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A051868 16-gonal (or hexadecagonal) numbers: a(n) = n*(7*n-6). 11

%I #77 Apr 28 2023 17:44:09

%S 0,1,16,45,88,145,216,301,400,513,640,781,936,1105,1288,1485,1696,

%T 1921,2160,2413,2680,2961,3256,3565,3888,4225,4576,4941,5320,5713,

%U 6120,6541,6976,7425,7888,8365,8856,9361,9880,10413,10960,11521

%N 16-gonal (or hexadecagonal) numbers: a(n) = n*(7*n-6).

%C Sequence found by reading the line from 0, in the direction 0, 16, ... and the parallel line from 1, in the direction 1, 45, ..., in the square spiral whose vertices are the generalized 16-gonal numbers. - _Omar E. Pol_, Jul 18 2012

%C This is also a star octagonal number: a(n) = A000567(n) + 8*A000217(n-1). - _Luciano Ancora_, Mar 29 2015

%C Let T(n) = A000217(n), the n-th triangular number. Then a(n) = T(n-1) + T(4n-3) - T(2n-4) + T(n-3). In general, let P(k,n) be the n-th k-gonal number. Then for k>1, P(T(k)+1,n) = T(n-1) + T((k-1)n-(k-2)) - T((k-3)n-2(k-3)) + T((k-4)n-3(k-4)) - ... + (-1)^(k+1)*T(n-(k-2)). - _Charlie Marion_, Dec 23 2019

%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 Ivan Panchenko, <a href="/A051868/b051868.txt">Table of n, a(n) for n = 0..1000</a>

%H John Elias, <a href="/A051868/a051868.png">Illustration: 16-gonal Star Configuration</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 a(n) = 14*n + a(n-1) - 13, with n>0, a(0)=0. - _Vincenzo Librandi_, Aug 06 2010

%F G.f.: x*(1+13*x)/(1-x)^3. - _Bruno Berselli_, Feb 04 2011

%F a(0)=0, a(1)=1, a(2)=16; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Harvey P. Dale_, May 07 2011

%F a(14*a(n) + 92*n + 1) = a(14*a(n) + 92*n) + a(14*n+1). - _Vladimir Shevelev_, Jan 24 2014

%F E.g.f.: exp(x)*x*(1 + 7*x). - _Stefano Spezia_, Dec 27 2019

%F a(n) = (4*n-3)^2 - (3*n-3)^2. In general, if we let P(k,n) be the n-th k-gonal number, then P(4k,n) = (k*n-k+1)^2 - ((k-1)*n-k+1)^2. In addition, {P(4k,n)} are the only polygonal number sequences each of whose terms can be written as the difference of two squares. - _Charlie Marion_, Feb 16 2020

%F Product_{n>=2} (1 - 1/a(n)) = 7/8. - _Amiram Eldar_, Jan 22 2021

%t Table[n(7n-6),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,1,16}, 51] (* _Harvey P. Dale_, May 07 2011 *)

%o (PARI) a(n)=n*(7*n-6) \\ _Charles R Greathouse IV_, Jan 24 2014

%Y Cf. A000290, A000566, A051682, A051874, A255187, A282852.

%K nonn,easy

%O 0,3

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

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