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%I #24 Mar 14 2018 21:04:17
%S 1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16383,32752,65399,
%T 130238,258096,507624,988116,1898712,3593934,6690448,12236830,
%U 21977516,38754732,67108864,114159428,190876696,313889477,508019104,809785133,1272196666
%N a(n) = Sum_{k=0..13} binomial(n, k).
%C a(n) is the number of compositions (ordered partitions) of n+1 into fourteen or fewer parts.
%C a(n) = sum(binomial(n+1, 2k-1), for k=1..7).
%C a(n) is the sum of the first fourteen terms in the n-th row of Pascal's triangle.
%H Robert Israel, <a href="/A219676/b219676.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = 1 +(n^13 -65*n^12 +2015*n^11 -37609*n^10 +470613*n^9 -4081935*n^8 +25378925*n^7 -110205667*n^6 +351042406*n^5 -657328100*n^4 +1303568760*n^3 +771653376*n^2 +4546558080*n)/13!. - corrected by Mokhtar Mohamed, Dec 01 2012
%F G.f.: (1 - 12*x + 67*x^2 - 230*x^3 + 541*x^4 - 920*x^5 + 1163*x^6 - 1106*x^7 + 791*x^8 - 420*x^9 + 161*x^10 - 42*x^11 + 7*x^12)/(1-x)^14.
%F a(n) = 2*a(n-1), for 1 <= n <= 13, with a(0) = 1, a(n) = 2*a(n-1) - C(n-1, 13), for n > 13.
%e a(14)= 16383 because there are 2^14 = 16384 compositions of 15 into any size parts but one of the compositions (1 + 1 + ... + 1 = 15) has more than fourteen parts.
%e When 1 <= n <= 13, a(7) = 2*a(6) = 2*64= 128, a(13) = 2*a(12) = 2*4096 = 8192.
%e When n > 13, a(14) = 2*a(13) - C(13, 13) = 2*8192 - 1 = 16383, a(15) = 2*a(14) - C(14, 13) = 2*16383 - 14 = 32766 - 14 = 32752.
%p f:= n -> add(binomial(n,k),k=0..13):
%p map(f, [$0..100]); # _Robert Israel_, Mar 14 2018
%t Table[Sum[Binomial[n, k], {k, 0, 13}], {n, 0, 40}] (* _T. D. Noe_, Nov 26 2012 *)
%Y Cf. A000127, A006261, A008859, A008860, A008861, A008862, A008863, A219531.
%K nonn,easy
%O 0,2
%A _Mokhtar Mohamed_, Nov 24 2012
%E Corrected and extended by _T. D. Noe_, Nov 26 2012
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