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A118372
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S-perfect numbers.
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7
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6, 24, 28, 96, 126, 224, 384, 496, 1536, 1792, 6144, 8128, 14336, 15872, 24576, 98304, 114688, 393216, 507904, 917504, 1040384, 1572864, 5540590, 6291456, 7340032, 9078520, 16252928, 22528935, 25165824, 33550336, 56918394, 58720256, 100663296, 133169152
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OFFSET
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1,1
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COMMENTS
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In base 12 the sequence becomes 6, 20, 24, 80, X6, 168, 280, 354, X80, 1054, 3680, 4854, 8368, 9228, 12280, 48X80, 56454, where X is 10 and E is 11. The perfect numbers (A000396) in this sequence in base 12 are 6, 24, 354, 4854. - Walter Kehowski, May 20 2006
Subsequence of A083207. [From Reinhard Zumkeller, Oct 28 2010]
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LINKS
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Donovan Johnson, Table of n, a(n) for n = 1..40 (terms < 4*10^9)
Jean-Marie De Koninck and Aleksandar Ivic, On a sum of divisors problem.
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FORMULA
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S={1}. Assume n>1 and that all numbers m<n have already been tested. Let s=sum{d: d|n, d<n and d in S}. If s<=n, then n is now in S. The paper linked to above has some characterization results. - Walter Kehowski, May 20 2006
I take the preceding comment to mean: S_0 = {1}. s_n = sum{ d: d | n & d < n & d in S_{n-1} }. Then S_n := S_{n-1} if s_n > n, and S_{n-1} U {n} if s_n <= n. - Hugo van der Sanden, Oct 28 2010.
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EXAMPLE
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2 is in S since s=sum{d: d|n, d<n and d in S} = sum{1} = 1 and 1<=2. Similarly, 3, 4, 5, 6 are in S with 6 as the first element such that s=n, that is, 6 is the first S-perfect number. - Walter Kehowski, May 20 2006
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MAPLE
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with(numtheory); S:={1}: SP:=[]: for w to 1 do for n from 1 to 2*10^5 do d:=select(proc(z) z in S and z<n end, divisors(n)); s:=convert(d, `+`); if s<=n then S:=S union {n} fi; if s=n then SP:=[op(SP), n]; print(n); fi; od; od; SP; - Walter Kehowski, May 20 2006
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MATHEMATICA
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S = {1}; SP = {}; Do[ s = Total[ Intersection[S , Divisors[n]]]; If[s <= n, S = Union[S, {n}]]; If[s == n, Print[n]; AppendTo[SP, n]] , {n, 2, 2*10^5} ]; SP (* Jean-François Alcover, Dec 06 2011, after Walter Kehowski *)
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PROG
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(C) #include <stdlib.h> #include <stdio.h> #define MAX_SIZE_SSET 1000000 int main(int argc, char*argv[]) { int Sset[MAX_SIZE_SSET] ; int Ssetsize= 1; Sset[0]=1 ; for(int n=2; n < MAX_SIZE_SSET; n++) { int dsum=0 ; for(int i=0; i< Ssetsize; i++) { if( n % Sset[i] ==0 && Sset[i] < n) dsum += Sset[i] ; if (dsum > n || Sset[i] >=n) break ; } if( dsum <= n) { if(dsum==n) printf("%d\n", n) ; Sset[Ssetsize++ ]= n ; } } } - R. J. Mathar, Oct 28 2010
(Haskell)
a118372_list = sPerfect 1 [] where
sPerfect x ss | v > x = sPerfect (x + 1) ss
| v < x = sPerfect (x + 1) (x : ss)
| otherwise = x : sPerfect (x + 1) (x : ss)
where v = sum (filter ((== 0) . mod x) ss)
-- Reinhard Zumkeller, Feb 25 2012, Oct 28 2010
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CROSSREFS
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Cf. A000396, A181487.
Sequence in context: A072710 A069235 A175200 * A219362 A226476 A216793
Adjacent sequences: A118369 A118370 A118371 * A118373 A118374 A118375
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KEYWORD
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nonn
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AUTHOR
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Vladeta Jovovic, May 15 2006
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EXTENSIONS
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More terms from R. J. Mathar, May 17 2006, a(18) and a(19) Oct 28 2010
Two more terms added and C-program reduced by R. J. Mathar, Oct 28 2010
More terms from William Rex Marshall, Oct 28 2010
Haskell program improved by Reinhard Zumkeller, Nov 02 2010
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STATUS
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approved
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