

A111802


n^2n1 for n>3; a(1)=1; a(2)=2; a(3)=3.


0



1, 2, 3, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549
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OFFSET

1,2


COMMENTS

Inspired by attempt to determine what incorrectlynamed A000519 may really be. Conjecture: Sequence is number of different maindiagonal sums among Latin squares of order n. Confirmed for first five terms. Guaranteed to be an upper bound as the diagonal sum can only be in the range from n to n^2 inclusive and it is impossible for the sum to be n+1 or n^21. There is probably an easy proof that all other sums in this interval can be realized as the only restriction seems to be that it is not permissible for exactly n1 numbers on a diagonal to be identical.


LINKS

Table of n, a(n) for n=1..51.
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

a(n) = n^2n1 = A028387(n2) for n>3; a(n) = n for 1<=n<=3.


EXAMPLE

a(3) = 3, the number of different diagonal sums of all order 3 Latin squares. Their diagonal sums can only be 3, 6 and 9.


MATHEMATICA

Table[If[n<4, n, n^2n1], {n, 60}] (* or *) LinearRecurrence[{3, 3, 1}, {1, 2, 3, 11, 19, 29}, 60] (* Harvey P. Dale, Sep 04 2018 *)


PROG

(PARI) a(n)=if(n>3, n^2n1, n) \\ Charles R Greathouse IV, Dec 20 2011


CROSSREFS

Cf. A028387.
Sequence in context: A306395 A135206 A235629 * A213894 A294668 A095282
Adjacent sequences: A111799 A111800 A111801 * A111803 A111804 A111805


KEYWORD

nonn,easy


AUTHOR

Rick L. Shepherd, Aug 17 2005


STATUS

approved



