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%I #19 Jan 11 2017 03:15:36
%S -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,
%T 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,
%U 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
%N a(n) = A076649(n) - A055642(n).
%C a(1) is the only negative term in this sequence. - _Ely Golden_, Jan 10 2017
%C a(n) = 0 if and only if n is a member of A109608. - _Ely Golden_, Jan 10 2017
%H Ely Golden, <a href="/A280827/b280827.txt">Table of n, a(n) for n = 1..10000</a>
%H Ely Golden, <a href="/A280827/a280827.txt">Proof that a(n)>=0 for all n>1</a>
%e a(10) = 0, as 2*5 have 2 digits total, and 10 has 2 digits. Thus a(10) = 2-2 = 0.
%e 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.
%o (SageMath)
%o def digits(x, n):
%o if(x<=0|n<2):
%o return []
%o li=[]
%o while(x>0):
%o d=divmod(x, n)
%o li.insert(0,d[1])
%o x=d[0]
%o return li;
%o def factorDigits(x, n):
%o if(x<=0|n<2):
%o return []
%o li=[]
%o f=list(factor(x))
%o for c in range(len(f)):
%o for d in range(f[c][1]):
%o ld=digits(f[c][0], n)
%o li+=ld
%o return li;
%o def digitDiff(x,n):
%o return len(factorDigits(x,n))-len(digits(x,n))
%o radix=10
%o index=1
%o while(index<=10000):
%o print(str(index)+" "+str(digitDiff(index,radix)))
%o index+=1
%Y Cf. A109608, A076649.
%K sign,base,easy
%O 1,8
%A _Ely Golden_, Jan 08 2017
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