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A130664 a(1)=1. a(n) = a(n-1) + (number of terms from among a(1) through a(n-1) which are factorials).

%I A130664 #31 Feb 19 2025 16:12:13
%S A130664 1,2,4,6,9,12,15,18,21,24,28,32,36,40,44,48,52,56,60,64,68,72,76,80,
%T A130664 84,88,92,96,100,104,108,112,116,120,125,130,135,140,145,150,155,160,
%U A130664 165,170,175,180,185,190,195,200,205,210,215,220,225,230,235,240,245,250
%N A130664 a(1)=1. a(n) = a(n-1) + (number of terms from among a(1) through a(n-1) which are factorials).
%C A130664 Also this is an irregular array where row n contains the n! consecutive multiples of n starting with n!.
%C A130664 For n >= 1, (a(n), A084555(n)) = (1,1), (2,3), (4,5), (6,8), (9,11), (12,14), ... gives the starting and ending offsets of the n-th permutation in the sequences like A030298 and A030496. Gives also the fixed points of A220662; we have A220662(a(n)) = a(n). - _Antti Karttunen_, Dec 18 2012
%H A130664 Antti Karttunen, <a href="/A130664/b130664.txt">Rows 1..7 of irregular table, flattened.</a>
%F A130664 a(n) = A084555(n-1) + 1.
%e A130664 When interpreted as an irregular table, the rows begin as:
%e A130664   1;
%e A130664   2, 4;
%e A130664   6, 9, 12, 15, 18, 21;
%p A130664 A[1]:= 1:
%p A130664 nextf:= 2!:
%p A130664 m:= 1:
%p A130664 for n from 2 to 100 do
%p A130664   A[n]:= A[n-1]+m;
%p A130664   if A[n] = nextf then
%p A130664      m:= m+1;
%p A130664       nextf:= (m+1)!;
%p A130664   fi;
%p A130664 od:
%p A130664 seq(A[i],i=1..100); # _Robert Israel_, Apr 28 2016
%t A130664 Table[Range[n!, (n + 1)! - 1, n], {n, 5}] // Flatten (* _Michael De Vlieger_, Aug 29 2017 *)
%o A130664 (Scheme) (define (A130664 n) (+ 1 (A084555(- n 1))))
%Y A130664 Cf. A000142, A084555, A220662.
%K A130664 easy,nonn
%O A130664 1,2
%A A130664 _Leroy Quet_, Jun 21 2007