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A290582
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Numbers m > 2 such that every divisor > 2 is the sum of two or more consecutive divisors.
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0
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6, 18, 54, 66, 162, 486, 726, 1458, 4374, 7986, 13122, 39366, 87846, 118098, 354294, 530226, 966306, 1062882, 3188646, 9565938, 10629366, 28697814, 43035786, 86093442, 116923026, 258280326, 578476566, 774840978, 1286153286, 2324522934, 6973568802, 14147686146
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OFFSET
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1,1
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COMMENTS
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a(n) is even because the divisors are {d(1), d(2), d(3), ...} with d(1) = 1, and if d(2) is odd then d(3) = d(2) + 1 is even, a contradiction. Therefore d(2) = 2 and d(3) = 3.
a(n) == 0 (mod 6).
The sequence is infinite because the numbers of the form 2*3^i (A025192) and the numbers of the form 6*11^i for i >= 1 are in the sequence.
The divisors of 2*3^i are {d(1), d(2), d(3), ...} = {1, 2, 3, 6, 9, 18, 27, 54, ...} where d(1 + 2i) = 3^i for i >= 0 and d(2i) = 2*3^(i-1) for i >= 1.
The divisors of the form 6*11^k are {d(1), d(2), d(3), ...} = {1, 2, 3, 6, 11, 22, 33, 66, 121, ...} where d(i + 4j) = i*11^j for i >= 1 and j >= 0 and where d(4j) = 6*11^(j-1) for j >= 1.
No term can be divisible by 4, 5, 7, 9, 13, 17, or 19. Up to 5*10^11, the only terms which are divisible by a prime > 11 are 530226 = 2*3^5*1091, 43035786 = 2*3^7*9839, 578476566 = 2*3^5*1091^2, and 2*3^7*9839^2. Larger such terms are 2*3*11^12*6904542428779 and 2*3^29*308836698141971. - Giovanni Resta, Aug 07 2017
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LINKS
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EXAMPLE
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66 is in the sequence because the divisors are {1, 2, 3, 6, 11, 22, 33, 66} and:
3 = 2 + 1;
6 = 3 + 2 + 1;
11 = 6 + 3 + 2;
22 = 11 + 6 + 3 + 2;
33 = 22 + 11;
66 = 33 + 22 + 11.
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MAPLE
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with(numtheory):nn:=10^5:
for n from 4 to nn do:
d:=divisors(n):n1:=nops(d):it:=0:
for k from 3 to n1 do:
for j from 1 to k-1 do:
s:=sum('d[k-i]', 'i'=1..j):
if s=d[k]
then
it:=it+1:
else
fi:
od:
od:
if n1>2 and it = n1-2
then
printf(`%d, `, n):
else
fi:
od:
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MATHEMATICA
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Select[Range[3, 10^5], Function[d, Function[t, AllTrue[ TakeWhile[ Reverse@ d, # > 2 &], MemberQ[t, #] &]]@ Union@ Flatten@ Array[Total /@ Partition[d, #, 1] &, Length@ d - 1, 2]]@ Divisors@ # &] (* Michael De Vlieger, Aug 07 2017 *)
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PROG
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(PARI) isokds(k, v) = {vsmall = select(x->(x < k), v); for (i=1, #vsmall, s = v[i]; if (s > k, break); for (j=i+1, #vsmall, s += vsmall[j]; if (s > k, break, if (k == s, return(1))); ); ); return (0); }
isok(n) = {if (n>2, my(d = divisors(n)); for (i=1, #d, if (d[i] > 2, if (! isokds(d[i], d), return (0)); ); ); return(1); )} \\ Michel Marcus, Aug 07 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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