cp's OEIS Frontend

A189391 The minimum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 0 to n, and each number in a higher row is the sum of the two numbers directly below it.

%I A189391 #67 Sep 08 2022 08:45:56
%S A189391 0,1,3,8,19,44,98,216,467,1004,2134,4520,9502,19928,41572,86576,
%T A189391 179587,372044,768398,1585416,3263210,6711176,13775068,28255568,
%U A189391 57863214,118430584,242061468,494523536,1009105372,2058327344,4194213448
%N A189391 The minimum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 0 to n, and each number in a higher row is the sum of the two numbers directly below it.
%C A189391 This is the Riordan transform of A000217 (triangular numbers) with the Riordan matrix (of the Bell type) A053121 (inverse of the Chebyshev S Bell matrix). See the resulting formulae below. - _Wolfdieter Lang_, Feb 18 2017.
%C A189391 Conjecture: a(n) is also half the sum of the "cuts-resistance" (see A319416, A319420, A319421) of all binary vectors of length n (see Lenormand, page 4). - _N. J. A. Sloane_, Sep 20 2018
%H A189391 Nathaniel Johnston, <a href="/A189391/b189391.txt">Table of n, a(n) for n = 0..1000</a>
%H A189391 F. Disanto and S. Rinaldi, <a href="http://www.mat.unisi.it/newsito/puma/public_html/22_1/2-disanto_rinaldi.pdf">Symmetric convex permutominoes and involutions</a>, PU. M. A., Vol. 22 (2011), No. 1, pp. 39-60. - From _N. J. A. Sloane_, May 04 2012
%H A189391 Steven R. Finch, <a href="https://arxiv.org/abs/1802.04615">How far might we walk at random?</a>, arXiv:1802.04615 [math.HO], 2018.
%H A189391 Claude Lenormand, <a href="/A318921/a318921.pdf">Deux transformations sur les mots</a>, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003. - _N. J. A. Sloane_, Sep 20 2018
%F A189391 If n even, a(n) = (n+1/2)*binomial(n,n/2) - 2^(n-1); if n odd, a(n) = ((n+1)/2)*binomial(n+1,(n+1)/2) - 2^(n-1). - _N. J. A. Sloane_, Nov 01 2018
%F A189391 a(n) = Sum_{k=0..floor((n-1)/2)} (2*n-4*k-1)*binomial(n,k).
%F A189391 G.f.: (2*x+sqrt(1-4*x^2)-1) / (2*(2*x-1)^2). - _Alois P. Heinz_, Feb 09 2012
%F A189391 a(n) ~ 2^n * (sqrt(2n/Pi)- 1/2). - _Vaclav Kotesovec_, Mar 16 2014 (formula simplified by _Lewis Chen_, May 25 2017)
%F A189391 D-finite with recurrence n*a(n) + (n-5)*a(n-1) + 2*(-5*n+6)*a(n-2) + 4*(-n+8)*a(n-3) + 24*(n-3)*a(n-4) = 0. - _R. J. Mathar_, Jan 04 2017
%F A189391 From _Wolfdieter Lang_, Feb 18 2017:(Start)
%F A189391 a(n) = Sum_{m=0..n} A053121(n, m)*A000217(m), n >= 0.
%F A189391 G.f.: c(x^2)*Tri(x*c(x^2)), with c and Tri the g.f. of A000108 and A000217, respectively. See the explicit form of the g.f. given above by _Alois P. Heinz_.
%F A189391 (End)
%F A189391 2*a(n) = A152548(n)-2^n. - _R. J. Mathar_, Jun 17 2021
%e A189391 For n = 4 consider the triangle:
%e A189391 ....19
%e A189391 ...8  11
%e A189391 ..5  3  8
%e A189391 .4  1 2  6
%e A189391 3  1 0 2  4
%e A189391 This triangle has 19 at its apex and no other such triangle with the numbers 0 - 4 on its base has a smaller apex value, so a(4) = 19.
%p A189391 a:=proc(n)return add((2*n-4*k-1)*binomial(n,k),k=0..floor((n-1)/2)): end:
%p A189391 seq(a(n),n=0..50);
%t A189391 CoefficientList[Series[(2*x+Sqrt[1-4*x^2]-1) / (2*(2*x-1)^2), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Mar 16 2014 *)
%o A189391 (PARI) A189391(n)=sum(i=0,(n-1)\2,(2*n-4*i-1)*binomial(n,i))  \\ _M. F. Hasler_, Jan 24 2012
%o A189391 (Magma) m:=30; R<x>:=PowerSeriesRing(Rationals(), m); [0] cat Coefficients(R!((2*x+Sqrt(1-4*x^2)-1)/(2*(2*x-1)^2))); // _G. C. Greubel_, Aug 24 2018
%Y A189391 Cf. A066411, A099325, A189390, A189162, A000217, A053121, A281862.
%Y A189391 Cf. also A319416, A319420, A319421.
%K A189391 easy,nonn
%O A189391 0,3
%A A189391 _Nathaniel Johnston_, Apr 20 2011