d) > d.
PROG
(PARI)
A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
CROSSREFS
Inverse Möbius transform of A252742.
Number of proper divisors of n that are abundant ( A005101).
+10
9
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 6
EXAMPLE
The proper divisors of 24 are 1, 2, 3, 4, 6, 8, 12. Only one of these, 12, is abundant (in A005101), thus a(24) = 1. The proper divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60. Six of these are abundant: 12, 20, 24, 30, 40, 60, thus a(120) = 6.
MATHEMATICA
a[n_] := Count[Most[Divisors[n]], _?(DivisorSigma[1, #] > 2*# &)]; Array[a, 100] (* Amiram Eldar, Mar 14 2024 *)
PROG
(PARI) A294929(n) = sumdiv(n, d, (d<n)*(sigma(d)>(2*d)));
Number of perfect divisors of n.
+10
8
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 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, 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, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0
COMMENTS
Number of divisors d of n with sigma(d) = 2*d (sigma = A000203).
FORMULA
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A335118 = 0.2045201... . - Amiram Eldar, Dec 31 2023
EXAMPLE
Divisors of n = 84: {1,2,3,4,6,7,12,14,21,24,28,42}, two of them are perfect: 6 = A000396(1) and 28 = A000396(2), therefore a(84) = 2.
MATHEMATICA
a[n_] := DivisorSum[n, 1 &, DivisorSigma[-1, #] == 2 &]; Array[a, 100] (* Amiram Eldar, Dec 31 2023 *)
PROG
(Haskell)
a080225 n = length [d | d <- takeWhile (<= n) a000396_list, mod n d == 0]
(PARI) a(n) = sumdiv(n, d, sigma(d, -1) == 2); \\ Amiram Eldar, Dec 31 2023
Number of nondeficient divisors of n.
+10
8
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 4, 0, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 5, 0, 0, 0, 0, 0, 2, 0, 0, 0, 1, 0, 6, 0, 0, 0, 0, 0, 2, 0, 3, 0, 0, 0, 5, 0, 0, 0, 1, 0, 4, 0, 0, 0, 0, 0, 5, 0, 0, 0, 2, 0, 2, 0, 1, 0, 0, 0, 6, 0, 0, 0, 3, 0, 2, 0, 0, 0, 0, 0, 8
COMMENTS
Number of nondeficient numbers ( A023196) dividing n.
MATHEMATICA
a[n_] := DivisorSum[n, 1 &, DivisorSigma[1, #] >= 2*# &]; Array[a, 120] (* Amiram Eldar, Feb 22 2021 *)
PROG
(PARI)
(PARI) A341620(n) = sumdiv(n, d, (sigma(d)>=(2*d)));
CROSSREFS
Differs from a derived sequence A341624 for the first time at n=120, where a(120)=8, while A341624(120)=1.
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