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%I #19 Oct 17 2022 10:47:29
%S 1,1,2,3,6,10,20,35,56,84,210,330,495,715,1001,3003,4368,6188,8568,
%T 11628,15504,54264,74613,100947,134596,177100,230230,296010,1184040,
%U 1560780,2035800,2629575,3365856,4272048,5379616,6724520,30260340,38608020,48903492
%N a(n) = binomial(n, A002024(n+1)-1) where A002024 is "n appears n times".
%C Number of subsets of [n] in which exactly half of the elements are triangular numbers: a(6) = 20: {}, {1,2}, {1,4}, {1,5}, {2,3}, {2,6}, {3,4}, {3,5}, {4,6}, {5,6}, {1,2,3,4}, {1,2,3,5}, {1,2,4,6}, {1,2,5,6}, {1,3,4,5}, {1,4,5,6}, {2,3,4,6}, {2,3,5,6}, {3,4,5,6}, {1,2,3,4,5,6}. - _Alois P. Heinz_, Oct 11 2022
%H Paul D. Hanna, <a href="/A180272/b180272.txt">Table of n, a(n) for n = 0..1000</a>
%e G.f.: A(x) = 1 + x + 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 20*x^6 +...
%e Terms are shown below in parenthesis as they appear in Pascals triangle:
%e (1);
%e 1,(1);
%e 1,(2),1;
%e 1,3,(3),1;
%e 1,4,(6),4,1;
%e 1,5,(10),5,1;
%e 1,6,15,(20),15,6,1;
%e 1,7,21,(35),35,21,7,1;
%e 1,8,28,(56),70,56,28,8,1;
%e 1,9,36,(84),126,126,84,36,9,1;
%e 1,10,45,120,(210),252,210,120,45,10,1; ...
%o (PARI) {a(n)=binomial(n,(sqrtint(8*n+1)-1)\2)}
%o (Python)
%o from math import comb, isqrt
%o def A180272(n): return comb(n,(isqrt(n+1<<3)+1>>1)-1) # _Chai Wah Wu_, Oct 17 2022
%Y Cf. A000217, A002024.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 17 2011
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