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Count up to n, n times.
8

%I #25 Nov 05 2024 12:17:58

%S 1,1,2,1,2,1,2,3,1,2,3,1,2,3,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4,

%T 5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,6,1,2,3,4,5,6,1,

%U 2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,1,2,3,4,5,6,7,1,2,3,4,5,6,7

%N Count up to n, n times.

%C The n-th block consists of n subblocks, each of which counts from 1 to n.

%C This a fractal sequence: removing the first instance of each value leaves the original sequence.

%C The first comment implies that this gives the column index of the n-th element of a sequence whose terms are coefficients, read by rows, of a sequence of matrices of size 1 X 1, 2 X 2, 3 X 3, etc.; cf. example. The row index is given by A238013(n), and the size of the matrix by A074279(n). - _M. F. Hasler_, Feb 16 2014

%e Sequence begins:

%e 1;

%e 1,2;

%e 1,2;

%e 1,2,3;

%e 1,2,3;

%e 1,2,3;

%e ...

%e The blocks of n subblocks of n terms (n=1,2,3,...) can be cast into a square matrices of order n; then the terms are equal to the index of the column they fall into.

%o (PARI) A121997(N=9)=concat(vector(N,i,concat(vector(i,j,vector(i,k,k))))) \\ Note: this creates a vector; use A121997()[n] to get the n-th term. - _M. F. Hasler_, Feb 16 2014

%o (Python)

%o from sympy import integer_nthroot

%o def A121997(n): return 1+(n-(k:=(m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1)))*(k-1)*((k<<1)-1)//6-1)%k # _Chai Wah Wu_, Nov 04 2024

%Y Cf. A081489 (locations of new values), A075349 (locations of 1's).

%Y Cf. A000290 (row lengths), A002411 (row sums), A036740 (row products).

%Y Cf. A002024 and references there, esp. in PROG section.

%Y Cf. A238013.

%K easy,nonn,tabf,changed

%O 1,3

%A _Franklin T. Adams-Watters_, Sep 11 2006