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A025012
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Central heptanomial coefficients: largest coefficient of (1+x+...+x^6)^n.
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42
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1, 1, 7, 37, 231, 1451, 9331, 60691, 398567, 2636263, 17538157, 117224317, 786588243, 5295520699, 35751527189, 241958082737, 1641010879207, 11150608945863, 75894449584849, 517331384963959, 3531097638576781
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..200
Doron Zeilberger, EKHAD
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FORMULA
| Conjecture: 3*n*(3*n-1)*(3*n-2)*a(n) +(41*n^3-723*n^2+1132*n-456)*a(n-1) +7*(-383*n^3+2607*n^2-5992*n+4608)*a(n-2) +49*(-83*n^3+1317*n^2-6706*n +10512)*a(n-3) +343*(199*n^3-2487*n^2+10394*n-14496)*a(n-4) +2401*(n-4)*(43*n^2-437*n+1116)*a(n-5) -16807*(n-4)*(n-5)*(5*n-24)*a(n-6) -117649*(n-5)*(n-6)*(n-4)*a(n-7) = 0. [From R. J. Mathar, Feb 21 2010]
I checked that Mathar's conjectured formula fits the 200 terms of the b-file. Note that the coefficients are powers of 7: 49=7^2, 343=7^3, 2401=7^4, 16807=7^5, 117649=7^6. - T. D. Noe, Nov 15 2010
Comment from Doron Zeilberger, Apr 12 2011: The conjectured recurrence can be rigorously proved (and also discovered at the same time, no need for a guessing program) automatically via the Almkvist-Zeilberger algorithm (procedure AZd in the EKHAD link). In fact the program finds a 4th order recurrence, not a 7th order one: 343/3*(4*n+15)*(n+3)*(n+2)*(n+1)/(3*n+7)/(4*n+7)/(11+3*n)/(n+4)*a(n)+98/3*(4*n+15)*(n+3)*(n+2)*(2*n+7) /(11+3*n)/(3*n+10)/(4*n+11)/(n+4)*a(n+1) -14/3*(n+3)*(92*n^3+713*n^2+1747*n+1372)/(3*n+7)/(4*n+7)/(11+3*n)/(n+4)*a(n+2) -2/3*(4*n+15)*(2*n+7)*(37*n^2 +222*n+332)/(11+3*n)/(3*n+10)/(4*n+1)/(n+4)*a(n+3) +a(n+4) = 0.
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MATHEMATICA
| With[{poly=Plus@@(x^Range[0, 6])}, Table[Max[CoefficientList[Expand[poly^i], x]] {i, 0, 20}]] (* From Harvey P. Dale, Mar 09 2011 *)
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CROSSREFS
| Cf. A001405, A002426, A005190, A005191, A018901, A025013, A025014.
Sequence in context: A077239 A046235 A144496 * A039938 A102760 A096965
Adjacent sequences: A025009 A025010 A025011 * A025013 A025014 A025015
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KEYWORD
| easy,nonn
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AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net)
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