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A328920 let f(N) be the smallest index m such that from the m-th term on, the sequence {k^k mod N: k >= 0} enters into a cycle, then a(n) is the smallest number N such that f(N) = n, or 0 if no such n exists. 1

%I #6 Nov 04 2019 11:17:32

%S 1,2,0,8,81,0,15625,128,5764801,0,0,2048,3138428376721,1594323,

%T 3937376385699289,32768,43046721,0,14063084452067724991009,524288,

%U 37589973457545958193355601,476837158203125,31381059609,8388608,480250763996501976790165756943041,847288609443,0,134217728,0,3219905755813179726837607

%N let f(N) be the smallest index m such that from the m-th term on, the sequence {k^k mod N: k >= 0} enters into a cycle, then a(n) is the smallest number N such that f(N) = n, or 0 if no such n exists.

%C By the formula for f(N) shown in A328914, a(n) = 0 if and only if n > 1 and n-1 is powerful (n-1 is in A001694). Note that if f(N) = n, where N = Product_{i=1..t} p_i^e_i, then f(p_i^e_i) <= n, and there exists some i such that f(p_i^e_i) = n.

%F a(0) = 1, a(1) = 2; for n > 1, let p be the smallest prime such that n-1 is divisible by p but not by p^2, then a(n) = p^n, or 0 if no such p exists.

%e By the formula shown in A328914, if f(N) = 7, then N must be divisible by either 2^7 = 128 or 3^7 = 729, so the smallest N is a(7) = 128.

%o (PARI) a(n) = if(n==0, 1, if(n==1, 2, my(v=factor(n-1)); for(i=1, omega(n-1), if(v[i,2]==1, return(v[i,1]^n))); return(0)))

%Y Cf. A328914, A001694.

%K nonn

%O 0,2

%A _Jianing Song_, Oct 31 2019

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Last modified April 16 03:28 EDT 2024. Contains 371696 sequences. (Running on oeis4.)