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%I #20 May 20 2017 10:22:39
%S 0,0,0,0,0,1,0,0,0,2,0,1,0,3,1,0,0,1,0,2,2,4,0,1,0,5,0,3,0,1,0,0,3,6,
%T 1,1,0,7,4,2,0,2,0,4,1,8,0,1,0,2,5,5,0,1,2,3,6,9,0,1,0,10,2,0,3,3,0,6,
%U 7,1,0,1,0,11,1,7,1,4,0,2,0,12,0,2,4,13,8,4,0
%N If n is a prime power, p_i^e, a(n) = 0, otherwise difference (i-j) of the indices of the two largest distinct primes p_i, p_j, i > j in the prime factorization of n: a(n) = A061395(n) - A061395(A051119(n)).
%H Antti Karttunen, <a href="/A242411/b242411.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F If A001221(n) = 1, then a(n) = 0, otherwise a(n) = A241919(n) = A061395(n) - A061395(A051119(n)).
%o (Scheme) (define (A242411 n) (if (= 1 (A001221 n)) 0 (- (A061395 n) (A061395 (A051119 n)))))
%o (Haskell)
%o a242411 1 = 0
%o a242411 n = i - j where
%o (i:j:_) = map a049084 $ ps ++ [p]
%o ps@(p:_) = reverse $ a027748_row n
%o -- _Reinhard Zumkeller_, May 15 2014
%o (Python)
%o from sympy import factorint, primefactors, primepi
%o def a061395(n): return 0 if n==1 else primepi(primefactors(n)[-1])
%o def a053585(n):
%o if n==1: return 1
%o p = primefactors(n)[-1]
%o return p**factorint(n)[p]
%o def a051119(n): return n/a053585(n)
%o def a(n): return 0 if n==1 or len(primefactors(n))==1 else a061395(n) - a061395(a051119(n)) # _Indranil Ghosh_, May 19 2017
%Y Cf. A000961 (positions of zeros).
%Y Cf. A241917, A241919, A051119, A061395.
%Y Cf. A049084, A027748.
%K nonn
%O 1,10
%A _Antti Karttunen_, May 13 2014
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