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Least positive k such that k^n is the sum of n consecutive integers, or 0 if no such k exists.
4

%I #27 Aug 28 2024 09:37:03

%S 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,

%T 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,

%U 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

%N Least positive k such that k^n is the sum of n consecutive integers, or 0 if no such k exists.

%C No k exists precisely when n == 0 (mod 4).

%H Amiram Eldar, <a href="/A076109/b076109.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = (n*A076107(n)+(n^2-n)/2)^(1/n) for n != 0 (mod 4).

%F a(n) = A076108^(1/n).

%F a(p) = p if p is a prime.

%F 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

%F 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

%F a(4k) = 0; otherwise a(n) = p1*...*pm where p1, ..., pm are all distinct odd primes dividing n. - _Max Alekseyev_, Jun 10 2005

%F 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

%t f[p_, e_] := If[p == 2, Boole[e == 1], p]; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Sep 09 2020 *)

%o (PARI) { A076109(n) = if(n%4==0,return(0)); if(n%2==0,n\=2); vecprod(factorint(n)[,1]); } \\ _Max Alekseyev_, Jun 10 2005

%Y Cf. A007947, A065463, A076107, A076108.

%K nonn,easy,mult

%O 1,3

%A _Amarnath Murthy_, Oct 08 2002

%E Corrected and extended by _Ralf Stephan_, Mar 30 2003

%E More terms from _Max Alekseyev_, Jun 10 2005