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A014370 If n = binomial(b,2)+binomial(c,1), b>c>=0 then a(n) = binomial(b+1,3)+binomial(c+1,2). 5

%I #19 Jan 29 2014 13:39:02

%S 1,2,4,5,7,10,11,13,16,20,21,23,26,30,35,36,38,41,45,50,56,57,59,62,

%T 66,71,77,84,85,87,90,94,99,105,112,120,121,123,126,130,135,141,148,

%U 156,165,166,168,171,175,180,186,193,201,210,220,221,223,226,230,235,241

%N If n = binomial(b,2)+binomial(c,1), b>c>=0 then a(n) = binomial(b+1,3)+binomial(c+1,2).

%C Triangle-tree numbers: a(n) = sum(b(m), m = 1..n), b(m) = 1,1,2,1,2,3,1,2,3,4,... = A002260. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)

%D W. Bruns and J. Herzog, Cohen-Macaulay Rings, Cambridge, 1993, p. 159.

%F a(n*(n+1)/2+m)=n*(n+1)*(n+2)/6 + m*(m+1)/2=A000292(n)+ A000217(m), m=0...n+1, n=1, 2, 3.. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de)

%F a(n) = a(n-1)+A002260(n). As a triangle with n >= k >= 1: a(n, k) =a(n-1, k)+(n-1)*n/2 =a(n, k-1)+k =(n^3-n+3k^2+3k)/6. - _Henry Bottomley_, Nov 14 2001

%F a(n) = b(n) * (b(n) + 1) * (b(n) + 2) / 6 + c(n) * (c(n) + 1) / 2, where b(n) = [sqrt(2 * n) - 1/2] and c(n) = n - b(n) * (b(n) + 1) / 2 - Robert A. Stump (bee_ess107(AT)msn.com), Sep 20 2002

%F As a triangle, T(n,k) = binomial(n+1, 3) + binomial(k+1,2). - _Franklin T. Adams-Watters_, Jan 27 2014

%e The triangle starts:

%e 1

%e 2 4

%e 5 7 10

%e 11 13 16 20

%e 21 23 26 30 35

%p a := 0: for i from 1 to 15 do for j from 1 to i do a := a+j: printf(`%d,`,a); od:od:

%Y Cf. A002260, A000292 (main diagonal), A000217, A014368, A014369, A006046, A050407 (1st column), A005581 (subdiagonal), A071239 (row sums), A212013.

%K nonn,easy,tabl

%O 1,2

%A _N. J. A. Sloane_.

%E More terms from _James A. Sellers_, Feb 05 2000

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Last modified April 24 04:14 EDT 2024. Contains 371918 sequences. (Running on oeis4.)