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A387654
Numbers k such that each of k and k+1 is either a practical number (A005153) or an almost practical number (A174533).
2
1, 4095, 5775, 5984, 11024, 21735, 21944, 26144, 39375, 49664, 58695, 61424, 69615, 76544, 79695, 89775, 91664, 98175, 104895, 106784, 111824, 116655, 116864, 121904, 134504, 135135, 144584, 152775, 155295, 160544, 165375, 170624, 174824, 180495, 185535, 192464
OFFSET
1,2
COMMENTS
Differs from A103289 by not having the terms 7424, 27404, 43064, 56924, 70784, ... . The first 344 terms of this sequence are in A103289. Is this sequence a subsequence of A103289?
Differs from A096399 by not having the terms 7424, 27404, 43064, 56924, 70784, ... . The first 342 terms after 1 and 4095 are in A096399. Is this sequence \ {1, 4095} a subsequence of A096399?
Terms k such that both k and k+1 are almost practical numbers are in A387653.
The only pair of consecutive integers that are both practical is 1 and 2, since 1 is the only odd practical number.
All the rest are pairs in which one member (the odd member) is almost practical and the second member (the even member) is practical.
Are there 3 consecutive numbers that are all either practical or almost practical? There are none below 2.8*10^6.
From David A. Corneth, Sep 06 2025: (Start)
There are no 3 consecutive numbers that are all either practical or almost practical.
Some definitions and properties:
Odd almost practical numbers are divisible by 15.
Practical numbers > 1 are even.
Almost practical numbers are not 5-rough (numbers with no prime factors < 5 (Cf. A007310)).
Proof:
If the middle number is even then the adjacent numbers are odd hence cannot be practical. At most one of them is a multiple of 3 so the other is 5-rough hence not almost practical. A contradiction.
If the middle number is odd then it must be almost practical hence divisible by 3 and by 5. The adjacent numbers are even.
One of them has multiplicity of 2 in prime factorization of 1. That number cannot be divisible by 3 or 5.
Therefore 4, 5 and 6 cannot be written as sums of distinct divisors of that number.
Thus the middle number cannot be odd and cannot be even hence it does not exist. (End)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000 (first 344 terms from Amiram Eldar)
MATHEMATICA
q[n_] := q[n] = Module[{d = Divisors[n], c, x}, c = CoefficientList[Product[1 + x^i, {i, d}], x]; MemberQ[{0, 2}, Total[d] - Count[Rest[c], _?(# > 0 &)]]];
Select[Range[40000], q[#] && q[#+1] &] (* warning: a slow program *)
PROG
(PARI) isp(k) = {my(d = divisors(k), nd = #d, s = vecsum(d), p = prod(i = 1, nd, 1 + 'x^d[i])); abs(#select(x -> x > 0, Col(p)) - s) == 1; }
list(kmax) = {my(is1 = 1, is2); for(k = 2, kmax, is2 = isp(k); if(is1 && is2, print1(k-1, ", ")); is1 = is2); } \\ warning: a slow program
CROSSREFS
A387653 is subsequence.
Sequence in context: A024022 A145591 A103289 * A038463 A359084 A043452
KEYWORD
nonn
AUTHOR
Amiram Eldar, Sep 05 2025
STATUS
approved