A128981 -id:A128981 - cp's OEIS frontend

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

José María Grau Ribas, Apr 10 2011

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

			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

Cf. A128981 (sum == 0 (mod n)), A188776 (sum == 1 (mod n)).
Cf. A057245.
Cf. A001923, A062970.

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

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

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

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

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

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

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

A188776 Numbers n such that Sum_{k=1..n} k^k == 1 (mod n).

Original entry on oeis.org

1, 2, 9, 30, 6871, 185779, 208541, 813162, 864355, 2573155

Offset: 1

José María Grau Ribas, Apr 10 2011

Numbers n such that A001923(n) == 1 (mod n).

a(11) > 10^7. - Hiroaki Yamanouchi, Aug 25 2015

Cf. A001923, A128981 (sum == 0 mod n), A188775 (sum == -1 mod n).

Union@Table[If[Mod[Sum[PowerMod[i,i,n],{i,1,n}],n]==1,Print[n];n],{n,1,20000}]

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

from itertools import accumulate, count, islice
def A188776_gen(): # generator of terms
    yield 1
    for i, j in enumerate(accumulate(k**k for k in count(2)),start=2):
        if not j % i:
            yield i
A188776_list = list(islice(A188776_gen(),5)) # Chai Wah Wu, Jun 18 2022

a(6)-a(9) from Lars Blomberg, May 10 2011

a(1) inserted and a(10) added by Hiroaki Yamanouchi, Aug 25 2015

A341437 Numbers k such that k divides Sum_{j=0..k} j^(k-j).

Original entry on oeis.org

1, 2, 6, 7, 9, 42, 46, 431, 1806, 2506, 11318, 16965, 25426, 33146, 33361, 37053, 49365, 99221, 224506, 359703, 436994

Offset: 1

Seiichi Manyama, Feb 11 2021

Numbers k such that k divides A026898(k-1).

a(19) > 10^5.

Cf. A026898, A128981, A188775, A341436.

Do[If[Mod[Sum[PowerMod[k, n - k, n], {k, 0, n}], n] == 0, Print[n]], {n, 1, 3000}] (* Vaclav Kotesovec, Feb 12 2021 *)

isok(n) = sum(k=0, n, Mod(k, n)^(n-k))==0;

0^6 + 1^5 + 2^4 + 3^3 + 4^2 + 5^1 + 6^0 = 66 = 6 * 11. So 6 is a term.

a(19) from Vaclav Kotesovec, Feb 14 2021

a(20)-a(21) from Chai Wah Wu, Feb 15 2021

A225727 Numbers k such that sum of first k primorials (A143293) is divisible by k.

Original entry on oeis.org

1, 3, 17, 51, 967, 2901, 16439, 49317, 147951, 1331559

Offset: 1

Alex Ratushnyak, May 13 2013

a(5) = 967 is a prime,
a(6) = a(5) * 3,
a(7) = a(5) * 17,
a(8) = a(5) * 51,
a(9) = a(5) * 51 * 3,
a(10) = a(5) * 51 * 27.
The next term, if it exists, is greater than 15600000. - Alex Ratushnyak, Jun 16 2013

			Sum of first 3 primorials is 1+2+6=9, because 9 is divisible by 3, the latter is in the sequence.
Sum of first 17 primorials is A143293(17) = 1955977793053588026279. Because A143293(17) is divisible by 17, the latter is in the sequence.

Cf. A143293, A002110, A057245, A128981.

primes = [2,3]
def addPrime(k):
  for p in primes:
    if k%p==0:  return
    if p*p > k:  break
  primes.append(k)
for n in range(5,1000000,6):
  addPrime(n)
  addPrime(n+2)
sum_ = 0
primorial = n = 1
for p in primes:
  sum_ += primorial
  primorial *= p
  if sum_ % n == 0:  print(n, end=', ')
  n += 1

from itertools import chain, accumulate, count, islice
from operator import mul
from sympy import prime
def A225727_gen(): return (i+1 for i, m in enumerate(accumulate(accumulate(chain((1,),(prime(n) for n in count(1))), mul))) if m % (i+1) == 0)
A225727_list = list(islice(A225727_gen(),6)) # Chai Wah Wu, Feb 23 2022

A343931 Numbers k such that Sum_{j=1..k} (-j)^j == 0 (mod k).

Original entry on oeis.org

1, 3, 4, 11, 131, 188, 324, 445, 3548, 8284, 201403, 253731, 564084, 1812500, 4599115

Offset: 1

Seiichi Manyama, May 04 2021

Also numbers k such that k divides A001099(k).

Cf. A001099, A128981, A341437, A343930, A343933.

q[n_] := Divisible[Sum[PowerMod[-k, k, n], {k, 1, n}], n]; Select[Range[8500], q] (* Amiram Eldar, May 04 2021 *)

isok(n) = sum(k=1, n, Mod(-k, n)^k)==0;

from itertools import accumulate, count, islice
def A343931_gen(): # generator of terms
    yield 1
    for i, j in enumerate(accumulate((-k)**k for k in count(1)),start=2):
        if j % i == 0:
            yield i
A343931_list = list(islice(A343931_gen(),10)) # Chai Wah Wu, Jun 18 2022

a(11)-a(13) from Chai Wah Wu, May 04 2021
a(14) from Martin Ehrenstein, May 05 2021
a(15) from Martin Ehrenstein, May 08 2021

A341436 Numbers k such that k divides Sum_{j=1..k} j^(k+1-j).

Original entry on oeis.org

1, 5, 16, 208, 688, 784, 2864, 9555, 17776, 81239

Offset: 1

Seiichi Manyama, Feb 11 2021

Numbers k such that k divides A003101(k).

a(11) > 10^5.

			1^5 + 2^4 + 3^3 + 4^2 + 5^1 = 65 = 5 * 13. So 5 is a term.

Mathematics Stack Exchange, Is it true that if p != 5 is a prime number then 1^p + 2^(p-1) + ... + (p-1)^2 + p^1 != 0 (mod p)?.

Cf. A003101, A128981, A188775, A341437.

Do[If[Mod[Sum[PowerMod[k, n + 1 - k, n], {k, 1, n}], n] == 0, Print[n]], {n, 1, 3000}] (* Vaclav Kotesovec, Feb 12 2021 *)

isok(n) = sum(k=1, n, Mod(k, n)^(n+1-k))==0;

A343932 a(n) = (Sum_{k=1..n} k^k) mod n.

Original entry on oeis.org

0, 1, 2, 0, 3, 5, 5, 4, 1, 7, 3, 4, 11, 13, 3, 4, 0, 15, 0, 4, 14, 13, 10, 20, 22, 11, 25, 20, 21, 1, 18, 4, 6, 17, 27, 12, 31, 27, 20, 28, 6, 41, 34, 32, 31, 45, 45, 4, 11, 25, 39, 48, 21, 45, 46, 12, 53, 47, 41, 32, 9, 5, 55, 4, 25, 7, 47, 8, 45, 19, 12, 60, 50, 43, 20, 60, 54, 29, 72, 36, 70, 31, 74, 40, 69, 7, 18, 20, 63, 3, 24, 32

Offset: 1

Seiichi Manyama, May 04 2021

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A001923, A128981, A182398, A188775, A188776.

a[n_] := Mod[Sum[PowerMod[k, k, n], {k, 1, n}], n]; Array[a, 100] (* Amiram Eldar, May 04 2021 *)

PARI
```
a(n) = sum(k=1, n, k^k)%n;
```

def A343932(n): return sum(pow(k,k,n) for k in range(1,n+1)) % n # Chai Wah Wu, Jun 18 2022

a(n) = A001923(n) mod n.

A383228 a(n) is the number of cases where both j and k (1 <= j < k <= n), are divisors of Sum_{i=j..k} i^i.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17

Offset: 1

Jean-Marc Rebert, Apr 20 2025

nonn

			1 and 4 divide Sum_{i = 1..4} i^i = 288, and
1 and 17 divide Sum_{i = 1..17} i^i = 846136323944176515621, and
1 and 19 divide Sum_{i = 1..19} i^i = 2018612200059554303215024, and
2 and 9 divide Sum_{i = 2..9} i^i = 405071316, and
2 and 30 divide Sum_{i = 2..30} i^i = 208492413443704093346554910065262730566475780, and
3 and 6 divide Sum_{i = 3..6} i^i = 50064, and
5 and 15 divide Sum_{i = 5..15} i^i = 449317984130199540, and
14 and 42 divide Sum_{i = 14..42} i^i = 151474018115847331407847533862930150275321445330384244581082952733296, and
22 and 26 divide Sum_{i = 22..26} i^i = 6246292379849897330286605654999947328, so a(42) = 9.

Jean-Marc Rebert, Table of n, a(n) for n = 1..1000

Cf. A000312, A001923, A128981.

a(n) = sum(j=1, n, sum(k=j+1, n, my(s=sum(i=j, k, i^i)); !(s % j) && !(s % k))); \\ Michel Marcus, Apr 20 2025

More terms from Michel Marcus, Apr 20 2025

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Python

Extensions

A188776 Numbers n such that Sum_{k=1..n} k^k == 1 (mod n).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Python

Extensions

A341437 Numbers k such that k divides Sum_{j=0..k} j^(k-j).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A225727 Numbers k such that sum of first k primorials (A143293) is divisible by k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Python

A343931 Numbers k such that Sum_{j=1..k} (-j)^j == 0 (mod k).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Python

Extensions

A341436 Numbers k such that k divides Sum_{j=1..k} j^(k+1-j).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A343932 a(n) = (Sum_{k=1..n} k^k) mod n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula