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A049920
a(n) = a(1) + a(2) + ... + a(n-1) - a(m) for n >= 4, where m = 2*n - 3 - 2^(p+1) and p is the unique integer such that 2^p < n-1 <= 2^(p+1), with a(1) = 1, a(2) = 3, and a(3) = 2.
2
1, 3, 2, 5, 9, 19, 37, 67, 106, 248, 495, 983, 1938, 3807, 7225, 13007, 20727, 48678, 97355, 194703, 389378, 778687, 1556985, 3112527, 6219767, 12426032, 24775436, 49258849, 97350091, 190037400, 361519131, 650463607, 1036758174
OFFSET
1,2
COMMENTS
The number m in the definition of the sequence equals 2*n - 3 - x, where x is the smallest power of 2 >= n-1. It turns out that m = A006257(n-2), where the sequence b(n) = A006257(n) satisfies b(2*n) = 2*b(n) - 1 and b(2*n + 1) = 2*b(n) + 1, and it is related to the so-called Josephus problem. - Petros Hadjicostas, Sep 25 2019
FORMULA
a(n) = -a(A006257(n-2)) + Sum_{i = 1..n-1} a(i) for n >= 4 with a(1) = 1, a(2) = 3, and a(3) = 2.
EXAMPLE
From Petros Hadjicostas, Sep 25 2019: (Start)
a(4) = -a(A006257(4-2)) + a(1) + a(2) + a(3) = -a(1) + a(1) + a(2) + a(3) = 5.
a(5) = -a(A006257(5-2)) + a(1) + a(2) + a(3) + a(4) = -a(3) + a(1) + a(2) + a(3) + a(4) = 9.
a(6) = -a(A006257(6-2)) + a(1) + a(2) + a(3) + a(4) + a(5) = 19.
a(7) = -a(A006257(7-2)) + a(1) + a(2) + a(3) + a(4) + a(5) + a(6) = 37.
(End)
MAPLE
A[1]:= 1: A[2]:= 3: A[3]:= 2:
for n from 4 to 100 do
q:= ceil(log[2](n-1));
m:= 2*n-3-2^q;
A[n]:= add(A[i], i=1..n-1)-A[m];
od:
seq(A[i], i=1..100); # Robert Israel, Feb 27 2017
CROSSREFS
KEYWORD
nonn
EXTENSIONS
Name edited by Petros Hadjicostas, Sep 25 2019
STATUS
approved