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%I #33 Sep 12 2017 21:32:28
%S 6,18,54,66,162,486,726,1458,4374,7986,13122,39366,87846,118098,
%T 354294,530226,966306,1062882,3188646,9565938,10629366,28697814,
%U 43035786,86093442,116923026,258280326,578476566,774840978,1286153286,2324522934,6973568802,14147686146
%N Numbers m > 2 such that every divisor > 2 is the sum of two or more consecutive divisors.
%C 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.
%C a(n) == 0 (mod 6).
%C 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.
%C 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.
%C 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.
%C 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
%e 66 is in the sequence because the divisors are {1, 2, 3, 6, 11, 22, 33, 66} and:
%e 3 = 2 + 1;
%e 6 = 3 + 2 + 1;
%e 11 = 6 + 3 + 2;
%e 22 = 11 + 6 + 3 + 2;
%e 33 = 22 + 11;
%e 66 = 33 + 22 + 11.
%p with(numtheory):nn:=10^5:
%p for n from 4 to nn do:
%p d:=divisors(n):n1:=nops(d):it:=0:
%p for k from 3 to n1 do:
%p for j from 1 to k-1 do:
%p s:=sum('d[k-i]', 'i'=1..j):
%p if s=d[k]
%p then
%p it:=it+1:
%p else
%p fi:
%p od:
%p od:
%p if n1>2 and it = n1-2
%p then
%p printf(`%d, `,n):
%p else
%p fi:
%p od:
%t 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 *)
%o (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);}
%o 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
%Y Cf. A008776, A025192.
%K nonn
%O 1,1
%A _Michel Lagneau_, Aug 07 2017
%E Name corrected by _Jon E. Schoenfield_, Sep 11 2017
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