A191279 -id:A191279 - cp's OEIS frontend

A191291 The number of bases >=2 in which n is 3-digit number.

0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5

Offset: 1

Vladimir Shevelev, May 29 2011

Cf. A191279.

Table[Floor[Sqrt[n]]-Floor[CubeRoot[n]],{n,90}] (* Harvey P. Dale, Mar 22 2023 *)

a(n)=sqrtint(n)-floor((n+.5)^(1/3)) \\ Charles R Greathouse IV, Jan 03 2013

a(n) = floor(sqrt(n))-floor(n^(1/3)) = A000196(n) - A048766(n).

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

Original entry on oeis.org

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

Vladimir Shevelev, May 30 2011

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.

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

Cf. A002113, A006995, A059809, A191279.

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

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

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A191291 The number of bases >=2 in which n is 3-digit number.

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