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A241915 After a(1)=1, numbers 1 .. A061395(n), followed by numbers 1 .. A061395(n+1), etc.

%I A241915 #7 May 15 2014 10:21:34
%S A241915 1,1,1,2,1,1,2,3,1,2,1,2,3,4,1,1,2,1,2,3,1,2,3,4,5,1,2,1,2,3,4,5,6,1,
%T A241915 2,3,4,1,2,3,1,1,2,3,4,5,6,7,1,2,1,2,3,4,5,6,7,8,1,2,3,1,2,3,4,1,2,3,
%U A241915 4,5,1,2,3,4,5,6,7,8,9,1,2,1,2,3,1,2,3,4,5,6,1,2,1,2,3,4,1,2,3,4,5,6,7,8,9,10,1,2,3,1,2,3,4,5,6,7,8,9,10,11,1
%N A241915 After a(1)=1, numbers 1 .. A061395(n), followed by numbers 1 .. A061395(n+1), etc.
%H A241915 Antti Karttunen, <a href="/A241915/b241915.txt">Table of n, a(n) for n = 1..10082</a>
%F A241915 a(1)=1, a(n) = n - A203623(A241920(n)-1) - 1.
%e A241915 Viewed as an irregular table, the sequence is constructed as:
%e A241915 "Row"
%e A241915   [1] 1; (by convention, a(1)=1)
%e A241915   [2] 1; (because A061395(2)=1 (the index of the largest prime factor), we have here terms from 1 to 1)
%e A241915   [3] 1, 2; (because A061395(3)=2, we have terms from 1 to 2)
%e A241915   [4] 1;
%e A241915   [5] 1, 2, 3; (because A061395(5)=3, we have terms from 1 to 3)
%e A241915   [6] 1, 2;    (because A061395(6)=2, we have terms from 1 to 2)
%e A241915   [7] 1, 2, 3, 4; (because A061395(7)=4, we have terms from 1 to 4)
%e A241915 etc.
%o A241915 (Scheme)
%o A241915 (define (A241915 n) (if (= n 1) 1 (- n (A203623 (- (A241920 n) 1)) 1)))
%Y A241915 One more than A241914.
%Y A241915 Cf. A203623, A241920, A241911.
%K A241915 nonn,tabf
%O A241915 1,4
%A A241915 _Antti Karttunen_, May 01 2014