cp's OEIS Frontend

A034699 Largest prime power factor of n.

%I A034699 #137 May 27 2025 18:57:34
%S A034699 1,2,3,4,5,3,7,8,9,5,11,4,13,7,5,16,17,9,19,5,7,11,23,8,25,13,27,7,29,
%T A034699 5,31,32,11,17,7,9,37,19,13,8,41,7,43,11,9,23,47,16,49,25,17,13,53,27,
%U A034699 11,8,19,29,59,5,61,31,9,64,13,11,67,17,23,7,71,9,73,37,25,19,11,13,79
%N A034699 Largest prime power factor of n.
%C A034699 n divides lcm(1, 2, ..., a(n)).
%C A034699 a(n) = A210208(n,A073093(n)) = largest term of n-th row in A210208. - _Reinhard Zumkeller_, Mar 18 2012
%C A034699 a(n) = smallest m > 0 such that n divides A003418(m). - _Thomas Ordowski_, Nov 15 2013
%C A034699 a(n) = n when n is a prime power (A000961). - _Michel Marcus_, Dec 03 2013
%C A034699 Conjecture: For all n between two consecutive prime numbers, all a(n) are different. - _I. V. Serov_, Jun 19 2019
%C A034699 Disproved with between p=prime(574) = 4177 and prime(575) = 4201, a(4180) = a(4199) = 19. See A308752. - _Michel Marcus_, Jun 19 2019
%C A034699 Conjecture: For any N > 0, there exist numbers n and m, N < n < n+a(n) <= m, such that all n..m are composite and a(n) = a(m). - _I. V. Serov_, Jun 21 2019
%C A034699 Conjecture: For all n between two consecutive prime numbers, all (-1)^n*a(n) are different. Checked up to 5*10^7. - _I. V. Serov_, Jun 23 2019
%C A034699 Disproved: between p = prime(460269635) = 10120168277 and p = prime(460269636) = 10120168507 the numbers n = 10120168284 and m = 10120168498 form a pair such that (-1)^n*a(n) = (-1)^m*a(m) = 107. - _L. Joris Perrenet_, Jan 05 2020
%C A034699 a(n) = cardinality of smallest set on which idempotence of order n+1 (f^{n+1} = f) differs from idempotence of order e for 2 <= e <= n (see von Eitzen link for proof); derivable from A245501. - _Mark Bowron_, May 22 2025
%H A034699 Antti Karttunen, <a href="/A034699/b034699.txt">Table of n, a(n) for n = 1..65537</a> (first 1000 terms from T. D. Noe)
%H A034699 Hagen von Eitzen, <a href="https://math.stackexchange.com/a/5069214">Least k to distinguish f^n = f on {1,...,k} from f^e = f for every 1 < e < n is largest prime power factor of n-1?</a>, Math Stack Exchange.
%F A034699 If n = p_1^e_1 *...* p_k^e_k, p_1 < ... < p_k primes, then a(n) = Max_i p_i^e_i.
%F A034699 a(n) = A088387(n)^A088388(n). - _Antti Karttunen_, Jul 22 2018
%F A034699 a(n) = n/A284600(n) = n - A081805(n) = A034684(n) + A100574(n). - _Antti Karttunen_, Aug 06 2018
%F A034699 a(n) = a(m) iff m = d*a(n), where d is a divisor of A038610(a(n)). - _I. V. Serov_, Jun 19 2019
%t A034699 f[n_] := If[n == 1, 1, Max[ #[[1]]^#[[2]] & /@ FactorInteger@n]]; Array[f, 79] (* _Robert G. Wilson v_, Sep 02 2006 *)
%t A034699 Array[Max[Power @@@ FactorInteger@ #] &, 79] (* _Michael De Vlieger_, Jul 26 2018 *)
%o A034699 (Haskell)
%o A034699 a034699 = last . a210208_row
%o A034699 -- _Reinhard Zumkeller_, Mar 18 2012, Feb 14 2012
%o A034699 (PARI) a(n) = if(1==n,n,my(f=factor(n)); vecmax(vector(#f[, 1], i, f[i, 1]^f[i, 2]))); \\ _Charles R Greathouse IV_, Nov 20 2012, check for a(1) added by _Antti Karttunen_, Aug 06 2018
%o A034699 (PARI) A034699(n) = if(1==n,n,fordiv(n, d, if(isprimepower(n/d), return(n/d)))); \\ _Antti Karttunen_, Aug 06 2018
%o A034699 (Python)
%o A034699 from sympy import factorint
%o A034699 def A034699(n): return max((p**e for p, e in factorint(n).items()), default=1) # _Chai Wah Wu_, Apr 17 2023
%Y A034699 Cf. A006530, A010055, A020639, A027750, A034684, A028233, A051283, A052128, A053585, A057110, A060818, A038610, A081805, A088387, A088388, A100574, A210208, A245501, A284600, A308752.
%Y A034699 Numbers n ordered by a(n): A305325.
%K A034699 nonn,easy,nice
%O A034699 1,2
%A A034699 _David W. Wilson_