Christopher Purcell - cp's OEIS frontend

User: Christopher Purcell

Christopher Purcell has authored 3 sequences.

A384868 a(n) = Sum_{i=1...|b|} i*(-1)^b_i where b is the lexicographically n-th binary string.

0, 1, -1, 3, -1, 1, -3, 6, 0, 2, -4, 4, -2, 0, -6, 10, 2, 4, -4, 6, -2, 0, -8, 8, 0, 2, -6, 4, -4, -2, -10, 15, 5, 7, -3, 9, -1, 1, -9, 11, 1, 3, -7, 5, -5, -3, -13, 13, 3, 5, -5, 7, -3, -1, -11, 9, -1, 1, -9, 3, -7, -5, -15, 21, 9, 11, -1, 13, 1, 3, -9, 15, 3, 5, -7, 7, -5, -3, -15, 17

Offset: 0

Christopher Purcell, Jun 11 2025

The first binary string is the empty string and is indexed n=0.

			The lexicographically 8th binary string is 001; therefore, a(8) = 1 + 2 - 3 = 0.
Sequence can be written as triangle T(n,k) with row lengths 2^n:
   0;
   1, -1;
   3, -1, 1, -3;
   6,  0, 2, -4, 4, -2, 0, -6;
  10,  2, 4, -4, 6, -2, 0, -8, 8, 0, 2, -6, 4, -4, -2, -10;
  ...

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Cf. A000079, A000217, A000225, A007931, A100575, A231204, A365968, A377170.

a(n) = my(b=[d|d<-binary(n+1)[^1]]); sum(i=1, #b, i*(-1)^b[i]); \\ Michel Marcus, Jun 11 2025

from math import comb
def A384868(n): return comb(len(s:=bin(n+1)[3:])+1,2)-(sum(i for i,j in enumerate(s,1) if j=='1')<<1) # Chai Wah Wu, Jun 13 2025

def a384868(n): return sum(i if b == '0' else -i for i, b in enumerate(bin(n + 1)[3:], 1)) # David Radcliffe, Jun 15 2025

From Alois P. Heinz, Jun 13 2025: (Start)
a(A000225(n)) = A000217(n).
a(2*(2^n-1)) = (-1)*A000217(n).
Sum_{i=0..2^n-1} a(i+2^n-1) = 0.
Sum_{i=0..2^n-1} i*a(i+2^n-1) = (-1)*A100575(n+1).
Sum_{i=0..2^n-1} abs(a(i+2^n-1)) = 2*A377170(n). (End)

A333717 a(n) is the minimal number of vertices in a simple graph with exactly n cycles.

Original entry on oeis.org

3, 5, 4, 6, 8, 5, 4, 6, 8, 6, 6, 5, 5, 6, 6, 8, 7, 6, 6, 6, 6, 5, 6, 6, 7, 7, 7, 7, 7, 6, 7, 6, 6, 7, 6, 6, 5, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 6, 6, 7, 7, 7, 7, 6, 7, 7, 7, 6, 7, 7, 7, 8, 8, 7, 7, 7, 8, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 7, 7

Offset: 1

Christopher Purcell, Sep 03 2020

nonn

a(n+1) is at most a(n) + 2 since we can add a triangle to a graph with a(n) vertices and increase the number of cycles by 1.

David Eppstein observed that an N-gon with each edge replaced by a triangle has 2^N + N cycles and 2N vertices, and gluing such graphs together greedily yields an upper bound on a(n) of O(log n).

			For n = 2, a pair of triangles sharing a vertex has five vertices; it is easy to check that no graph on three or four vertices has exactly two cycles, so a(2) = 5.

David Eppstein, Mathstodon post about O(log n) upper bound, 26 August 2020.

A300758 a(n) = 2n(n+1)(2n+1).

Original entry on oeis.org

0, 12, 60, 168, 360, 660, 1092, 1680, 2448, 3420, 4620, 6072, 7800, 9828, 12180, 14880, 17952, 21420, 25308, 29640, 34440, 39732, 45540, 51888, 58800, 66300, 74412, 83160, 92568, 102660, 113460, 124992, 137280, 150348, 164220, 178920, 194472, 210900, 228228

Offset: 0

Christopher Purcell, Mar 12 2018

The altitude h(n) = a(n)/A001844(n) of the (A005408(n), A046092(n) and A001844(n)) rectangular triangle is an irreducible fraction. - Ralf Steiner, Feb 25 2020

In this case, area A = a(n)/2 = A055112(n). - Bernard Schott, Feb 27 2020

S. P. Borgatti and M. G. Everett, Notions of Position in Social Network Analysis, Sociological Methodology, 22 (1992), 1-35.
C. Purcell and P. Rombach, Role colouring graphs in hereditary classes, arXiv:1802.10180 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A002492, A006331, A055112, A059270, A005408, A046092, A001844, A000384, A016754, A060300.

Cf. A027480.

a(n) = 12*A000330(n).
G.f.: 12*x*(1+x)/(1-x)^4. - Colin Barker, Mar 12 2018
a(n) = 6*A006331(n) = 4*A059270(n) = 3*A002492(n) = 2*A055112(n). - Omar E. Pol, Apr 04 2018
From Ralf Steiner, Feb 27 2020: (Start)
a(n) = 2*n*A000384(n+1).
a(n) = sqrt(A016754(n)*A060300(n)).
(End)
a(n) = A005408(n) * A046092(n). - Bruce J. Nicholson, Apr 24 2020

Edited by N. J. A. Sloane, Aug 01 2019

cp's OEIS Frontend

User: Christopher Purcell

A384868 a(n) = Sum_{i=1...|b|} i*(-1)^b_i where b is the lexicographically n-th binary string.

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A333717 a(n) is the minimal number of vertices in a simple graph with exactly n cycles.

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A300758 a(n) = 2n(n+1)(2n+1).

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cp's OEIS Frontend

User: Christopher Purcell

A384868 a(n) = Sum_{i=1...|b|} i*(-1)^b_i where b is the lexicographically n-th binary string.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Python

Python

Formula

A333717 a(n) is the minimal number of vertices in a simple graph with exactly n cycles.

Views

Author

Keywords

Comments

Examples

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A300758 a(n) = 2n*(n+1)*(2n+1).

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Author

Keywords

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Links

Crossrefs

Formula

Extensions

A300758 a(n) = 2n(n+1)(2n+1).