A109608 -id:A109608 - cp's OEIS frontend

A280827 a(n) = A076649(n) - A055642(n).

-1, 0, 0, 1, 0, 1, 0, 2, 1, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 1, 0, 1, 0, 2, 0, 1, 1, 1, 0, 1, 0, 3, 1, 1, 0, 2, 0, 1, 1, 2, 0, 1, 0, 2, 1, 1, 0, 3, 0, 1, 1, 2, 0, 2, 1, 2, 1, 1, 0, 2, 0, 1, 1, 4, 1, 2, 0, 2, 1, 1, 0, 3, 0, 1, 1, 2, 1, 2, 0, 3, 2, 1, 0, 2, 1, 1, 1, 3, 0, 2, 1, 2, 1, 1, 1, 4, 0, 1, 2, 1

Offset: 1

Ely Golden, Jan 08 2017

a(1) is the only negative term in this sequence. - Ely Golden, Jan 10 2017

a(n) = 0 if and only if n is a member of A109608. - Ely Golden, Jan 10 2017

			a(10) = 0, as 2*5 have 2 digits total, and 10 has 2 digits. Thus a(10) = 2-2 = 0.
a(1) is defined to be -1, as the empty product has 0 digits, and 1 has 1 digit. Thus a(1) = 0-1 = -1.

Ely Golden, Table of n, a(n) for n = 1..10000
Ely Golden, Proof that a(n)>=0 for all n>1

Cf. A109608, A076649.

def digits(x, n):
    if(x<=0|n<2):
        return []
    li=[]
    while(x>0):
        d=divmod(x, n)
        li.insert(0,d[1])
        x=d[0]
    return li;
def factorDigits(x, n):
    if(x<=0|n<2):
        return []
    li=[]
    f=list(factor(x))
    for c in range(len(f)):
        for d in range(f[c][1]):
            ld=digits(f[c][0], n)
            li+=ld
    return li;
def digitDiff(x,n):
    return len(factorDigits(x,n))-len(digits(x,n))
radix=10
index=1
while(index<=10000):
    print(str(index)+" "+str(digitDiff(index,radix)))
    index+=1

A348306 List of Agathokakological Numbers "k": string of digits of the juxtaposition of the prime factors of k has the same length as k but these digits do not appear in k.

Original entry on oeis.org

10, 14, 21, 49, 106, 111, 118, 129, 134, 146, 158, 161, 166, 177, 201, 219, 249, 259, 267, 329, 343, 413, 511, 553, 623, 1011, 1029, 1046, 1077, 1081, 1101, 1106, 1114, 1119, 1138, 1149, 1167, 1186, 1227, 1299, 1318, 1354, 1358, 1363, 1418, 1454, 1466, 1538, 1541, 1546, 1561, 1589, 1591

Offset: 1

Samuel Harkness, Oct 11 2021

Theorem: (See PDF "PROOFS" in Links)
Of Agathokakological Numbers k,
No k have a leading 9.
No k end in 2 or 5.
10 is the only k to end in 0. It is also the only k with 5 as a prime factor.
Can only be square terms when k is of the order 10^m where m is odd.
For k written as a*10^m, k can only be even when 1<=a<1.888...
Empirical observation: When graphed with the log of the n-th term on x axis and the log of the n-th term's value on the y axis a pattern appears with a similar shape for each new power of ten (see figure "LogLogGraph" in Links)
Special cases 28651 = 7*4093 and 65821 = 7*9043 use all digits 0-9 once.
"Agathokakological" is a Greek word meaning "composed of both good and evil." (Merriam-Webster) The composition (prime factorization) of Agathokakological Numbers is both good (same length) and evil (no common digits).

			158 = 2 * 79 since {2,7,9} do not appear in {1,5,8} and both have 3 digits.

Samuel Harkness, Table of n, a(n) for n = 1..6388
Samuel Harkness, MATLAB
Samuel Harkness, LogLogGraph
Samuel Harkness, PROOFS

Cf. A055642, A076649, A280928.
Intersection of A035139 and A109608.
Subsequence of A047201 from n=2.

q[n_] := Module[{d = IntegerDigits[n], f = FactorInteger[n]}, Length[d] == Plus @@ ((Last[#]*IntegerLength[First[#]]) & /@ f ) && Intersection[d, Join @@ IntegerDigits[f[[;; , 1]]]] == {}]; Select[Range[1600], q] (* Amiram Eldar, Oct 12 2021 *)

digsf(n) = my(f=factor(n), list=List()); for (k=1, #f~, my(dk=digits(f[k,1])); for (i=1, f[k,2], for (j=1, #dk, listput(list, dk[j])))); Vec(list);
isokd(m) = my(df=digsf(m), d=digits(m)); (#df == #d) && (#setintersect(Set(df), Set(d)) == 0); \\ Michel Marcus, Oct 11 2021

from sympy import factorint
def ok(n):
    s, f = str(n), factorint(n)
    pfd = set("".join(str(p) for p in f))
    if set(s) & pfd != set(): return False
    return len(s) == sum(len(str(p))*f[p] for p in f)
print(list(filter(ok, range(1601)))) # Michael S. Branicky, Oct 11 2021

cp's OEIS Frontend

A280827 a(n) = A076649(n) - A055642(n).

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