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0, 1, 9, 40, 120, 280, 552, 968, 1560, 2360, 3400, 4712, 6328, 8280, 10600, 13320, 16472, 20088, 24200, 28840, 34040, 39832, 46248, 53320, 61080, 69560, 78792, 88808, 99640, 111320, 123880, 137352, 151768, 167160, 183560, 201000, 219512, 239128, 259880, 281800
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OFFSET
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0,3
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COMMENTS
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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.
a(n) = f(9,n), where f(k,n) is the set of all series derived from the anchored series.
k = (All whole numbers (including negative values))
n = (All whole numbers >= 1)
The anchored series is f(0,n).
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.
A Visual Representation of the summation process is as follows:
a.) f(7,n) + f(7,n-1) = f(8,n)
b.) f(8,n) + f(8,n-1) = f(9,n)
a.) b.)
1 + 0 = 1 1 + 0 = 1
7 + 1 = 8 8 + 1 = 9
25 + 7 = 32 32 + 8 = 40
63 + 25 = 88 88 + 32 = 120
129 + 63 = 192 192 + 88 = 280
231 + 129 = 360 360 + 192 = 552
377 + 231 = 608 608 + 360 = 968
575 + 377 = 952 952 + 608 = 1560
... iterate infinitely many times. (End)
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REFERENCES
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William Dunham, Euler The Master of Us All, The Mathematical Association of America, 1999 p. 40.
Joseph and Frances Gies, Leonard of Pisa and the New Mathematics of the Middle Ages, Thomas Y. Crowell Company New York, 1969, p. 78.
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LINKS
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FORMULA
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G.f. : x*(x + 1)^5 / (x - 1)^4.
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.
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
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PROG
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(PARI) concat(0, Vec(x*(x+1)^5/(x-1)^4 + O(x^30))) \\ Michel Marcus, May 24 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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