This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A082369 #58 Jun 02 2025 04:58:44
%S A082369 13,43,73,103,133,163,193,223,253,283,313,343,373,403,433,463,493,523,
%T A082369 553,583,613,643,673,703,733,763,793,823,853,883,913,943,973,1003,
%U A082369 1033,1063,1093,1123,1153,1183,1213,1243,1273,1303,1333,1363,1393,1423,1453
%N A082369 Numbers congruent to 13 mod 30.
%C A082369 Solutions to 19^x + 23^x == 29 mod 31.
%C A082369 The form of these numbers is obviously 30X + 13. 3^x + 5^x == 7 mod 11 and 17^x + 19^x == 23 mod 29 have no solutions. In fact, 3^x + 5^x == m mod 11 is only solvable for m < 11 = 1, 2, 8, 9. Similarly, 17^x + 19^x == m mod 29 is not solvable for m < 29 = 6, 11, 13, 14, 15, 16, 18, 23. I can't even prove 3^x + 5^x-7 <> 11k for all integers x, k. Anyone have a general proof of these statements say, a^x + -b^x == m mod k true or false for certain a, b, m, k, x combinations?
%C A082369 a^x + b^x == m (mod k) is periodic mod phi(k), so it suffices to check x = 1, 2, ..., phi(k). - _Charles R Greathouse IV_, Nov 19 2013
%H A082369 Michael G. Kaarhus, <a href="/A082369/b082369.txt">Table of n, a(n) for n = 1..10000</a>
%H A082369 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A082369 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F A082369 a(n) = 30n + 13.
%F A082369 G.f.: x*(13+17*x)/(1-x)^2. - _Colin Barker_, Jan 11 2012
%t A082369 Range[13, 7000, 30] (* _Vladimir Joseph Stephan Orlovsky_, Jul 13 2011 *)
%t A082369 LinearRecurrence[{2,-1},{13,43},50] (* _Harvey P. Dale_, Mar 02 2023 *)
%o A082369 (PARI) anpbn(n)= for(x=1,n, if((19^x+23^x-29)%31==0,print1(x, ", "))) \\ solutions to 19^x+23^x == 29 mod 31
%o A082369 (Maxima) f(a):= mod((19^a + 23^a),31)$ a:-1$ for n:1 thru 3000 step 0 do(a:a+1, if f(a)=29 then (if mod(a,30)=13 then (print(n," ",a), n:n+1) else (print("Exception at ",a,", ",f(a)), n:3001))); /* f(a)==29 only when a is cong. to 13 (mod 30). No exceptions thru a=89983, n=3000. _Michael G. Kaarhus_, Nov 18 2013 */
%K A082369 nonn,easy
%O A082369 1,1
%A A082369 _Cino Hilliard_, May 11 2003
%E A082369 Simpler name from _Charles R Greathouse IV_, Nov 19 2013