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 A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n*(n + 1)*(8*n - 5)/6. (Formerly M4791) 15
 0, 1, 11, 38, 90, 175, 301, 476, 708, 1005, 1375, 1826, 2366, 3003, 3745, 4600, 5576, 6681, 7923, 9310, 10850, 12551, 14421, 16468, 18700, 21125, 23751, 26586, 29638, 32915, 36425, 40176, 44176, 48433, 52955, 57750 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Binomial transform of [1, 10, 17, 8, 0, 0, 0,...] = (1, 11, 38, 90,...). - Gary W. Adamson, Mar 18 2009 This sequence is related to A000384 by a(n) = n*A000384(n) - Sum_{i=0..n-1} A000384(i) and this is the case d=4 in the identity n*(n*(d*n-d+2)/2) - Sum_{k=0..n-1} k*(d*k-d+2)/2 = n*(n+1)*(2*d*n - 2*d + 3)/6. - Bruno Berselli, Apr 21 2010 For n>0, (a(n)) is the principal diagonal of the convolution array A213750. - Clark Kimberling, Jun 20 2012 From Ant King, Oct 30 2012: (Start) The partial sums of the figurate decagonal numbers A001107. For n>1, the digital roots of this sequence A010888(A007585(n)) form the purely periodic 27-cycle {1,2,2,9,4,4,8,6,6,7,8,8,6,1,1,5,3,3,4,5,5,3,7,7,2,9,9}. For n>1, the unitsâ€™ digits of this sequence A010879(A007585(n)) form the purely periodic 20-cycle {1,1,8,0,5,1,6,8,5,5,6,6,3,5,0,6,1,3,0,0}. (End) REFERENCES A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194. E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93. N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Harvey P. Dale, Table of n, a(n) for n = 0..1000 B. Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian). Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1) FORMULA a(n) = (8*n-5)*binomial(n+1, 2)/3. G.f.: x*(1+7*x)/(1-x)^4. a(n) = (8*n^3 + 3*n^2 - 5*n)/6. - Vincenzo Librandi, Aug 01 2010 a(0)=0, a(1)=1, a(2)=11, a(3)=38, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 20 2011 From Ant King, Oct 30 2012: (Start) a(n) = a(n-1) + n*(4*n-3). a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 8. a(n) = (n+1)*(2*A001107(n) + n)/6. a(n) = A000292(n) + 7*A000292(n-1). a(n) = A007584(n) + A000292(n-1). a(n) = A000217(n) + 8*A000292(n-1). a(n) = binomial(n+2,3) + 7*binomial(n+1,3). Sum_{n>=1} 1/a(n) = 6*(4*pi*(sqrt(2)-1) + 4*(8-sqrt(2))*log(2) + 8*sqrt(2)*log(2-sqrt(2))-5)/65 =  1.145932345... (End) a(n) = Sum_{i=0..n-1} (n-i)*(8*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014 E.g.f.: x*(6 + 27*x + 8*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017 MAPLE seq(n*(n+1)*(8*n-5)/6, n=0..40); # G. C. Greubel, Aug 30 2019 MATHEMATICA Table[n(n+1)(8n-5)/6, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *) LinearRecurrence[{4, -6, 4, -1}, {0, 1, 11, 38}, 40] (* Harvey P. Dale, Dec 20 2011 *) PROG (PARI) a(n)=(8*n^3+3*n^2-5*n)/6 \\ Charles R Greathouse IV, Sep 24 2015 (MAGMA) [n*(n+1)*(8*n-5)/6: n in [0..40]]; // G. C. Greubel, Aug 30 2019 (Sage) [n*(n+1)*(8*n-5)/6 for n in (0..40)] # G. C. Greubel, Aug 30 2019 (GAP) List([0..40], n-> n*(n+1)*(8*n-5)/6); # G. C. Greubel, Aug 30 2019 CROSSREFS Cf. A000384. Cf. A093565 ((8, 1) Pascal, column m=3). Partial sums of A001107. Cf. similar sequences listed in A237616. Sequence in context: A139276 A010002 A143109 * A024202 A213775 A133258 Adjacent sequences:  A007582 A007583 A007584 * A007586 A007587 A007588 KEYWORD nonn,easy,nice AUTHOR STATUS approved

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Last modified October 14 00:10 EDT 2019. Contains 327990 sequences. (Running on oeis4.)