1,2

Also least odd number with exactly n divisors. - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 30 2006

If n is prime, a(n)=3^(n-1). - Zak Seidov (zakseidov(AT)yahoo.com), Apr 18 2006

a(2n-1) = {1,9,81,729,225,59049,...} are the perfect squares. A122842[n] = Sqrt[ a(2n-1) ] = {1,3,9,27,15,243,729,45,6561,19683,135,177147,225,105,4782969,14348907,1215,...}. - Alexander Adamchuk (alex(AT)kolmogorov.com), Sep 13 2006

Also the least number k such that there are n partitions of k whose elements are consecutive integers. i.e.; 1=1, 3=1+2=3, 9=2+3+4=4+5=9, 15=1+2+3+4+5=4+5+6=7+8=15, etc. - Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 02 2007

Don Reble, Table of n, a(n) for n = 1..2000

T. Verhoeff, Rectangular and Trapezoidal Arrangements, J. Integer Sequences, Vol. 2, 1999, #99.1.6.

a(n) = A119265(n,n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 11 2006

It was sugested by Alexander Adamchuk that for all n >= 1, we have a(3^(n-1)) = (p(n)#/2)^2 = (A002110(n)/2)^2 = A070826(n)^2. But this is false! E.g. (p(n)#/2)^2=3^2 5^2 7^2 ...23^2 29^2 does indeed have 3^9 odd factors, but it is greater than 3^8*5^2 7^2 ...23^2 which has 9*3*3*3*3*3*3*3 = 9*3^7=3^9 odd factors. - Richard Sabey, Oct 06 2007.

a(A053640(m)) = a(A000005(A053624(m))) = A053624(m). - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Apr 20 2008

A122842 = Sqrt[ a(2n-1) ].

Cf. A001227, A005179, A002110, A070826.

Cf. A000005, A053640, A053624.

Sequence in context: A062804 A110960 A050869 this_sequence A083556 A015664 A134137

Adjacent sequences: A038544 A038545 A038546 this_sequence A038548 A038549 A038550

nonn,nice

Tom Verhoeff (Tom.Verhoeff(AT)acm.org)

Corrected by Ron Knott (R.Knott(AT)altavista.net), Feb 22, 2001.

a(30) from Zak Seidov (zakseidov(AT)yahoo.com), Apr 18 2006

a(32)-a(34) from Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 30 2006