A204987 Least k such that n divides 2^k - 2^j for some j satisfying 1 <= j < k.
2, 2, 3, 3, 5, 3, 4, 4, 7, 5, 11, 4, 13, 4, 5, 5, 9, 7, 19, 6, 7, 11, 12, 5, 21, 13, 19, 5, 29, 5, 6, 6, 11, 9, 13, 8, 37, 19, 13, 7, 21, 7, 15, 12, 13, 12, 24, 6, 22, 21, 9, 14, 53, 19, 21, 6, 19, 29, 59, 6, 61, 6, 7, 7, 13, 11, 67, 10, 23, 13, 36, 9, 10, 37, 21, 20, 31, 13, 40, 8, 55, 21, 83, 8, 9, 15, 29, 13
Offset: 1
Keywords
Examples
1 divides 2^2 - 2^1, so a(1)=2; 2 divides 2^2 - 2^1, so a(2)=2; 3 divides 2^3 - 2^1, so a(3)=3; 4 divides 2^3 - 2^2, so a(4)=3; 5 divides 2^5 - 2^1, so a(5)=5.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..6556
Programs
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Mathematica
s[n_] := s[n] = 2^n; z1 = 1000; z2 = 50; Table[s[n], {n, 1, 30}] (* A000079 *) u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]] Table[u[m], {m, 1, z1}] (* A204985 *) v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0] w[n_] := w[n] = Table[v[n, h], {h, 1, z1}] d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]] Table[d[n], {n, 1, z2}] (* A204986 *) k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2] m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2] j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2 Table[k[n], {n, 1, z2}] (* A204987 *) Table[j[n], {n, 1, z2}] (* A204988 *) Table[s[k[n]], {n, 1, z2}] (* A204989 *) Table[s[j[n]], {n, 1, z2}] (* A140670 ? *) Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A204991 *) Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204992 *) %%/2 (* A204990=(1/2)*A204991 *) -
PARI
A204987etA204988(n) = { my(k=2); while(1,for(j=1,k-1,if(!(((2^k)-(2^j))%n),return([k,j]))); k++); }; \\ (Computes also A204988 at the same time) - Antti Karttunen, Nov 19 2017
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PARI
a(n)={my(k=valuation(n,2)); max(k, 1) + znorder(Mod(2, n>>k))} \\ Andrew Howroyd, Aug 08 2018
Formula
Extensions
More terms from Antti Karttunen, Nov 19 2017
Comments