A191303 An infinite sequence of 4-digit half-palindromes.

52029, 316725, 1093345, 2811129, 6031029, 11445709, 19879545, 32288625, 49760749, 73515429, 104903889, 145409065, 196645605, 260359869, 338429929, 432865569, 545808285, 679531285, 836439489, 1019069529, 1230089749, 1472300205, 1748632665, 2062150609

Offset: 1

Author

Vladimir Shevelev, May 30 2011

Keywords

• nonn
• easy
• base

Comments

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.

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.

Conjecture. For every k>=2, there exists a polynomial of degree k taking only k-digit half-palindrome values.

Links

• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A002113, A006995, A059809, A191279.

Programs

Mathematica

LinearRecurrence[{5,-10,10,-5,1},{52029,316725,1093345,2811129,6031029},20] (* Harvey P. Dale, Sep 19 2018 *)

Formula

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.

G.f. -x*(52029+56580*x+30010*x^2-8636*x^3+1729*x^4) / (x-1)^5. - R. J. Mathar, Jul 01 2012