%I #20 Sep 01 2024 09:54:30
%S 1,7,33,124,397,1125,2893,6871,15269,32065,64130,122916,226922,405218,
%T 702378,1185263,1952198,3145208,4966118,7697483,11729498,17594247,
%U 26008887,37929627,54618663,77726559,109392935,152368731,210163767,287223815,389141943
%N a(n) = 1/11*(binomial(n,11) - floor(n/11)).
%C Let p be a prime. Saikia and Vogrinc have proved that 1/p*{binomial(n,p) - floor(n/p)} is an integer sequence. The present sequence is the case p = 11. Other cases are A002620 (p = 2), A014125 (p = 3), A215052 (p = 5) and A215053 (p = 7).
%H M. P. Saikia and J. Vogrinc, <a href="http://arxiv.org/abs/1207.6707">A simple number theoretic result</a> arxiv.1207.6707v1 [mathNT]
%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 2, -11, 55, -165, 330, -462, 462, -330, 165, -55, 11, -1).
%F a(n) = 1/11*(binomial(n,11) - floor(n/11)).
%F O.g.f.: sum_{n>=0} a(n)*x^n = x^12*(1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8)/((1-x^11)*(1-x)^11) = x^12*(1 + 7*x + 33*x^2 + 124*x^3 + ...). The numerator polynomial 1 - 4*x + 11*x^2 - 19*x^3 + 23*x^4 - 19*x^5 + 11*x^6 - 4*x^7 + x^8 is the negative of the row generating polynomial for row 11 of A178904.
%t Table[(Binomial[n,11]-Floor[n/11])/11,{n,12,50}] (* _Harvey P. Dale_, Aug 06 2012 *)
%o (Maxima) A215054(n):=1/11*(binomial(n,11) - floor(n/11))$ makelist(A215054(n),n,12,30); /* _Martin Ettl_, Oct 25 2012 */
%Y A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215053(p = 7).
%Y Partial sums of A032169.
%K nonn,easy
%O 12,2
%A _Peter Bala_, Aug 01 2012