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A215054
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a(n) = 1/11*(binomial(n,11) - floor(n/11)).
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3
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1, 7, 33, 124, 397, 1125, 2893, 6871, 15269, 32065, 64130, 122916, 226922, 405218, 702378, 1185263, 1952198, 3145208, 4966118, 7697483, 11729498, 17594247, 26008887, 37929627, 54618663, 77726559, 109392935, 152368731, 210163767, 287223815, 389141943
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OFFSET
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12,2
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COMMENTS
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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).
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LINKS
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FORMULA
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a(n) = 1/11*(binomial(n,11) - floor(n/11)).
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.
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MATHEMATICA
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Table[(Binomial[n, 11]-Floor[n/11])/11, {n, 12, 50}] (* Harvey P. Dale, Aug 06 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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