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1, 64, 343, 1000, 2197, 4096, 6859, 10648, 15625, 21952, 29791, 39304, 50653, 64000, 79507, 97336, 117649, 140608, 166375, 195112, 226981, 262144, 300763, 343000, 389017, 438976, 493039, 551368, 614125
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| The inverse binomial transform is 1, 63, 216, 162, 0, 0, 0 (0 continued). R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2008
Perfect cubes with digital root 1 in base 10. Proof: perfect cubes are one of (3*s)^3, (3*s+1)^3 or (3*s+2)^3. Digital roots of (3*s)^3 are 0, digital roots of (3*s+1)^3 are 1, and digital roots of (3*s+2)^3 are 8, using trinomial expansion and the multiplicative property of digits roots. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 31 2010]
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REFERENCES
| S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.6.3.
Amarnath Murthy, Fabricating a perfect cube with a given valid digit sum (to be published)
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LINKS
| Harry J. Smith, Table of n, a(n) for n=0,...,1000
Index to sequences with linear recurrences with constant coefficients, signature (4,-6,4,-1).
M. L. Perez et al., eds., Smarandache Notions Journal
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FORMULA
| Sum_{n>=0} 1/a(n) = 2*pi^2 / (81*sqrt(3)) + 13*zeta(3)/27.
O.g.f.: (1+60*x+93*x^2+8*x^3)/(1-x)^4. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2008
Empirical G.f.: (1+60*x+93*x^2+8*x^3)/(1-4*x+6*x^2-4*x^3+x^4) [Colin Barker, Jan 02 2012]
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EXAMPLE
| 343 = (3*2+1)^3
6859 = (3*6+1)^3
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PROG
| (PARI) { b=0; for (n=0, 1000, until (s==1, b++; s=b^3; s-=9*(s\9)); write("b016779.txt", n, " ", b^3) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Jul 18 2009]
(MAGMA) [(3*n+1)^3: n in [0..30]]; // Vincenzo Librandi, May 09 2011
(PARI) a(n)=(3*n+1)^3 \\ Charles R Greathouse IV, Jan 02 2012
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CROSSREFS
| Cf. A054966, A016791.
Sequence in context: A017258 A017366 A186441 * A061102 A137739 A017486
Adjacent sequences: A016776 A016777 A016778 * A016780 A016781 A016782
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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