A006002 a(n) = n*(n+1)^2/2.
0, 2, 9, 24, 50, 90, 147, 224, 324, 450, 605, 792, 1014, 1274, 1575, 1920, 2312, 2754, 3249, 3800, 4410, 5082, 5819, 6624, 7500, 8450, 9477, 10584, 11774, 13050, 14415, 15872, 17424, 19074, 20825, 22680, 24642, 26714, 28899, 31200, 33620, 36162, 38829, 41624
Offset: 0
Examples
Let P^n=6^2. The product of the divisors of 36 = 10077796 = 6^9, i.e., for n=2, k=9. - _James A. Raymond_ & Douglas Raymond, Dec 04 2011
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- T. D. Noe, Table of n, a(n) for n = 0..1000
- Quang T. Bach, Roshil Paudyal and Jeffrey B. Remmel, A Fibonacci analogue of Stirling numbers, arXiv preprint arXiv:1510.04310 [math.CO], 2015.
- Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.
- Dexter Jane L. Indong and Gilbert R. Peralta, Inversions of permutations in Symmetric, Alternating, and Dihedral Gropus, JIS, Vol. 11 (2008), Article 08.4.3.
- S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
- S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
- Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
- Index entries for two-way infinite sequences.
Crossrefs
Programs
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GAP
List([0..10^3],n->n*(n+1)^2/2); # Muniru A Asiru, Feb 04 2018
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Haskell
a006002 n = n * (n + 1) ^ 2 `div` 2 -- Reinhard Zumkeller, Dec 12 2012
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Magma
[n*(n+1)^2/2 : n in [0..50]]; // Wesley Ivan Hurt, Nov 17 2014
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Maple
seq(binomial(n+1,2)*(n+1), n=0..36); # Zerinvary Lajos, Apr 25 2007
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Mathematica
Table[(n^3-n^2)/2, {n, 41}] (* Zerinvary Lajos, Mar 21 2007 *) LinearRecurrence[{4,-6,4,-1}, {0,2,9,24}, 40] (* Harvey P. Dale, Aug 14 2012 *) Accumulate @ # (# + 1) & [Range[0,50]] (* Waldemar Puszkarz, Jan 24 2015 *)
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PARI
a(n)=n*(n+1)^2/2
Formula
G.f.: x*(x + 2)/(1 - x)^4. - Michael Somos, Jan 30 2004
a(n) = (n + 1) * binomial(n+1, 2). - Zerinvary Lajos, Jan 10 2006
a(n) = A035006(n+1)/4. - Johannes W. Meijer, Feb 04 2010
a(n) = 2*binomial(n+1, 2) + 3*binomial(n+1, 3). - Gary Detlefs, Jun 06 2010
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Aug 14 2012
a(n) = A045991(n+1)/2. - J. M. Bergot, Aug 10 2013
a(n) = Sum_{j=1..n} Sum_{i=1..j} (2*j - i + 1). - Wesley Ivan Hurt, Nov 17 2014
a(n) = Sum_{i=0..n} n*(n - i) + i. - Bruno Berselli, Jan 13 2016
Sum_{n>0} 1/a(n) = 4 - Pi^2/3. - Jaume Oliver Lafont, Jul 11 2017 [corrected by Amiram Eldar, Jan 28 2022]
E.g.f.: exp(x)*x*(4 + 5*x + x^2)/2. - Stefano Spezia, May 21 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/6 + 4*log(2) - 4. - Amiram Eldar, Jan 28 2022
From J.S. Seneschal, Jun 27 2024: (Start)
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