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a(n) = A008578(n+2) - A008578(n+1).
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%I #24 Dec 12 2021 16:06:43

%S 1,1,2,2,4,2,4,2,4,6,2,6,4,2,4,6,6,2,6,4,2,6,4,6,8,4,2,4,2,4,14,4,6,2,

%T 10,2,6,6,4,6,6,2,10,2,4,2,12,12,4,2,4,6,2,10,6,6,6,2,6,4,2,10,14,4,2,

%U 4,14,6,10,2,4,6,8,6,6,4,6,8,4,8,10,2,10,2,6,4,6,8,4,2,4,12,8,4,8,4,6

%N a(n) = A008578(n+2) - A008578(n+1).

%C n appears this number of times in A000720. - _Lekraj Beedassy_, Jun 19 2006

%C a(0) = 1, for n >= 1: a(n) = differences between consecutive primes (A001223(n)) = A158611(n+2) - A158611(n+1). Partial sums give A006093 (shifted). - _Jaroslav Krizek_, Aug 04 2009

%C First differences of noncomposite numbers. - _Juri-Stepan Gerasimov_, Feb 17 2010

%C This is 1 together with A001223. A054541 is 2 together with A001223. A125266 is 3 together with A001223. - _Omar E. Pol_, Nov 01 2013

%F a(n) = A001223(n) for n>0.

%t Prime[Range[100]] // Differences // Prepend[#, 1]& (* _Jean-François Alcover_, Dec 11 2021 *)

%Y Cf. A000040, A001223, A075527, A158611.

%K nonn

%O 0,3

%A _Reinhard Zumkeller_, Sep 22 2002

%E Correction for change of offset in A158611 and A008578 in Aug 2009 _Jaroslav Krizek_, Jan 27 2010