A049979 -id:A049979 - cp's OEIS frontend

A049963 a(n) = a(1) + a(2) + ... + a(n-1) + a(m) for n >= 4, where m = 2*n - 2 - 2^(p+1) and p is the unique integer such that 2^p < n-1 <= 2^(p+1), with a(1) = 1, a(2) = 2 and a(3) = 4.

1, 2, 4, 9, 25, 43, 93, 220, 617, 1016, 2039, 4112, 8401, 17598, 38292, 90070, 252612, 415156, 830319, 1660672, 3321521, 6643838, 13290772, 26595030, 53262532, 106850150, 214945816, 434874798, 889700788, 1859656696

Offset: 1

Clark Kimberling

nonn

The number m in the definition of the sequence equals 2*n - 2 - x, where x is the smallest power of 2 >= n-1. It turns out that m = 1 + 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

			From _Petros Hadjicostas_, Sep 25 2019: (Start)
a(4) = a(1 + A006257(4-2)) + a(1) + a(2) + a(3) = a(2) + a(1) + a(2) + a(3) = 9.
a(7) = a(1 + A006257(7-2)) + a(1) + a(2) + a(3) + a(4) + a(5) + a(6) = a(4) + a(1) + a(2) + a(3) + a(4) + a(5) + a(6) = 93.
(End)

Index entries for sequences related to the Josephus Problem

Cf. A006257, A049920, A049939, A049960, A049964, A049979.

Cf. A049914 (similar, but with minus a(m/2)), A049915 (similar, but with minus a(m)), A049962 (similar, but with plus a(m/2)).

a := proc(n) local i; option remember; if n < 4 then return [1, 2, 4][n]; end if; add(a(i), i = 1 .. n - 1) + a(2*n - 3 - Bits:-Iff(n - 2, n - 2)); end proc;
seq(a(n), n = 1..40); # Petros Hadjicostas, Sep 25 2019, courtesy of Peter Luschny

a(n) = a(1 + A006257(n-2)) + Sum_{i = 1..n-1} a(i) for n >= 4 with a(1) = 1, a(2) = 2 and a(3) = 4. - Petros Hadjicostas, Sep 25 2019

Name edited by Petros Hadjicostas, Sep 25 2019

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.

Original entry on oeis.org

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

Clark Kimberling

nonn

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

			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)

Robert Israel, Table of n, a(n) for n = 1..3323
Index entries for sequences related to the Josephus Problem

Cf. A006257, A049939, A049960, A049964, A049979.

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

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.

Name edited by Petros Hadjicostas, Sep 25 2019

cp's OEIS Frontend

A049963 a(n) = a(1) + a(2) + ... + a(n-1) + a(m) for n >= 4, where m = 2*n - 2 - 2^(p+1) and p is the unique integer such that 2^p < n-1 <= 2^(p+1), with a(1) = 1, a(2) = 2 and a(3) = 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

Extensions