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A256251
First differences of A256250.
4
1, 4, 4, 12, 4, 12, 20, 28, 4, 12, 20, 28, 36, 44, 52, 60, 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108, 116, 124, 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100, 108, 116, 124, 132, 140, 148, 156, 164, 172, 180, 188, 196, 204, 212, 220, 228, 236, 244, 252, 4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, 100
OFFSET
0,2
COMMENTS
Number of cells turned ON at n-th stage in the structure of A256250.
Apart from the initial 1, four times A006257 (Josephus problem).
FORMULA
a(0) = 1. For n >= 1; a(n) = 4*A006257(n).
For n>0, a(n) = 8*(n - 2^floor(log_2(n))) + 4 (by the formula of Gregory Pat Scandalis in A006257). - Danny Rorabaugh, Apr 20 2015
EXAMPLE
Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
1;
4;
4,12;
4,12,20,28;
4,12,20,28,36,44,52,60;
4,12,20,28,36,44,52,60,68,76,84,92,100,108,116,124;
4,12,20,28,36,44,52,60,68,76,84,92,100,108,116,124,132,140,148,156,164,172,180,188,196,204,212,220,228,236,244,252;
...
Row sums give A000302.
Right border gives A173033.
PROG
(Sage) [1] + [8*(n - 2^floor(log(n, base=2))) + 4 for n in range(1, 77)] # Danny Rorabaugh, Apr 20 2015
(PARI) a(n) = if(n, 8*(n - 2^logint(n, 2)) + 4, 1)
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Mar 20 2015
STATUS
approved