A303643 Numbers k such that k and phi(k) are in A292544.
1, 1664, 6815744, 10092544, 27917287424, 4707284156416, 5506148073472, 7060926234624, 8259222110208, 114349209288704, 108649341010313216, 468374361246531584, 1918461383665793368064, 7858017827495089635590144, 11635911013790805806546944, 183907840308875463202177024
Offset: 1
Keywords
Examples
1 is in A292544, and eulerphi(1)=1, so 1 is a term. 1664 and 768=eulerphi(1668) are both in A292544, so 1664 is a term.
Programs
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PARI
isA292544(n) = Mod(2, n)^eulerphi(n)==eulerphi(n); isok(n) = isA292544(n) && isA292544(eulerphi(n));
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PARI
{ ZK(m) = my(z,k); z=znorder(Mod(2,m)); k=znlog(eulerphi(m),Mod(2,m)); if(type(k)!="t_INT",return()); [z,k]; } { getpowerof2(m) = my(m2,t,zk,zk2,r); m2 = eulerphi(m); t = valuation(m2,2); m2 \= 2^t; if( m2==1, return(0)); zk=ZK(m); zk2=ZK(m2); if(!zk || !zk2, return()); r = [zk[1],zk2[1],zk[2]-t-zk2[2]+1]; \\ solving r[1] * i = r[2] * j + r[3] r /= content(r); if( gcd(r[1],r[2])>1, return()); ((r[2]*lift(Mod(-r[3]/r[2],r[1])) + r[3])/r[1] + r[2]*x)*zk[1] - zk[2] + 1;} \\ getpowerof2(m) returns z*i - k + 1 with x parameter (see formula section), i.e., getpowerof2(13) returns 12*x+7, that is, 13*2^(12*x+7) is a term for all x >= 0.
Formula
Following steps can be used in order to produce terms of this sequence.
(1) Take odd m, find z and k (see formula section of A292544).
(2) Represent phi(m) = 2^t*m', where m' is odd (i.e., m' = A053575(m)).
(3) For this m', find z' and k'.
(4) Solve z*i - k + t = z'*j - k' + 1 for positive i, j.
(5) Each such solution gives a term m*2^(z*i - k + 1) of this sequence.
For all x >= 0, 13*2^(12*x+7), 77*2^(60*x+17), 137*2^(136*x+35), 173*2^(1204*x+259), 193*2^(96*x+49), 269*2^(8844*x+6567), 411*2^(136*x+34), 519*2^(1204*x+258), 557*2^(38364*x+28635), 563*2^(19670*x+9836), 581*2^(2460*x+789), 641*2^(64*x+33), 653*2^(52812*x+39447), 667*2^(4620*x+3405), 769*2^(384*x+193), 807*2^(8844*x+6566) are terms of this sequence (m < 10^3 where m*2^(z*i - k + 1) is the corresponding form).