A214621 Numbers n such that at least one other integer m exists with the same smallest and same largest prime factors, and same multisets of decimal and binary digits as n.

1080, 1260, 1800, 2016, 2673, 3024, 3267, 3402, 4032, 4500, 4653, 4950, 5346, 5400, 5670, 5757, 5940, 6048, 6345, 6534, 6804, 7056, 7560, 7575, 8064, 11084, 11542, 12654, 12915, 13026, 13068, 13260, 13860, 14018, 14490, 14652, 14904, 15124, 15129, 16032, 16320

Offset

1,1

Comments

Decimal digits of m are a permutation of decimal digits of n,

binary digits of m are a permutation of binary digits of a.

Conjecture: there is X such that among integers bigger than X more than 50% are in the sequence.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Example

1080 and 1800 have the same set of decimal digits, same set of binary digits (10000111000 versus 11100001000), same smallest prime factor 2, and same largest prime factor 5.

Maple

Res:= {}:
for n from 1 to 2^16-1 do
  f:= numtheory:-factorset(n);
  v:= [min(f), max(f), ilog2(n), convert(convert(n, base, 2), `+`), sort(convert(n, base, 10))];
  if assigned(R[v]) then
     Res:= Res union {n, R[v]}
  else
     R[v]:= n
  fi
od:
sort(convert(Res, list)); # Robert Israel, Sep 28 2018

Prog

(Python)
# primes = [2...99991]
# ~15 minutes
TOP = 100000
smallest = [0]*TOP
largest  = [0]*TOP
decimal = [0]*TOP
binary  = [0]*TOP
flags = [0]*TOP
for n in range(1, TOP):
    curSm = curLa = curDec = curBin = 0
    t = b = d = n
    while b:
        curBin += 1000**( b&1 )
        b /= 2
    while d:
        curDec += 10**( d%10 )
        d /= 10
    for p in primes:
        if t%p==0:
            if curSm==0:
                curSm = p
            curLa = p
            t/=p
            while t%p==0:
                t/=p
            if t==1:
                break
    binary[n] = curBin
    decimal[n] = curDec
    smallest[n] = curSm
    largest[n]  = curLa
    for k in range(1, n):
        if smallest[k]==curSm and largest[k]==curLa:
            if decimal[k]==curDec and binary[k]==curBin:
                flags[k]+=1
                flags[n]+=1
for n in range(1, TOP):
    if flags[n]>0:
        print n,

Crossrefs

Cf. A214619, A214620.

Sequence in context: A023085 A351672 A163562 * A179688 A159210 A250538

Adjacent sequences: A214618 A214619 A214620 * A214622 A214623 A214624

Keyword

nonn,base

Author

Alex Ratushnyak, Jul 23 2012

Status

approved