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%I #108 May 06 2025 11:26:10
%S 0,1,9,40,120,280,552,968,1560,2360,3400,4712,6328,8280,10600,13320,
%T 16472,20088,24200,28840,34040,39832,46248,53320,61080,69560,78792,
%U 88808,99640,111320,123880,137352,151768,167160,183560,201000,219512,239128,259880,281800
%N a(n) = A008412(n-1) + A008412(n-2) for n>1, a(0)=0, a(1)=1.
%C Let's iteratively apply the summation of two consecutive terms to A000292. It generates A000330, then A005900, then A001845, then A008412, then this sequence. Every sequence in this series starts with 1 followed by the sum of 1 and the next term in the previous sequence; because of that, for A008412 and this sequence, the initial term(s) are exceptions from the general formula.
%C From _Leo James Borcherding_, May 23 2017: (Start)
%C a(n) = f(9,n), where f(k,n) is the set of all series derived from the anchored series.
%C k = (All whole numbers (including negative values))
%C n = (All whole numbers >= 1)
%C The anchored series is f(0,n).
%C See the attached file for an in-depth explanation of the family of tetrahedron sequences that f(9,n) (this sequence) is a part of.
%C A Visual Representation of the summation process is as follows:
%C a.) f(7,n) + f(7,n-1) = f(8,n)
%C b.) f(8,n) + f(8,n-1) = f(9,n)
%C a.) b.)
%C 1 + 0 = 1 1 + 0 = 1
%C 7 + 1 = 8 8 + 1 = 9
%C 25 + 7 = 32 32 + 8 = 40
%C 63 + 25 = 88 88 + 32 = 120
%C 129 + 63 = 192 192 + 88 = 280
%C 231 + 129 = 360 360 + 192 = 552
%C 377 + 231 = 608 608 + 360 = 968
%C 575 + 377 = 952 952 + 608 = 1560
%C ... iterate infinitely many times. (End)
%D William Dunham, Euler The Master of Us All, The Mathematical Association of America, 1999 p. 40.
%D Joseph and Frances Gies, Leonard of Pisa and the New Mathematics of the Middle Ages, Thomas Y. Crowell Company New York, 1969, p. 78.
%H Colin Barker, <a href="/A287324/b287324.txt">Table of n, a(n) for n = 0..1000</a>
%H Leo James Borcherding, <a href="/A287324/a287324_2.pdf">Tetrahedron Family of f(k,n)</a>
%H Leo James Borcherding, <a href="/A287324/a287324_4.pdf">The Unified Tetrahedral Family Defined by Pascal's Triangle</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: x*(x + 1)^5 / (x - 1)^4.
%F a(n) = 8*(n - 1)*((n - 1)^2 + 2)/3 + 8*(n - 2)*((n - 2)^2 + 2)/3 = 8*(2*n - 3)*(n^2 - 3*n + 5)/3 for n>2, a(0)=0, a(1)=1, a(2)=9.
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>6. - _Colin Barker_, Jun 05 2017
%o (PARI) concat(0, Vec(x*(x+1)^5/(x-1)^4 + O(x^30))) \\ _Michel Marcus_, May 24 2017
%Y Cf. A000292, A000330, A001845, A002623, A005900, A006918, A008412.
%K nonn,easy
%O 0,3
%A _Leo James Borcherding_, May 23 2017