%I #53 Jan 21 2025 09:02:56
%S 1,5,7,15,35,51,57,77,117,145,155,187,247,287,301,345,425,477,495,551,
%T 651,715,737,805,925,1001,1027,1107,1247,1335,1365,1457,1617,1717,
%U 1751,1855,2035,2147,2185,2301,2501,2625,2667,2795,3015,3151,3197,3337
%N Numbers k such that A067588(k) is an odd number.
%C The terms are exactly the odd pentagonal numbers; that is, they are all the odd numbers of the form k*(3*k-1)/2 where k is an integer. - _James A. Sellers_, Jun 09 2007
%C Apparently groups of two odd pentagonal numbers (A000326, A014632) followed by two odd 2nd pentagonal numbers (A005449), which leads to the conjectured generating function x*(x^2+4*x+1)*(x^4-2*x^3+4*x^2-2*x+1)/((x^2+1)^2*(1-x)^3). - _R. J. Mathar_, Jul 26 2009
%C Odd generalized pentagonal numbers. - _Omar E. Pol_, Aug 19 2011
%C From _Peter Bala_, Jan 10 2025: (Start)
%C The sequence terms are the exponents in the expansion of Sum_{n >= 0} x^(2*n+1)/(Product_{k = 1..2*n+1} 1 + x^(2*k+1)) = x + x^5 - x^7 - x^15 + x^35 + x^51 - x^57 - x^77 + + - - ... (follows from Berndt et al., Theorem 3.3). Cf. A193828.
%C For positive integer m, define b_m(n) = Sum_{k = 1..n} k^(2*m+1)*A000009(k)*A000009(n-k). We conjecture that
%C i) for odd n, b(n)/ n is an integer
%C ii) b(2*n)/n is an integer, which is odd iff n is a member of this sequence.
%C Cf. A067567. (End)
%H Vincenzo Librandi, <a href="/A067589/b067589.txt">Table of n, a(n) for n = 1..1000</a>
%H B. C. Berndt, B. Kim, and A. J. Yee, <a href="http://dx.doi.org/10.1016/j.jcta.2009.07.005">Ramanujan's lost notebook: Combinatorial proofs of identities associated with Heine's transformation or partial theta functions</a>, J. Comb. Thy. Ser. A, 117 (2010), 957-973.
%F Sum_{n>=1} 1/a(n) = Pi/2. - _Amiram Eldar_, Aug 18 2022
%t With[{nn=50},Sort[Select[Table[(n(3n-1))/2,{n,-nn,nn}],OddQ]]] (* _Harvey P. Dale_, Feb 16 2014 *)
%Y Cf. A067567, A067588, A000009.
%Y Cf. A001318, A193828.
%Y Cf. A014493, A128880, A154293.
%K nonn,easy
%O 1,2
%A _Naohiro Nomoto_, Jan 31 2002
%E Corrected by _T. D. Noe_, Oct 25 2006