A152619 -id:A152619 - cp's OEIS frontend

A214731 a(n) = n^3 - 2*n^2 - 1.

-2, -1, 8, 31, 74, 143, 244, 383, 566, 799, 1088, 1439, 1858, 2351, 2924, 3583, 4334, 5183, 6136, 7199, 8378, 9679, 11108, 12671, 14374, 16223, 18224, 20383, 22706, 25199, 27868, 30719, 33758, 36991, 40424, 44063, 47914, 51983, 56276, 60799, 65558, 70559

Offset: 1

Marco Piazzalunga, Jul 27 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A080859, A085490, A144390 (first differences), A152619.

Similar sequences: A152015 (of the type m^3+2m^2-1), A081437 (m^3-2m^2+1).

[n^3-2*n^2-1: n in [1..50]]; // Vincenzo Librandi, Jul 29 2012

A214731:=n->n^3-2*n^2-1: seq(A214731(n), n=1..60); # Wesley Ivan Hurt, Apr 18 2017

Table[n^3 - 2 n^2 - 1, {n, 50}] (* Vincenzo Librandi, Jul 29 2012 *)

a(n)=n^3-2*n^2-1 \\ Charles R Greathouse IV, Jul 27 2012

[n^2*(n-2)-1 for n in range(1,51)] # G. C. Greubel, Dec 31 2023

From Bruno Berselli, Jul 27 2012: (Start)
G.f.: -x*(2-7*x-x^3)/(1-x)^4.
a(n) = A085490(n-1) + 2.
a(n) = A152619(n-2) - 1 for n>1.
a(n) - a(n-2) = A080859(n-2) - 1 for n>2. (End)
E.g.f.: 1 - (1-x)*(1+x)^2*exp(x). - G. C. Greubel, Dec 31 2023

a(3) corrected by Charles R Greathouse IV, Jul 27 2012

A224220 a(n) = smallest number k with property that if the base-n expansion of k is reversed, the result is a nontrivial multiple of k.

Original entry on oeis.org

32, 75, 8, 245, 12, 21, 16, 1089, 15, 1859, 21, 39, 28, 4335, 24, 6137, 24, 57, 40, 11109, 33, 115, 39, 45, 52, 22707, 35, 27869, 40, 93, 64, 55, 51, 47915, 57, 111, 76, 65559, 48, 75809, 56, 129, 88, 99405, 69, 329, 60, 119, 65, 143259, 72, 265, 63, 95, 112, 198417, 87, 219539

Offset: 3

N. J. A. Sloane, Apr 01 2013

In other words, k divides (reversal of k in base n), and (k-reversed)/k > 1.
The numbers are written in base 10.
Theorem: The length of k (in base n) is 2 iff n>=5 and n+1 is composite, otherwise 4.

			The numbers a(n) for n = 3, ..., 11 written in base n are 1012, 1023, 13, 1045, 15, 25, 17, 1089, 14.
For example, 1012 (base 3) = 32 (base 10), and 2101 (base 3) = 64 (base 10) = 2*32.

N. J. A. Sloane, paper in preparation.
See A214927 for further references and links.

Michel Marcus, Table of n, a(n) for n = 3..400
T. J. Kaczynski, Note on a Problem of Alan Sutcliffe, Math. Mag., 41 (1968), 84-86.
Lara Pudwell, Digit Reversal Without Apology, Mathematics Magazine, Vol. 80 (2007), pp. 129-132. Also arXiv:math/0511366 [math.HO], 2005.
Alan Sutcliffe, Integers That Are Multiplied When Their Digits Are Reversed, Mathematics Magazine, 39 (1966), 282-287.

Cf. A214927, A006093, A072668, A152619.

Table[k = 2; While[Nand[IntegerQ@ #, # != 1] &[FromDigits[#, n]/k] &@ Reverse@ IntegerDigits[k, n], k++]; k, {n, 3, 60}] (* Michael De Vlieger, Feb 26 2017 *)

isok(k, n) = {my(rk = fromdigits(Vecrev(digits(k, n)), n)); !(rk % k) && (rk > k);}
a(n) = {my(k = 1); while (!isok(k, n), k++); k;} \\ Michel Marcus, Feb 26 2017

If n=3 or n>3 and n+1 is prime, a(n) = (n^2-1)(n+1) (cf. A152619).

cp's OEIS Frontend

A214731 a(n) = n^3 - 2*n^2 - 1.

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