|
|
A076107
|
|
First of n consecutive integers whose sum is a positive n-th power, or 0 if no such integers exist.
|
|
3
|
|
|
1, 0, 8, 0, 623, 119, 117646, 0, 2183, 976558, 25937424596, 0, 23298085122475, 48444505197, 29192926025390618, 0, 48661191875666868473, 21523352, 104127350297911241532832, 0, 278218429446951548637196391
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
No sum exists precisely when n == 0 (mod 4). a(2) = 0 is a legitimate value.
a(p) = p^(p-1) - (p-1)/2 for prime p.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = A076108(n)/n - (n-1)/2 for n != 0 (mod 4).
a(4k)=0; otherwise a(n) = (2*A076108(n)/n - n + 1)/2 = (2*p1^n*...*pm^n/n - n + 1)/2 where p1, ..., pm are all distinct odd primes dividing n. - Max Alekseyev, Jun 10 2005
|
|
EXAMPLE
|
a(3) = 8 as 8+9+10 = 27 = 3^3. a(6) = 119 as 119+120+...+124 = 729 = 3^6.
|
|
PROG
|
(PARI) for(n=1, 30, t=n*(n-1)/2:f=0:for(r=1, 10^4, if((r^n-t)%n==0, f=(r^n-t)/n:break)):print1(f", "))
(PARI) {A076107(n)=if(n%4==0, return(0)); m=n; if(m%2==0, m\=2); f=factorint(m)[, 1]; p=1; (2*prod(i=1, length(f), f[i])^n/n-n+1)/2} (Alekseyev)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|