A082900 a(n) = A082894(n)-A000079(n), the difference of 2^n and the number closest to it and divisible by n.

0, 0, 1, 0, -2, 2, -2, 0, 1, -4, -2, -4, -2, -4, 7, 0, -2, 8, -2, 4, -8, -4, -2, 8, -7, -4, 1, 12, -2, -4, -2, 0, -8, -4, 17, 8, -2, -4, -8, -16, -2, 20, -2, -16, -17, -4, -2, -16, 19, -24, -8, -16, -2, 26, 12, 24, -8, -4, -2, -16, -2, -4, -8, 0, -32, 2, -2, -16, -8, 26, -2, 8, -2, -4, 7, -16, -18, 14, -2, -16, 1, -4, -2, 20, -32, -4, -8

Offset: 1

Author

Labos Elemer, Apr 22 2003

Keywords

• sign
• look

Comments

n=5:2^5=32 and number divisible by n=5 and closest to 32 is 30 = A082894(5), a(5)=30-32=-2 is the corresponding difference.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A082893, A082894, A082895, A082896, A082898, A082899, A082901.

Programs

Mathematica

Table[n*Floor[(Floor[n/2]+2^n)/n]-2^n, {n, 1, 100}]

PARI

A082894(n) = (n*(((n\2)+2^n)\n));
A082900(n) = (A082894(n)-(2^n)); \\ Antti Karttunen, Feb 18 2023

PARI

A082900(n) = { my(x=2^n); for(k=0,oo,if(!((x-k)%n),return(-k), if(!((x+k)%n),return(k)))); }; \\ Antti Karttunen, Feb 18 2023

Python

def A082900(n): return (m:=(k:=1<>1))-m%n-k # Chai Wah Wu, Apr 23 2025

Formula

a(n)=n*floor[(floor(n/2)+2^n)/n]-2^n