login
Primes p such that p^2 is the sum of a prime and its reverse.
2

%I #23 Dec 30 2023 23:34:20

%S 2,11,307,35419,347651,3091643,3417569,30001253,34158919,35515619,

%T 305524927,312123463,313513517,327371987,337660679,348898811,

%U 352023571,3013005397,3051026827,3147298717,3149171717,3171167353,3175236553,3226951193,3248169343,3306563683,3350101739,3366748421,3403341569

%N Primes p such that p^2 is the sum of a prime and its reverse.

%C Primes p such that p^2 = A056964(q) for some term q of A367796.

%C Do all terms except for 2 and 11 start with 3?

%C From _Ivan N. Ianakiev_, Dec 16 2023: (Start)

%C To prove that for all n > 2 the first digit of a(n) is 3 is easy if the number of digits of q is odd. Sketch of a proof: Let p^2 = q + rev(q). We observe that:

%C a) the last digit of q must be 1, 3, 7, or 9;

%C b) the last digit of rev(q) cannot be zero, since the first digit of q cannot be zero;

%C c) the last digit of rev(q) cannot be odd, since the last digit of p^2 cannot be even (if it were, that would imply that p is even).

%C The rest is just a matter of bookkeeping.

%C To prove that for n > 2 the number of digits of q cannot be even is probably much more difficult. (End)

%e a(1) = 2 is a term because 2^2 = 4 = 2 + 2 with 2 prime.

%e a(2) = 11 is a term because 11^2 = 121 = 29 + 92 with 11 and 29 prime.

%e a(3) = 307 is a term because 307^2 = 94249 = 20147 + 74102 with 307 and 20147 prime.

%e a(4) = 35419 is a term because 35419^2 = 1254505561 = 261104399 + 993401162 with 35419 and 261104399 prime.

%e a(5) = 347651 is a term because 347651^2 = 120861217801 = 20870609999 + 99990607802 with 347651 and 20870609999 prime.

%e a(6) = 3091643 is a term because 3091643^2 = 9558256439449 = 2059108419947 + 7499148019502 with 3091643 and 2059108419947 prime.

%e a(7) = 3417569 is a term because 3417569^2 = 11679777869761 = 2080783998959 + 9598993870802 with 3417569 and 2080783998959 prime.

%e a(8) = 30001253 is a term because 30001253^2 = 900075181570009 = 200000140570007 + 700075041000002 with 30001253 and 200000140570007 prime.

%e a(9) = 34158919 is a term because 34158919^2 = 1166831747248561 = 206841324099959 + 959990423148602 with 34158919 and 206841324099959 prime.

%e a(10) = 35515619 is a term because 35515619^2 = 1261359192953161 = 261359249999999 + 999999942953162 with 35515619 and 261359249999999 prime.

%p f:= proc(n) local y,c,d,dp,i,delta,m;

%p y:= convert(n^2,base,10);

%p d:= nops(y);

%p if d::even then

%p if y[-1] <> 1 then return false fi;

%p dp:= d-1;

%p y:= y[1..-2];

%p c[dp]:= 1;

%p else

%p dp:= d;

%p c[dp]:= 0;

%p fi;

%p c[0]:= 0;

%p for i from 1 to floor(dp/2) do

%p delta:= y[i] - y[dp+1-i] - c[i-1] - 10*c[dp+1-i];

%p if delta = 0 then c[dp-i]:= 0; c[i]:= 0;

%p elif delta = -1 then c[dp-i]:= 1; c[i]:= 0;

%p elif delta = -10 then c[dp-i]:= 0 ; c[i]:= 1;

%p elif delta = -11 then c[dp-i]:= 1; c[i]:= 1;

%p else return false

%p fi;

%p if y[i] + 10*c[i] - c[i-1] < 0 or (i=1 and y[i]+10*c[i]-c[i-1]=1) then return false fi;

%p od;

%p m:= (dp+1)/2;

%p delta:= y[m] + 10*c[m] - c[m-1];

%p if not member(delta, [seq(i,i=0..18,2)]) then return false fi;

%p [seq(y[i]+ 10*c[i]-c[i-1],i=1..m)]

%p end proc:

%p g:= proc(L) local T,d,t,p, x, i; uses combinat;

%p d:= nops(L);

%p T:= cartprod([select(t -> t[1]::odd, [seq([L[1]-x,x],x=max(1,L[1]-9)..min(L[1],9))]),

%p seq([seq([L[i]-x,x],x=max(0,L[i]-9)..min(9, L[i]))],i=2..d-1)]);

%p while not T[finished] do

%p t:= T[nextvalue]();

%p p:= add(t[i][1]*10^(i-1),i=1..d-1) + L[-1]/2 * 10^(d-1) +

%p add(t[i][2]*10^(2*d-i-1),i=1..d-1);

%p if isprime(p) then return p fi;

%p od;

%p -1

%p end proc:

%p p:= 11: R:= 2, 11:

%p while p < 10^8 do

%p p:= nextprime(p);

%p d:= 1+ilog10(p^2);

%p if d::even and p^2 >= 2*10^(d-1) then p:= nextprime(floor(10^(d/2))); fi;

%p v:= f(p);

%p if v = false then next fi;

%p q:= g(v);

%p if q = -1 then next fi;

%p R:= R, p;

%p od:

%p R;

%Y Cf. A056964, A367796, A367900, A367871.

%K nonn,base

%O 1,1

%A _Robert Israel_, Nov 30 2023