A203908 -id:A203908 - cp's OEIS frontend

A051674 a(n) = prime(n)^prime(n).

4, 27, 3125, 823543, 285311670611, 302875106592253, 827240261886336764177, 1978419655660313589123979, 20880467999847912034355032910567, 2567686153161211134561828214731016126483469, 17069174130723235958610643029059314756044734431

Offset: 1

Asher Auel

Numbers k such that bigomega(k)^(bigomega(k)) = k, where bigomega = A001222. - Lekraj Beedassy, Aug 21 2004
Positive k such that k' = k, where k' is the arithmetic derivative of k. - T. D. Noe, Oct 12 2004
David Beckwith proposes (in the AMM reference): "Let n be a positive integer and let p be a prime number. Prove that (p^p) | n! implies that (p^(p + 1)) | n!". - Jonathan Vos Post, Feb 20 2006
Subsequence of A100716; A003415(m*a(n)) = A129283(m)*a(n), especially A003415(a(n)) = a(n). - Reinhard Zumkeller, Apr 07 2007
A168036(a(n)) = 0. - Reinhard Zumkeller, May 22 2015

			a(1) = 2^2 = 4.
a(2) = 3^3 = 27.
a(3) = 5^5 = 3125.

J.-M. De Koninck & A. Mercier, 1001 Problemes en Theorie Classique Des Nombres, Problem 740 pp. 95; 312, Ellipses Paris 2004.

T. D. Noe, Table of n, a(n) for n = 1..40
David Beckwith, Problem 11158, American Mathematical Monthly, Vol. 112, No. 5 (May 2005), p. 468.
Jurij Kovic, The Arithmetic Derivative and Antiderivative, Journal of Integer Sequences, Vol. 15 (2012), #12.3.8.

Cf. A000040, A000312, A003415 (arithmetic derivative of n), A129150, A129151, A129152, A048102, A072873 (multiplicative closure), A104126.
Subsequence of A100717; A203908(a(n)) = 0.
Subsequence of A097764.
Cf. A168036, A094289 (decimal expansion of Sum(1/p^p)).

a051674_list = map (\p -> p ^ p) a000040_list
-- Reinhard Zumkeller, Jan 21 2012

[p^p: p in PrimesUpTo(30)]; // Vincenzo Librandi, Mar 27 2014

A051674:=n->ithprime(n)^ithprime(n): seq(A051674(n), n=1..10); # Wesley Ivan Hurt, Jun 25 2016

Array[Prime[ # ]^Prime[ # ] &, 12] (* Vladimir Joseph Stephan Orlovsky, May 01 2008 *)
#^#&/@Prime[Range[10]] (* Harvey P. Dale, May 17 2024 *)

a(n)=n=prime(n);n^n \\ Charles R Greathouse IV, Mar 20 2013

from gmpy2 import mpz
[mpz(prime(n))**mpz(prime(n)) for n in range(1,100)] # Chai Wah Wu, Jul 28 2014

a(n) = A000312(A000040(n)). - Altug Alkan, Sep 01 2016

Sum_{n>=1} 1/a(n) = A094289. - Amiram Eldar, Oct 13 2020

A008473 If n = Product (p_j^k_j) then a(n) = Product (p_j + k_j).

Original entry on oeis.org

1, 3, 4, 4, 6, 12, 8, 5, 5, 18, 12, 16, 14, 24, 24, 6, 18, 15, 20, 24, 32, 36, 24, 20, 7, 42, 6, 32, 30, 72, 32, 7, 48, 54, 48, 20, 38, 60, 56, 30, 42, 96, 44, 48, 30, 72, 48, 24, 9, 21, 72, 56, 54, 18, 72, 40, 80, 90, 60, 96, 62, 96, 40, 8, 84, 144, 68, 72, 96, 144, 72, 25, 74, 114

Offset: 1

Olivier Gérard

Coincides with sigma (A000203) for squarefree n. - Franklin T. Adams-Watters, Jan 31 2016
Every positive integer except 2 occurs in this sequence, but none occur infinitely often. For m > 4, there are n > m with a(n) = m. This implies that every integer greater than 4 occurs in the iterated sequence infinitely often. For example, 5 <- 8 <- 125 <- 113^12 <- .... - Franklin T. Adams-Watters, Jan 31 2016
Sum of the powerfree parts of the divisors of n. - Wesley Ivan Hurt, Jun 13 2021

Vincenzo Librandi, Table of n, a(n) for n = 1..10000

Cf. A027748, A124010, A001221, A203908, A000203, A322965, A322966.

Cf. A055231 (powerfree part of n).

a008473 n = product $ zipWith (+) (a027748_row n) (a124010_row n)
-- Reinhard Zumkeller, Jul 17 2014

A008473 := proc(n) local e,j; e := ifactors(n)[2]:
mul (e[j][1]+e[j][2], j=1..nops(e)) end:
seq (A008473(n), n=1..80);
# Peter Luschny, Jan 17 2011

Array[Times @@ Total /@ FactorInteger[ # ] &, 80] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Jul 28 2006 *)

a(n)=my(f = factor(n)); for (k=1, #f~, f[k, 1] = f[k, 1] + f[k, 2]; f[k, 2] = 1;); factorback(f); \\ Michel Marcus, Jan 31 2016

Multiplicative with a(p^e) = p+e. - David W. Wilson, Aug 01 2001
a(n) = Product_{k=1..A001221(n)} (A027748(n,k) + A124010(n,k)). - Reinhard Zumkeller, Jul 17 2014
a(n)/A007947(n) = A322965(n)/A322966(n). - David S. Metzler, Jan 01 2019
a(n) = Sum_{d|n} A055231(d). - Wesley Ivan Hurt, Jun 13 2021
Sum_{k=1..n} a(k) ~ c * n^2, where c = (Pi^4/72) * Product_{p prime} (1 - 2/p^2 + 2/p^4 - 1/p^5) = 0.5342800948... . - Amiram Eldar, Dec 08 2022

More terms from Joseph Biberstine (jrbibers(AT)indiana.edu), Jul 28 2006

A100717 Numbers k having a prime divisor p such that p^p is the highest power of p that divides k.

Original entry on oeis.org

4, 12, 20, 27, 28, 36, 44, 52, 54, 60, 68, 76, 84, 92, 100, 108, 116, 124, 132, 135, 140, 148, 156, 164, 172, 180, 188, 189, 196, 204, 212, 216, 220, 228, 236, 244, 252, 260, 268, 270, 276, 284, 292, 297, 300, 308, 316, 324, 332, 340, 348, 351, 356, 364, 372

Offset: 1

Leroy Quet, Dec 10 2004

nonn

For each prime p, the sequence includes all k*p^p for k such that gcd(k,p)=1. - T. D. Noe

The asymptotic density of this sequence is 1 - Product_{p prime} (1 - 1/p^p + 1/p^(p+1)) = 0.14682429539560371215... . - Amiram Eldar, Jun 25 2022

			54 is included because 3^3, but not 3^4, divides 54.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A100716, A203908.

Subsequences: A051674, A048102 \ {1}.

a100717 n = a100717_list !! (n-1)
a100717_list = filter ((== 0) . a203908) [1..]
-- Reinhard Zumkeller, Dec 24 2013

fQ[n_] := Union[ Table[ #[[1]] == #[[2]]] & /@ FactorInteger[n]][[ -1]] == True; Select[ Range[2, 375], fQ[ # ] &] (* Robert G. Wilson v, Dec 14 2004 *)

A203908(a(n)) = 0. - Reinhard Zumkeller, Dec 24 2013

More terms from T. D. Noe and Robert G. Wilson v, Dec 14 2004

cp's OEIS Frontend

A051674 a(n) = prime(n)^prime(n).

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A008473 If n = Product (p_j^k_j) then a(n) = Product (p_j + k_j).

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Haskell

Maple

Mathematica

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A100717 Numbers k having a prime divisor p such that p^p is the highest power of p that divides k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Formula

Extensions