Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #28 May 10 2019 14:42:11
%S 52029,316725,1093345,2811129,6031029,11445709,19879545,32288625,
%T 49760749,73515429,104903889,145409065,196645605,260359869,338429929,
%U 432865569,545808285,679531285,836439489,1019069529,1230089749,1472300205,1748632665,2062150609
%N An infinite sequence of 4-digit half-palindromes.
%C For the definition of k-digit half-palindromes, see A191279. Although there exist infinitely many polynomials taking only 3-digit half-palindrome values (see comment to A191279), only two polynomials are known with all values 4-digit half-palindromes. They are the polynomials P(n) which were discovered in SeqFan Discussion list from Mar 14 2011 and its double.
%C The sequence lists values of P(n). All these are odd. For a given k>=5, up to now it is unknown if there are polynomials taking only k-digit half-palindrome values and it is unknown whether there exist infinitely many such numbers.
%C Conjecture. For every k>=2, there exists a polynomial of degree k taking only k-digit half-palindrome values.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = (2*n+2)(14*n+9)^3+(3*n+3)*(14*n+9)^2+(5*n+3)*(14*n+9)+(2*n+1) = (2*n+1)*(14*n+11)^3 +(5*n+3)*(14*n+11)^2 +(3*n+3)*(14*n+11) +(2*n+2), such that the bases b < c are b=14*n+9, c=14*n+11.
%F G.f. -x*(52029+56580*x+30010*x^2-8636*x^3+1729*x^4) / (x-1)^5. - _R. J. Mathar_, Jul 01 2012
%t LinearRecurrence[{5,-10,10,-5,1},{52029,316725,1093345,2811129,6031029},20] (* _Harvey P. Dale_, Sep 19 2018 *)
%Y Cf. A002113, A006995, A059809, A191279.
%K nonn,easy,base
%O 1,1
%A _Vladimir Shevelev_, May 30 2011