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 A242411 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)). 8
 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, 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, 7, 1, 0, 1, 0, 11, 1, 7, 1, 4, 0, 2, 0, 12, 0, 2, 4, 13, 8, 4, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,10 LINKS Antti Karttunen, Table of n, a(n) for n = 1..10000 FORMULA If A001221(n) = 1, then a(n) = 0, otherwise a(n) = A241919(n) = A061395(n) - A061395(A051119(n)). PROG (Scheme) (define (A242411 n) (if (= 1 (A001221 n)) 0 (- (A061395 n) (A061395 (A051119 n))))) (Haskell) a242411 1 = 0 a242411 n = i - j where             (i:j:_) = map a049084 \$ ps ++ [p]             ps@(p:_) = reverse \$ a027748_row n -- Reinhard Zumkeller, May 15 2014 (Python) from sympy import factorint, primefactors, primepi def a061395(n): return 0 if n==1 else primepi(primefactors(n)[-1]) def a053585(n):     if n==1: return 1     p = primefactors(n)[-1]     return p**factorint(n)[p] def a051119(n): return n/a053585(n) def a(n): return 0 if n==1 or len(primefactors(n))==1 else a061395(n) - a061395(a051119(n)) # Indranil Ghosh, May 19 2017 CROSSREFS Cf. A000961 (positions of zeros). Cf. A241917, A241919, A051119, A061395. Cf. A049084, A027748. Sequence in context: A175790 A124305 A087073 * A286470 A243055 A245151 Adjacent sequences:  A242408 A242409 A242410 * A242412 A242413 A242414 KEYWORD nonn AUTHOR Antti Karttunen, May 13 2014 STATUS approved

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