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

Author

Ely Golden, Jan 08 2017

Keywords

• sign
• base
• easy

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A109608, A076649.

Programs

SageMath

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