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Numbers n for which abs((-1)^k*Sum_{k=1..n} ((n-k+1) mod k)) = 0.
1

%I #17 Feb 04 2018 03:19:29

%S 1,4,8,25,27,75,209,3507,8466,16179,29285,33987,175904,326764,1161207

%N Numbers n for which abs((-1)^k*Sum_{k=1..n} ((n-k+1) mod k)) = 0.

%C Subset of A154585.

%C a(16) > 10^7. - _Donovan Johnson_, Oct 03 2011

%F abs{(-1)^k*A004125} = 0

%F {a(n): A154585(a(n))=0}. - _R. J. Mathar_, Jan 14 2009

%e n=8 -> abs(-(8 mod 1) + (7 mod 2) - (6 mod 3) + (5 mod 4) - (4 mod 5) + (3 mod 6) - (2 mod 7) + (1 mod 8)) = abs(-0 + 1 - 0 + 1 - 4 + 3 - 2 + 1) = abs(0) = 0.

%p P:=proc(i) local a,n; for n from 1 to i do a:=abs(add((-1)^k*((n-k+1) mod k),k=1..n)); if a=0 then print(n); fi; od; end: P(100);

%o (C) #include <stdio.h> int main(int argc, char * argv[]) { for(int n=1;;n++) { unsigned long long a = 0; for(int k=1;k <=n;k += 2) a -= (n-k+1) % k ; for(int k=2;k <=n;k += 2) a += (n-k+1) % k ; if ( a == 0) printf("%d,\n",n) ; } } /* _R. J. Mathar_, Jan 14 2009 */

%Y Cf. A004125, A154585.

%K more,nonn

%O 1,2

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Jan 12 2009

%E 8466 inserted, and sequence extended up to a(13), by _R. J. Mathar_, Jan 14 2009

%E a(14)-a(15) from _Donovan Johnson_, Oct 03 2011