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%I #15 Apr 29 2023 21:09:39
%S 0,0,1,1,3,4,2,6,2,6,1,2,1,10,5,7,1,12,1,15,18,12,1,12,21,14,4,12,1,
%T 28,1,29,1,18,6,5,1,20,14,10,1,14,1,34,15,24,1,3,8,16,18,27,1,34,23,
%U 16,1,30,1,16,1,32,17,57,40,56,1,1,47,60,1,54,1,38,36,58,12,66,1,63,10,42,1
%N Binomial(n+p,n) mod n where p=10.
%C Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.
%H Ray Chandler, <a href="/A133620/b133620.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_58060800">Index entries for linear recurrences with constant coefficients</a>, order 58060800.
%F a(n) = binomial(n+p,p) mod n.
%F a(n) = 1 if n is a prime > p, since binomial(n+p,n)==(1+floor(p/n))(mod n), provided n is a prime.
%F a(n) = A001287(n+10) mod n. - _Michel Marcus_, Jul 15 2013; corrected by _Michel Marcus_, Jan 27 2020
%F For n > 58060802, a(n) = 2*a(n-29030400) - a(n-58060800). - _Ray Chandler_, Apr 29 2023
%t Table[Mod[Binomial[n + 10, n], n], {n, 90}] (* _Harvey P. Dale_, Apr 04 2015 *)
%o (PARI) a(n) = binomial(n+10, n) % n \\ _Michel Marcus_, Jul 15 2013
%Y Cf. A000040, A133621-A133625, A133630, A038509, A133634, A133635, A133636.
%Y Cf. A133880, A133890, A133900, A133910, A362686, A362687, A362688, A362689.
%K nonn
%O 1,5
%A _Hieronymus Fischer_, Sep 30 2007
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