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A200525 Zeisel numbers with p(0)=4.

%I A200525 #31 Feb 16 2025 08:33:16
%S A200525 385,2585,7315,8911,27001,39905,48391,87283,192211,196285,319705,
%T A200525 410089,425585,441091,624605,679855,1310185,1899163,2460439,2586971,
%U A200525 2777041,6654005,7042411,7501261,8291459,9516637,10484585,11141671,12527281,13015891,13788319
%N A200525 Zeisel numbers with p(0)=4.
%C A200525 Pick any integers A and B and consider the linear recurrence relation given by p(0) = 4, p(i + 1) = A*p(i) + B. If for some n > 2, p(1), p(2), ..., p(n) are distinct primes, then the product of these primes is called a Zeisel number.
%H A200525 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ZeiselNumber.html">Zeisel Number.</a>
%H A200525 Wikipedia, <a href="http://en.wikipedia.org/wiki/Zeisel_number">Zeisel number</a>
%e A200525 a=2, b=-3 => p(1) = (4*2)+(-3) = 5; p(2) = (5*2)+(-3) = (7); p(3) = (7*2)+(-3) = 11 => 5*7*11 = 385.
%e A200525 a=2, b=5 => p(1) = (4*2)+5 = 13; p(2) = (13*2)+5 = 31; p(3) = (31*2)+5 = 67 => 13*31*67 = 27001.
%o A200525 (Rexx)
%o A200525 n0=4
%o A200525 do m=1 to 53
%o A200525   a=2*m
%o A200525   do b=(1-(4*a)) to 999
%o A200525     n1=(n0*a)+b
%o A200525     n2=(n1*a)+b
%o A200525     n3=(n2*a)+b
%o A200525     z=n1*n2*n3
%o A200525     say n0 a b
%o A200525     lineout("zeisel_4.txt",z||" = "||n1||"*"||n2||"*"||n3||"      "||a||" "||b||" n0="||n0)
%o A200525     end
%o A200525   end
%Y A200525 Cf. A051015.
%K A200525 nonn
%O A200525 1,1
%A A200525 _Karsten Meyer_, Nov 18 2011