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A237719
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Numbers n such that k(n) = (n(n+1)/2 mod n) = (antisigma(n) mod n) + (sigma(n) mod n).
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1
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1, 2, 6, 12, 18, 20, 24, 28, 30, 40, 42, 54, 56, 66, 70, 78, 80, 88, 100, 102, 104, 112, 114, 120, 126, 138, 140, 150, 160, 162, 174, 176, 180, 186, 196, 198, 200, 204, 208, 220, 222, 224, 228, 234, 240, 246, 258, 260, 272, 276, 282, 294, 304, 306, 308, 318, 320
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OFFSET
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1,2
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COMMENTS
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k(n) = 0 for odd n, k(n) = n/2 for even n.
If there are any odd multiply-perfect numbers, they are members of this sequence.
If there is no odd multiply-perfect number, then:
(1) the only odd number in this sequence is 1,
(2) corresponding sequence of numbers k(n): {0; a(n) / 2 for n > 1}.
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LINKS
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EXAMPLE
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12 is in the sequence because k(12) = (12*(12+1)/2) mod 12 = antisigma(12) mod 12 + sigma(12) mod 12; k(12) = 6 = 4 + 2 = n/2.
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PROG
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(Magma) [n: n in [1..320] | IsZero(n*(n+1)div 2 mod n - SumOfDivisors(n) mod n - (n*(n+1)div 2-SumOfDivisors(n)) mod n)]
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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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