A124923 -id:A124923 - cp's OEIS frontend

A124924 Primes p such that p^2 divides A124923((3p-1)/2) = ((3p-1)/2)^(3(p-1)/2) + 1.

5, 13, 173, 5501

Offset: 1

Alexander Adamchuk, Nov 12 2006

p divides A124923((3p-1)/2) for primes p in A003628. Hence this sequence is a subsequence of A003628.
Also, primes p such that (-2)^((p-1)/2) == -1-3p/2 (mod p^2).
No other terms below 10^11.

			5 is in this sequence because A124923((3*5-1)/2) = A124923(7) = 7^8 + 1 = 117650 is divisible by 5^2 = 25.

Cf. A124923, A003628.

Do[ p = Prime[n]; m = (3p-1)/2; f = PowerMod[ m, m-1, p^2 ] + 1; If[ IntegerQ[ f/p^2 ], Print[p] ], {n,2,10000} ]

Edited by Max Alekseyev, Jan 28 2012

A066068 a(n) = n^n + n.

Original entry on oeis.org

1, 2, 6, 30, 260, 3130, 46662, 823550, 16777224, 387420498, 10000000010, 285311670622, 8916100448268, 302875106592266, 11112006825558030, 437893890380859390, 18446744073709551632, 827240261886336764194

Offset: 0

George E. Antoniou, Dec 02 2001

Harry J. Smith, Table of n, a(n) for n=0..100

Cf. A000169, A000312, A124923, A286193.

[ n^n+n: n in [0..100] ]; // Vincenzo Librandi, Apr 15 2011

Table[ n^n + n, {n, 1, 18} ]
Table[If[n==0,1,n^n+n],{n,0,18}] (* Vladimir Joseph Stephan Orlovsky, Apr 14 2011 *)

a(n) = n^n + n; \\ Harry J. Smith, Nov 09 2009

E.g.f.: (1-x*e^x*T(x)+x*e^x)/(1-T(x)), where T(x) is Euler's tree function (see A000169). - Len Smiley, Dec 04 2001
Resultant of nx^n+1 and nx-1. - Ralf Stephan, Nov 20 2004
a(n) = n*A124923(n), n>0. - R. J. Mathar, Oct 31 2015
Sum_{n>=1} 1/a(n) = A286193. - Amiram Eldar, Nov 19 2020

A303370 Least integer k such that (k+1)^k >= n.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 0

M. F. Hasler, Apr 26 2018

nonn

n -> a(n+1) is a left inverse for A152917 and also, just like n -> a(n)+1, a left inverse to A000169(n) = n^(n-1), with codomain = the positive integers.
a(n+1) is also the smallest m such that n can be written with at most m digits in base m+1. For example, the largest number that can be written with 2 digits in base 3 is 8 = 22[3], to write 9 we need to switch to a(9+1) = 3 digits in base 4. Similarly, 63 is the largest number that can be written with a(63+1) = 3 digits in base 4, and from 64 on we need to go to a(64+1) = 4 digits in base 5.
There is no doubt that k must be nonnegative, except for the case n = 0, for which we do not want to consider all negative even k which would also satisfy the inequality, and therefore require k >= 0 to get a well defined solution.
First occurrence of k = 0, 1, 2, ...: 0, 2, 3, 10, 65, 626, 7777, 117650, ... = A124923. - Robert G. Wilson v, Apr 29 2018

			(0+1)^0 = 1 >= 0, therefore a(0) = 0.
(0+1)^0 = 1 >= 1, therefore a(1) = 0.
(0+1)^0 = 1 < 2, but (1+1)^1 = 2 >= 2, therefore a(2) = 1.
(1+1)^1 = 2 < 3, but (2+1)^2 = 9 >= 3, therefore a(n) = 2 for 3 <= n <= 9.
(2+1)^2 = 9 < 10, but (3+1)^3 = 64 >= 10, therefore a(n) = 3 for 10 <= k <= 64.
(3+1)^3 = 64 < 65, but (4+1)^4 = 625 >= 65, therefore a(n) = 4 for 65 <= k <= 625.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A000169, A152917.

A:= Array(0..5^4):
w:= 0:
for k from 0 to 4 do
  v:= (k+1)^k;
  A[w..v]:= k;
  w:= v+1
od:
seq(A[i],i=0..5^4); # Robert Israel, Apr 30 2018

f[n_] := Block[{k = 0}, While[(k + 1)^k < n, k++]; k]; Array[f, 105, 0] (* Robert G. Wilson v, Apr 29 2018 *)

a(n,k=ceil(solve(k=0,log(n+1),(k+1)^k-n)))=k-(k&&k^(k-1)>=n) \\ Corrective term in case ceil() incorrectly rounded up.

A359701 a(n) = Sum_{d|n} d^(d + n/d - 2).

Original entry on oeis.org

1, 3, 10, 69, 626, 7812, 117650, 2097425, 43046803, 1000003158, 25937424602, 743008418676, 23298085122482, 793714774077816, 29192926025406980, 1152921504623628545, 48661191875666868482, 2185911559739084235093, 104127350297911241532842

Offset: 1

Seiichi Manyama, Jan 11 2023

nonn

Cf. A082245, A124923, A262843, A294956.

a[n_] := DivisorSum[n, #^(# + n/# - 2) &]; Array[a, 20] (* Amiram Eldar, Aug 14 2023 *)

PARI
```
a(n) = sumdiv(n, d, d^(d+n/d-2));
```

my(N=20, x='x+O('x^N)); Vec(sum(k=1, N, k^(k-1)*x^k/(1-k*x^k)))

G.f.: Sum_{k>0} k^(k-1) * x^k / (1 - k * x^k).

If p is prime, a(p) = 1 + p^(p-1).

cp's OEIS Frontend

A124924 Primes p such that p^2 divides A124923((3p-1)/2) = ((3p-1)/2)^(3(p-1)/2) + 1.

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A066068 a(n) = n^n + n.

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A303370 Least integer k such that (k+1)^k >= n.

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Maple

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A359701 a(n) = Sum_{d|n} d^(d + n/d - 2).

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Mathematica

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