A188775 Numbers k such that Sum_{j=1..k} j^j == -1 (mod k).

1, 2, 3, 6, 14, 42, 46, 1806, 2185, 4758, 5266, 10895, 24342, 26495, 44063, 52793, 381826, 543026, 547311, 805002

Offset: 1

Author

José María Grau Ribas, Apr 10 2011

Keywords

• nonn
• hard
• more

Comments

Numbers k such that A001923(k) == -1 (mod k).

a(21) > 10^7. - Hiroaki Yamanouchi, Aug 25 2015

Numbers k such that k divides A062970(k). - Jianing Song, Feb 03 2019

Examples

			6 is a term because 1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6 = 50069 and 50069 + 1 = 6 * 8345. - _Bernard Schott_, Feb 03 2019
		

Crossrefs

Cf. A128981 (sum == 0 (mod n)), A188776 (sum == 1 (mod n)).

Cf. A057245.

Cf. A001923, A062970.

Programs

Maple

isA188775 := proc(n) add( modp(k &^ k,n),k=1..n) ; if modp(%,n) = n-1 then true; else false; end if; end proc:
for n from 1 do if isA188775(n) then printf("%d\n",n) ; end if; end do: # R. J. Mathar, Apr 10 2011

Mathematica

Union@Table[If[Mod[Sum[PowerMod[i,i,n],{i,1,n}],n]==n-1,Print[n];n],{n,1,10000}]

PARI

f(n)=lift(sum(k=1,n,Mod(k,n)^k));
for(n=1,10^6,if(f(n)==n-1,print1(n,", "))) \\ Joerg Arndt, Apr 10 2011

PARI

m=0;for(n=1,1000,m=m+n^n;if((m+1)%n==0,print1(n,", "))) \\ Jinyuan Wang, Feb 04 2019

Python

sum = 0
for n in range(10000):
    sum += n**n
    if sum % (n+1) == 0:
        print(n+1, end=',')
# Alex Ratushnyak, May 13 2013

Extensions

a(12)-a(16) from Joerg Arndt, Apr 10 2011

a(17)-a(20) from Lars Blomberg, May 10 2011