A079009 Least k such that the 2^n successive values of phi(k+j) (j=0..2^n-1) are all distinct.
1, 2, 11, 37, 149, 1359, 14130, 175327, 1218073, 108387730, 14305141265
Offset: 0
Examples
For n = 7: a(7) = 175327 because phi(175327+j), for j=0..127 are all distinct: {175326, 87648, ..., 175452, 85320}.
Programs
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PARI
isdist(v) = forstep(i = #v, 1, -1, forstep(j = i - 1, 1, -1, if(v[i] == v[j], return(j)))); 0; a(n) = {my(m = 2^n, phis = vector(m, i, eulerphi(i)), k = m, ind = isdist(phis)); while(ind != 0, phis = concat(vecextract(phis, Str("^1.."ind)), vector(ind, i, eulerphi(k+i))); k += ind; ind = isdist(phis)); k - m + 1;} \\ Amiram Eldar, Aug 27 2024
Formula
a(n) = A079008(2^n).
Extensions
a(8)-a(9) from Donovan Johnson, Oct 27 2008
a(10) from Donovan Johnson, Nov 13 2010