Proposed update for A076107-9

[David Wilson]

I wrote a program for A076107, and in so doing, found workable definitions 
that preserve the existing sequences and proved A076109 is in fact 
multiplicative.  Propose to replace the existing sequences with the 
following:

In order to work out the proper definitions for A076107-9, I wrote a program 
to generate them.

In working on that program, I arrived at a function for A076109 which was 
manifestly

multiplicative.

The proposed changes are:

%I A076107
%S A076107 
1,0,8,0,623,119,117646,0,2183,976558,25937424596,0,23298085122475,
%T A076107 48444505197,29192926025390618,0,48661191875666868473,21523352,
%U A076107 104127350297911241532832,0,278218429446951548637196391
%N A076107 First of n consecutive integers whose sum is a positive nth 
power, or 0 if no such integers exist.
%C A076107 No sum exists precisely when n == 0 (mod 4).  a(2) = 0 is a 
legitimate value.
%C A076107 The sum is given by A076108(n) = A076109(n)^n for n != 0 (mod 4).
%C A076107 a(p) = p^(p-1)-(p-1)/2 for prime p.
%F A076107 a(n) = A076108(n)/n-(n-1)/2 for n != 0 (mod 4).
%e A076107 a(3) = 8 as 8+9+10 = 27 = 3^3. a(6) = 119 as 119+120+..+124 = 729 
= 3^6.
%o A076107 (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","))
%Y A076107 Cf. A076108, A076109.
%K A076107 nonn
%O A076107 1,3
%A A076107 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Oct 08 2002
%E A076107 Corrected and extended by Ralf Stephan 
(ralf(AT)ark.in-berlin.de), Mar 30 2003

%I A076108
%S A076108 1,1,27,0,3125,729,823543,0,19683,9765625,285311670611,0,
%T A076108 302875106592253,678223072849,437893890380859375,0,
%U A076108 827240261886336764177,387420489,1978419655660313589123979,0
%N A076108 Least positive nth power that is the sum of n consecutive 
integers, or 0 if no such nth power exists.
%C A076108 No nth power exists precisely when n == 0 (mod 4).
%C A076108 The first term of the sum is A076107(n) for n != 0 (mod 4).
%C A076108 a(p) = p^p for prime p.
%F A076108 a(n) = n*A076107(n)+(n^2-n)/2 for n != 0 (mod 4).
%F A076108 a(n) = A076109(n)^n.
%e A076108 27 = 3^3 = 8+9+10 is least positive cube that is sum of 3 
consecutive integers, hence a(3) = 27.
%o A076108 (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:break)):print1(f","))
%Y A076108 Cf. A076107, A076109.
%K A076108 nonn
%O A076108 1,3
%A A076108 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Oct 08 2002
%E A076108 Corrected and extended by Ralf Stephan 
(ralf(AT)ark.in-berlin.de), Mar 30 2003

%I A076109
%S A076109 
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,
%T A076109 
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,
%U A076109 29,59,0,61,31,21,0,65,33,67,0,69,35,71,0,73,37,15,0,77,39,79,0
%N A076109 Least positive k such that k^n is the sum of n consecutive 
integers, or 0 if no such k exists.
%C A076109 No k exists precisely when n == 0 (mod 4).
%C A076109 a(p) = p if p is a prime.
%F A076109 a(n) = (n*A076107(n)+(n^2-n)/2)^(1/n) for n != 0 (mod 4).
%F A076109 a(n) = A076108^(1/n).
%F A076109 Multiplicative with a(p^e) = A007947(n) if n == 1 (mod 2); 
A007947(n/2) if n == 2 (mod 4); 0 if n == 0 (mod 4).
%o A076109 (PARI) 
for(n=1,100,t=n*(n-1)/2:f=0:for(r=1,10^4,if((r^n-t)%n==0,f=r:break)):print1(f","))
%Y A076109 Cf. A076107, A076108.
%K A076109 nonn,mult
%O A076109 1,3
%A A076109 Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Oct 08 2002
%E A076109 Corrected and extended by Ralf Stephan 
(ralf(AT)ark.in-berlin.de), Mar 30 2003


- David W. Wilson

"Truth is just truth -- You can't have opinions about the truth."
   - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"