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A152454
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Irregular triangle in which row n lists the numbers whose proper divisors sum to n.
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16
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4, 9, 6, 25, 8, 10, 49, 15, 14, 21, 121, 27, 35, 22, 169, 16, 33, 12, 26, 39, 55, 289, 65, 77, 34, 361, 18, 51, 91, 20, 38, 57, 85, 529, 95, 119, 143, 46, 69, 133, 28, 115, 187, 841, 32, 125, 161, 209, 221, 58, 961, 45, 87, 247, 62, 93, 145, 253, 24, 155, 203, 299, 323, 1369
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OFFSET
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2,1
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COMMENTS
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In an aliquot sequence, all numbers in row n can be predecessors of n. This sequence is a permutation of the composite numbers; number k appears in row A001065(k). We start with n=2 because every prime would be in row 1. Note that row 2 is empty -- as are all the rows listed in A005114. Row n contains A048138(n) numbers. When n is prime, the largest number in row n+1 is n^2. When n>7 is odd, the largest number in row n is less than ((n-1)/2)^2 and (if a strong form of the Goldbach conjecture is true) has the form pq, with primes p<q and p+q=n-1.
The first row with several terms is row(6), where the difference between extreme terms is 25-6=19. The next row with a smaller difference is row(13) with a difference 35-27=8. And the next one is row(454) with a difference 602-596=6. Is there a next row with a smaller difference? - Michel Marcus, Nov 11 2014
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LINKS
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EXAMPLE
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Irregular triangle starts:
; (empty row at n=2)
4;
9;
; (empty row at n=5)
6, 25;
8;
10, 49;
15;
14;
21;
121;
27, 35;
22, 169;
16, 33;
12, 26;
39, 55;
289;
...
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MAPLE
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N:= 100: # for rows 2 to N, flattened
for s from 2 to N do B[s]:= NULL od:
for k from 1 to N^2 do
if not isprime(k) then
s:= numtheory:-sigma(k)-k;
if s <= N then
B[s]:= B[s], k;
fi
fi
od:
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MATHEMATICA
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nn=100; s=Table[{}, {nn}]; Do[k=DivisorSigma[1, n]-n; If[1<k<=nn, AppendTo[s[[k]], n]], {n, nn^2}]; Flatten[s]
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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