A215053 a(n) = 1/7*( binomial(n,7) - floor(n/7) ).

1, 5, 17, 47, 113, 245, 490, 919, 1634, 2778, 4546, 7198, 11074, 16611, 24363, 35022, 49443, 68671, 93971, 126861, 169148, 222968, 290828, 375653, 480836, 610292, 768516, 960645, 1192525, 1470781, 1802893, 2197276, 2663365, 3211705, 3854046

Offset: 8

Author

Peter Bala, Aug 01 2012

Keywords

• nonn
• easy

Comments

Not the same as A011853.

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).

Links

• Alexandre Laugier and Manjil Saikia, A characterization of a prime p from the binomial coefficient binomial(n,p), arXiv:1209.2373 [math.NT], The Mathematics Student, Vol. 83 (1-4), pp 1-7, 2014
• M. P. Saikia and J. Vogrinc, A simple number theoretic result, arXiv:1207.6707v1 [math.NT]
• Index entries for linear recurrences with constant coefficients, signature (7, -21, 35, -35, 21, -7, 2, -7, 21, -35, 35, -21, 7, -1).

Crossrefs

Cf. A002620 (p = 2), A014125 (p = 3), A178904, A215052 (p = 5), A215054 (p = 11).

Programs

Magma

[(Binomial(n, 7)-Floor(n/7))/7: n in [8..50]]; // Vincenzo Librandi, Jun 23 2015

Mathematica

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 *)

PARI

a(n) = (binomial(n, 7) - n\7) / 7; \\ Michel Marcus, Jan 23 2014

Formula

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.