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%I #26 Sep 08 2022 08:46:03
%S 1,5,17,47,113,245,490,919,1634,2778,4546,7198,11074,16611,24363,
%T 35022,49443,68671,93971,126861,169148,222968,290828,375653,480836,
%U 610292,768516,960645,1192525,1470781,1802893,2197276,2663365,3211705,3854046
%N a(n) = 1/7*( binomial(n,7) - floor(n/7) ).
%C Not the same as A011853.
%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 = 7. Other cases are A002620 (p = 2), A014125 (p = 3), A215052 (p = 5) and A215054 (p = 11).
%H Alexandre Laugier and Manjil Saikia, <a href="http://arxiv.org/abs/1209.2373">A characterization of a prime p from the binomial coefficient binomial(n,p)</a>, arXiv:1209.2373 [math.NT], The Mathematics Student, Vol. 83 (1-4), pp 1-7, 2014
%H M. P. Saikia and J. Vogrinc, <a href="http://arxiv.org/abs/1207.6707">A simple number theoretic result</a>, arXiv:1207.6707v1 [math.NT]
%H <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (7, -21, 35, -35, 21, -7, 2, -7, 21, -35, 35, -21, 7, -1).
%F O.g.f.: sum {n>=0} a(n)*x^n = x^8*(1 - 2*x + 3*x^2 - 2*x^3 + x^4)/((1-x^7)*(1-x)^7) = x^8*(1 + 5*x + 17*x^2 + 47*x^3 + ...). The numerator polynomial 1 - 2*x + 3*x^2 - 2*x^3 + x^4 is the negative of the row generating polynomial for row 7 of A178904.
%t Table[(Binomial[n,7]-Floor[n/7])/7,{n,8,50}] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,2,-7,21,-35,35,-21,7,-1},{1,5,17,47,113,245,490,919,1634,2778,4546,7198,11074,16611},40] (* _Harvey P. Dale_, Dec 23 2014 *)
%o (PARI) a(n) = (binomial(n, 7) - n\7) / 7; \\ _Michel Marcus_, Jan 23 2014
%o (Magma) [(Binomial(n, 7)-Floor(n/7))/7: n in [8..50]]; // _Vincenzo Librandi_, Jun 23 2015
%Y Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215054 (p = 11).
%K nonn,easy
%O 8,2
%A _Peter Bala_, Aug 01 2012
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