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A343824
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Sum of the elements in all pairs (d1, d2) of divisors of n such that d1<=d2, d1|n, d2|n, and d1 + d2 <= n.
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1
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0, 2, 2, 9, 2, 24, 2, 28, 12, 32, 2, 96, 2, 40, 36, 75, 2, 126, 2, 132, 44, 56, 2, 288, 18, 64, 52, 168, 2, 336, 2, 186, 60, 80, 52, 495, 2, 88, 68, 400, 2, 432, 2, 240, 198, 104, 2, 760, 24, 258, 84, 276, 2, 528, 68, 512, 92, 128, 2, 1296, 2, 136, 246, 441, 76, 624, 2, 348
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OFFSET
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1,2
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COMMENTS
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If n is prime, then a(n) = 2.
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LINKS
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FORMULA
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a(n) = Sum_{k=1..floor(n/2)} Sum_{i=1..k} c(n/k) * c(n/i) * (i+k), where c(n) = 1 - ceiling(n) + floor(n).
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EXAMPLE
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a(7) = 2; There is one divisor pair of 7 whose sum is less than or equal to 7: (1,1). The sum is then 1+1 = 2.
a(9) = 12; The divisor pairs of 9 whose sum is less than or equal to 9 are: (1,1), (1,3) and (3,3). The sum of the coordinates is then (1+1) + (1+3) + (3+3) = 12.
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MATHEMATICA
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Table[Sum[Sum[(i + k) (1 - Ceiling[n/k] + Floor[n/k]) (1 - Ceiling[n/i] + Floor[n/i]), {i, k}], {k, Floor[n/2]}], {n, 80}]
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PROG
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(PARI) a(n) = sumdiv(n, d1, sumdiv(n, d2, if ((d1 <= d2) && (d1+d2 <= n), d1+d2))); \\ Michel Marcus, May 01 2021
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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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