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A050492 Thickened cube numbers: a(n) = n*(n^2 + (n-1)^2) + (n-1)*2*n*(n-1). 7
1, 14, 63, 172, 365, 666, 1099, 1688, 2457, 3430, 4631, 6084, 7813, 9842, 12195, 14896, 17969, 21438, 25327, 29660, 34461, 39754, 45563, 51912, 58825, 66326, 74439, 83188, 92597, 102690, 113491, 125024, 137313, 150382, 164255, 178956 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
In other words, positive integers k such that 2*k - 1 is a perfect cube. - Altug Alkan, Apr 15 2016
a(n) represents the first term in a sum of (2*n - 1)^3 consecutive integers which equals (2*n - 1)^6. - Patrick J. McNab, Dec 24 2016
LINKS
FORMULA
a(n) = n*(4*n^2-6*n+3).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(1)=1, a(2)=14, a(3)=63, a(4)=172. - Harvey P. Dale, Oct 02 2011
G.f.: x*(1+10*x+13*x^2)/(1-4*x+6*x^2-4*x^3+x^4). - Colin Barker, Jan 04 2012
a(n) = ((2n-1)^3 + 1)/2. - Dave Durgin, May 07 2014
E.g.f.: x*(4*x^2 + 6*x + 1)*exp(x). - G. C. Greubel, Apr 15 2016
EXAMPLE
* * * * *
a(2) = * + * * + * = 14.
* * * * *
MATHEMATICA
Table[n(n^2+(n-1)^2)+(n-1)2n(n-1), {n, 40}] (* or *) LinearRecurrence[ {4, -6, 4, -1}, {1, 14, 63, 172}, 40] (* Harvey P. Dale, Oct 02 2011 *)
PROG
(Magma) [n*(4*n^2-6*n+3): n in [1..40]]; // Vincenzo Librandi, Oct 03 2011
(PARI) a(n)=n*(4*n^2-6*n+3) \\ Charles R Greathouse IV, Nov 10 2015
CROSSREFS
Sequence in context: A022674 A044152 A044533 * A255499 A229739 A339136
KEYWORD
nonn,easy,nice
AUTHOR
Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 27 1999
STATUS
approved

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Last modified April 24 02:28 EDT 2024. Contains 371917 sequences. (Running on oeis4.)