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Numbers, a(n) where binomial(a(n), 5n-1) == 0 (mod 5) and binomial(a(n), k) != 0 (mod 5) for k != 5n - 1.
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%I #34 Feb 12 2024 13:29:01

%S 8,13,18,23,48,73,98,123,248,373,498,623,1248,1873,2498,3123,6248,

%T 9373,12498,15623,31248,46873,62498,78123,156248,234373,312498,390623,

%U 781248,1171873,1562498,1953123,3906248,5859373,7812498,9765623,19531248,29296873

%N Numbers, a(n) where binomial(a(n), 5n-1) == 0 (mod 5) and binomial(a(n), k) != 0 (mod 5) for k != 5n - 1.

%C Row numbers of Pascal's triangle where only every 5th binomial coefficient in the row is divisible by 5.

%C Numbers of the form (i*5^j) - 2 (i = 2,3,4,5, j >= 1).

%C The binomial coefficients which are multiples of 5 are easily seen in the display of Pascal's triangle given in the link below by _N. J. A. Sloane_.

%D Thomas M. Green, Prime Patterns in Pascal's Triangle, paper in review process, 2013.

%H Harvey P. Dale, <a href="/A224251/b224251.txt">Table of n, a(n) for n = 1..1000</a>

%H N. J. A. Sloane, <a href="/A007318/b007318.txt">First 141 rows of Pascal's triangle, formatted as a simple linear sequence n, a(n)</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,5,-5).

%F a(n) = 5*(n + 1)- 2, for n <= 4; a(5) = (5^2)*2 - 2; a(n)= a(n-1)+ 5*a(n-4)- 5*a(n-5) for n>=6.

%F G.f.: -x*(15*x^4-5*x^3-5*x^2-5*x-8) / ((x-1)*(5*x^4-1)). - _Colin Barker_, Apr 02 2013

%e a(4) = 23. In the 23rd row of Pascal's triangle, the binomial coefficients C(23, 4), C(23, 9), C(23, 14) and C(23, 19) are divisible by 5 and none of the others are.

%e C(23, 4) = 8855 = C(23, 19) and C(23, 9) = 817,190 = C(23, 14).

%t LinearRecurrence[{1,0,0,5,-5},{8,13,18,23,48},40] (* _Harvey P. Dale_, May 15 2016 *)

%Y Cf. A007318 (Pascal's triangle), A181287.

%K nonn,easy

%O 1,1

%A _Thomas M. Green_, Apr 02 2013

%E More terms from _Colin Barker_, Apr 02 2013