A230492 Numbers such that the sequence of all possible sums of divisors of n is increasing but not strictly so, the sums being ordered by their characteristic functions, seen as binary numbers (see example).

6, 28, 117, 475, 496, 775, 2009, 8128, 13079, 13189, 14663, 16211, 23903, 26675, 30503, 35425, 37271, 41123, 43681, 44591, 46163, 47519, 55991, 59831, 63971, 66263, 66785, 73511, 76751, 78319, 81923, 88559, 88723, 107423, 112631, 127571, 130271, 140825

Offset: 1

Author

Michel Marcus, Oct 20 2013

Keywords

• nonn

Comments

Consider the sums obtained by adding up divisors of n, ordered by the decimal value of the binary representation of the divisors used for a particular sum. Three cases can occur.

On one hand, the sums that are obtained may be strictly increasing. For instance, with n=3, divisors are (1,3), sums are 1,3,4. It appears that this case might give distended numbers (A051772).

On the other hand, the sequence of sums may not be monotonic. For instance, with n=12, divisors are (1,2,3,4,6,12) and the sums are 1,2,3,3,4,5,6,4,...

In between, there is a third case, this sequence, in which the sums are increasing but not strictly so (see example for n=6). It appears that perfect numbers (A000396) belong to this sequence, and when not perfect, then terms of the sequence are odd.

More information and proofs can be found in the SeqFan thread, "Questions on A230492", see links below. - Michel Marcus, Dec 21 2013

Examples

			For n=6, the divisors of 6 are (1,2,3,6) and the sums of distinct divisors are:
0001:             1 = 1
0010:         2     = 2
0011:         2 + 1 = 3
0100:     3         = 3
0101:     3     + 1 = 4
0110:     3 + 2     = 5
0111:     3 + 2 + 1 = 6
1000: 6             = 6
1001: 6 +         1 = 7
1010: 6 +     2     = 8
1011: 6 +     2 + 1 = 9
1100: 6 + 3         = 9
1101: 6 + 3     + 1 = 10
1110: 6 + 3 + 2     = 11
1111: 6 + 3 + 2 + 1 = 12
The numbers in the right column are increasing but not strictly so, thus 6 belongs to the sequence.
Missing row 1000 added following remark by _Vladimir Shevelev_. - _Michel Marcus_, Dec 20 2013
		

Links

• Michel Marcus, Table of n, a(n) for n = 1..250
• Jack Brennen, Re: Questions on A230492, Dec 20 2013
• Donovan Johnson, Re: Questions on A230492, Dec 20 2013
• Vladimir Shevelev, Re: Questions on A230492, Dec 20 2013

Crossrefs

Cf. A030057.

Programs

Mathematica

inOrderQ[n_] := Module[{d, len, hasZero, last, b, p}, d = Reverse[Divisors[n]]; len = Length[d]; hasZero = False; last = 1;b = 2; While[p = Inner[Times, d, IntegerDigits[b, 2, len], Plus]; If[p == last, hasZero = True]; p >= last && b < 2^len - 1, b++; last = p]; hasZero && p >= last && b == 2^len - 1]; Select[Range[2, 150000], inOrderQ] (* T. D. Noe, Oct 23 2013 *)

PARI

padbin(n, len) = {b = binary(n); while(length(b) < len, b = concat(0, b);); b;}
tds(n) = {/* returns -1, if sums go up and down; returns 0 if sums are increasing but not strictly ; returns 1 if sums are strictly increasing */ divs = divisors(n); nbdivs = #divs; rdivs = vector(nbdivs, i, divs[nbdivs-i+1]); nb = 2^nbdivs-1; iseq = 0; precs = 0 ; for (i=1, nb, vb = padbin(i, nbdivs); nexts = sum(j=1, nbdivs, rdivs[j]*vb[j]); diff = nexts - precs; if (diff < 0, return (-1)); if (diff == 0, iseq = 1); precs = nexts; ); return (1 - iseq);}
isok(n) = tds(n) == 0; \\ Michel Marcus, Oct 20 2013

PARI

divsums(n) = {/* returns vector of ordered sums of divisors */ divs = divisors(n); nbdivs = #divs; rdivs = vector(nbdivs, i, divs[nbdivs-i+1]); nb = 2^nbdivs-1; vsd = vector(nb); for (i=1, nb, vb = padbin(i, nbdivs); vsd[i] = sum(j=1, nbdivs, rdivs[j]*vb[j]);); vsd;} \\ Michel Marcus, Oct 20 2013

PARI

is_A230492(n)={my(s=0,t,v,ok=0); for(i=2, 2^(#n=vecextract( divisors(n),"^1"))-1, s+1>( t=sum(j=1, #v=vecextract(n,i), v[j])) && return; s+1==(s=t) && ok=1);ok} \\ M. F. Hasler, Oct 23 2013

Extensions

Values double-checked by M. F. Hasler, Oct 23 2013

b-file extended by Michel Marcus, Dec 18 2013