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15-gonal (or pentadecagonal) numbers: n*(13n-11)/2.
16

%I #57 Jun 11 2024 09:33:19

%S 0,1,15,42,82,135,201,280,372,477,595,726,870,1027,1197,1380,1576,

%T 1785,2007,2242,2490,2751,3025,3312,3612,3925,4251,4590,4942,5307,

%U 5685,6076,6480,6897,7327,7770,8226,8695,9177,9672,10180,10701

%N 15-gonal (or pentadecagonal) numbers: n*(13n-11)/2.

%C Sequence found by reading the line from 0, in the direction 0, 15,... and the parallel line from 1, in the direction 1, 42,..., in the square spiral whose vertices are the generalized 15-gonal numbers. - _Omar E. Pol_, Jul 18 2012

%D Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 189.

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.

%H Harvey P. Dale, <a href="/A051867/b051867.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Pol#polygonal_numbers">Index to sequences related to polygonal numbers</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(1+12*x)/(1-x)^3. - _Bruno Berselli_, Feb 04 2011

%F a(n) = 13*n+a(n-1)-12 (with a(0)=0) - _Vincenzo Librandi_, Aug 06 2010

%F a(0)=0, a(1)=1, a(2)=15, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). - _Harvey P. Dale_, Feb 29 2012

%F a(13*a(n)+79*n+1) = a(13*a(n)+79*n) + a(13*n+1). - _Vladimir Shevelev_, Jan 24 2014

%F Product_{n>=2} (1 - 1/a(n)) = 13/15. - _Amiram Eldar_, Jan 21 2021

%F E.g.f.: exp(x)*(x + 13*x^2/2). - _Nikolaos Pantelidis_, Feb 06 2023

%F a(n) = A000326(3*n-2) - 7*(n-1)^2. In general, if we let P(k,n) = the n-th k-gonal number, then P(5*k,n) = P(5,k*n-k+1) - A005449(k-1)*(n-1)^2. More generally, if we let SP(k,n) = the n-th second k-gonal number, then for m>2 and k>0, P(m*k,n) = P(m,k*n-k+1) - SP(m,k-1)*(n-1)^2. - _Charlie Marion_, May 21 2024

%t Table[n (13n-11)/2,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,1,15},50] (* _Harvey P. Dale_, Feb 29 2012 *)

%o (PARI) a(n)=n*(13*n-11)/2 \\ _Charles R Greathouse IV_, Oct 07 2015

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, Dec 15 1999