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 A258840 a(n) is the least integer k such that there are n values of i <= k for which gpf(i^2 + 1) = gpf(k^2 + 1), where gpf(x) is the greatest prime factor of x. 0

%I

%S 1,3,7,38,47,157,302,327,515,616,697,798,818,1303,2818,3141,3323,5648,

%T 6962,9193,9872,13213,13747,15445,16271,17149,18263,20491,20727,24389,

%U 26915,29078,31867,37848,38007,40182,41508,43328,46349,55025,62258,63133,66893

%N a(n) is the least integer k such that there are n values of i <= k for which gpf(i^2 + 1) = gpf(k^2 + 1), where gpf(x) is the greatest prime factor of x.

%C A014442(n) gives the largest prime factor of n^2 + 1.

%C The primes of the sequence are 3, 7, 47, 157, 1303, 3323, 46349, ...

%C The corresponding sequence Gpf(a(n)^2+1) is 2, 5, 5, 17, 17, 29, 37, 37, 101, 101, 101, 101, 101, 101, 101, 101, 101, 97, 97, 97, 97, 401, 349, 389, 557, 557, 557, 557, 557, 421, 421, 421, 557, ... and it is interesting to observe the frequency of repetitions for the numbers 5, 17, 37, 97, 101, 557, ...

%e a(3) = 7 because gpf(7^2 + 1) = gpf(3^2 + 1) = gpf(2^2 + 1) = 5 => 3 occurrences.

%e a(4) = 38 because gpf(38^2 + 1) = gpf(21^2 + 1) = gpf(13^2 + 1) = gpf(4^2 + 1) = 17 => 4 occurrences.

%p with(numtheory):nn:=70000:T:=array(1..nn):k:=0:kk:=1:

%p for m from 1 to nn do:

%p x:=factorset(m^2+1):n1:=nops(x):p:=x[n1]:k:=k+1:T[k]:=p:

%p od:

%p for n from 1 to 43 do:jj:=0:for k from kk to nn while(jj=0) do:

%p q:=T[k]:ii:=0:jj:=0:

%p for i from 1 to k do:

%p if T[i]=q then ii:=ii+1:

%p else

%p fi:

%p od:if ii=n then jj:=1:kk:=k:

%p printf ( "%d %d \n",n,k):else fi:

%p od:od:

%o (PARI) gpf(n) = my(f=factor(n^2+1)); f[#f~,1];

%o nboc(k) = my(gpfk = gpf(k)); sum(i=1, k, gpf(i) == gpfk);

%o a(n) = my(k = 1); while (nbo(k) != n, k++); k; \\ _Michel Marcus_, Jun 12 2015

%Y Cf. A014442, A242012.

%K nonn

%O 1,2

%A _Michel Lagneau_, Jun 12 2015

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Last modified April 23 09:35 EDT 2019. Contains 322385 sequences. (Running on oeis4.)