A076109 Least positive k such that k^n is the sum of n consecutive integers, or 0 if no such k exists.
1, 1, 3, 0, 5, 3, 7, 0, 3, 5, 11, 0, 13, 7, 15, 0, 17, 3, 19, 0, 21, 11, 23, 0, 5, 13, 3, 0, 29, 15, 31, 0, 33, 17, 35, 0, 37, 19, 39, 0, 41, 21, 43, 0, 15, 23, 47, 0, 7, 5, 51, 0, 53, 3, 55, 0, 57, 29, 59, 0, 61, 31, 21, 0, 65, 33, 67, 0, 69, 35, 71, 0, 73, 37, 15, 0, 77, 39, 79, 0
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
f[p_, e_] := If[p == 2, Boole[e == 1], p]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 09 2020 *)
-
PARI
{ A076109(n) = if(n%4==0,return(0)); if(n%2==0,n\=2); vecprod(factorint(n)[,1]); } \\ Max Alekseyev, Jun 10 2005
Formula
a(n) = (n*A076107(n)+(n^2-n)/2)^(1/n) for n != 0 (mod 4).
a(n) = A076108^(1/n).
a(p) = p if p is a prime.
Multiplicative with a(2^1) = 1; a(2^e) = 0 if e >= 2; a(p^e) = p if p >= 3. - David W. Wilson, Jun 10 2005
a(n) = A007947(n) if n == 1 (mod 2); A007947(n/2) if n == 2 (mod 4); 0 if n == 0 (mod 4). - David W. Wilson, Jun 10 2005
a(4k) = 0; otherwise a(n) = p1*...*pm where p1, ..., pm are all distinct odd primes dividing n. - Max Alekseyev, Jun 10 2005
Sum_{k=1..n} a(k) ~ c * n^2, where c = (3/8) * Product_{p prime} (1 - 1/(p*(p+1))) = (3/8) * A065463 = 0.264165... . - Amiram Eldar, Oct 28 2022
Extensions
Corrected and extended by Ralf Stephan, Mar 30 2003
More terms from Max Alekseyev, Jun 10 2005
Comments