cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

Original entry on oeis.org

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335
Offset: 0

Views

Author

Keywords

Comments

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012
From R. K. Guy, Dec 28 2005: (Start)
"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):
0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...
0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...
.....-.-.....+..+.....-..-.....+..+......-...-.......+....
"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.
"Append signs according as the pair have the same (+) or opposite (-) parity.
"Then Euler's pentagonal number theorem is easy to remember:
"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n
where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.
"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."
(End)
The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003
Number of levels in the partitions of n + 1 with parts in {1,2}.
a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007
Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008
From Matthew Vandermast, Oct 28 2008: (Start)
Numbers n for which A000326(n) is a member of A000332. Cf. A145920.
This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)
Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009
Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010
Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011
A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012
a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013
Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013
Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014
(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014
Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016
The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018
Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018
Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018
The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018
Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022
8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022
a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022
Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023
From Peter Bala, Jan 06 2025: (Start)
The sequence terms are the exponents in the expansions of the following infinite products:
1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).
2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....
3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....
4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

Examples

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...
		

References

  • Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
  • Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
  • Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
  • Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).
Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).
Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.
Union of A036498 and A036499.
Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).
Column 1 of A195152.
Squares in APs: A221671, A221672.
Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.
Cf. A002620.

Programs

  • GAP
    a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018
    
  • Haskell
    a001318 n = a001318_list !! n
    a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015
    
  • Magma
    [(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014
    
  • Magma
    [(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014
    
  • Maple
    A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
    A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011
  • Mathematica
    Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
    Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
    CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
    LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
    a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)
  • PARI
    {a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */
    
  • PARI
    {a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */
    
  • Python
    def a(n):
        p = n % 2
        return (n + p)*(3*n + 2 - p) >> 3
    print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022
    
  • Python
    def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024
  • Sage
    @CachedFunction
    def A001318(n):
        if n == 0 : return 0
        inc = n//2 if is_even(n) else n
        return inc + A001318(n-1)
    [A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012
    

Formula

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).
A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006
Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011
a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]
a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011
Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011
G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams
a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003
Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003
a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005
a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005
a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007
If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007
a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011
a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012
a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013
a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A026741(n),
a(n) = a(n+2) - A001651(n),
a(n) = a(n+3) - A184418(n),
a(n) = a(n+4) - A007310(n),
a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),
a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),
a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),
a(n) = a(n+12)- A007310(n)*3,
a(n) = a(n+14)- A001651(n)*7. (End)
a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018
a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013
G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016
a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018
a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018
a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019
Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021
a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022
a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024

A002817 Doubly triangular numbers: a(n) = n*(n+1)*(n^2+n+2)/8.

Original entry on oeis.org

0, 1, 6, 21, 55, 120, 231, 406, 666, 1035, 1540, 2211, 3081, 4186, 5565, 7260, 9316, 11781, 14706, 18145, 22155, 26796, 32131, 38226, 45150, 52975, 61776, 71631, 82621, 94830, 108345, 123256, 139656, 157641, 177310, 198765, 222111, 247456, 274911, 304590
Offset: 0

Views

Author

Keywords

Comments

Number of inequivalent ways to color vertices of a square using <= n colors, allowing rotations and reflections. Group is dihedral group D_8 of order 8 with cycle index (1/8)*(x1^4 + 2*x4 + 3*x2^2 + 2*x1^2*x2); setting all x_i = n gives the formula a(n) = (1/8)*(n^4 + 2*n + 3*n^2 + 2*n^3).
Number of semi-magic 3 X 3 squares with a line sum of n-1. That is, 3 X 3 matrices of nonnegative integers such that row sums and column sums are all equal to n-1. - [Gupta, 1968, page 653; Bell, 1970, page 279]. - Peter Bertok (peter(AT)bertok.com), Jan 12 2002. See A005045 for another version.
Also the coefficient h_2 of x^{n-3} in the shelling polynomial h(x)=h_0*x^n-1 + h_1*x^n-2 + h_2*x^n-3 + ... + h_n-1 for the independence complex of the cycle matroid of the complete graph K_n on n vertices (n>=2) - Woong Kook (andrewk(AT)math.uri.edu), Nov 01 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-4) is equal to the number of 5-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
Starting with offset 1 = binomial transform of [1, 5, 10, 9, 3, 0, 0, 0, ...]. - Gary W. Adamson, Aug 05 2009
Starting with "1" = row sums of triangle A178238. - Gary W. Adamson, May 23 2010
The equation n*(n+1)*(n^2 + n + 2)/8 may be arrived at by solving for x in the following equality: (n^2+n)/2 = (sqrt(8x+1)-1)/2. - William A. Tedeschi, Aug 18 2010
Partial sums of A006003. - Jeremy Gardiner, Jun 23 2013
Doubly triangular numbers are revealed in the sums of row sums of Floyd's triangle.
1, 1+5, 1+5+15, ...
1
2 3
4 5 6
7 8 9 10
11 12 13 14 15
- Tony Foster III, Nov 14 2015
From Jaroslav Krizek, Mar 04 2017: (Start)
For n>=1; a(n) = sum of the different sums of elements of all the nonempty subsets of the sets of numbers from 1 to n.
Example: for n = 6; nonempty subsets of the set of numbers from 1 to 3: {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}; sums of elements of these subsets: 1, 2, 3, 3, 4, 5, 6; different sums of elements of these subsets: 1, 2, 3, 4, 5, 6; a(3) = (1+2+3+4+5+6) = 21, ... (End)
a(n) is also the number of 4-cycles in the (n+4)-path complement graph. - Eric W. Weisstein, Apr 11 2018

Examples

			G.f. = x + 6*x^2 + 21*x^3 + 55*x^4 + 120*x^5 + 231*x^6 + 406*x^7 + 666*x^8 + ...
		

References

  • A. Björner, The homology and shellability of matroids and geometric lattices, in Matroid Applications (ed. N. White), Encyclopedia of Mathematics and Its Applications, 40, Cambridge Univ. Press 1992.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 124, #25, Q(3,r).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. P. Stanley, Enumerative Combinatorics I, p. 292.

Crossrefs

Cf. A006003 (first differences), A165211 (mod 2).
Multiple triangular: A000217, A064322, A066370.
Cf. A006528 (square colorings).
Cf. A236770 (see crossrefs).
Row n=3 of A257493 and row n=2 of A331436 and A343097.
Cf. A000332.
Cf. A000292 (3-cycle count of \bar P_{n+4}), A060446 (5-cycle count of \bar P_{n+3}), A302695 (6-cycle count of \bar P_{n+5}).

Programs

  • Maple
    A002817 := n->n*(n+1)*(n^2+n+2)/8;
  • Mathematica
    a[ n_] := n (n + 1) (n^2 + n + 2) / 8; (* Michael Somos, Jul 24 2002 *)
    LinearRecurrence[{5,-10,10,-5,1}, {0,1,6,21,55},40] (* Harvey P. Dale, Jul 18 2011 *)
    nn=50;Join[{0},With[{c=(n(n+1))/2},Flatten[Table[Take[Accumulate[Range[ (nn(nn+1))/2]], {c,c}],{n,nn}]]]] (* Harvey P. Dale, Mar 19 2013 *)
  • PARI
    {a(n) = n * (n+1) * (n^2 + n + 2) / 8}; /* Michael Somos, Jul 24 2002 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(1-x)^5 + O(x^50))) \\ Altug Alkan, Nov 15 2015
    
  • Python
    def A002817(n): return (m:=n*(n+1))*(m+2)>>3 # Chai Wah Wu, Aug 30 2024

Formula

a(n) = 3*binomial(n+2, 4) + binomial(n+1, 2).
G.f.: x*(1 + x + x^2)/(1-x)^5. - Simon Plouffe (in his 1992 dissertation); edited by N. J. A. Sloane, May 13 2008
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Warut Roonguthai, Dec 13 1999
a(n) = 5a(n-1) - 10a(n-2) + 10a(n-3) - 5a(n-4) + a(n-5) = A000217(A000217(n)). - Ant King, Nov 18 2010
a(n) = Sum(Sum(1 + Sum(3*n))). - Xavier Acloque, Jan 21 2003
a(n) = A000332(n+1) + A000332(n+2) + A000332(n+3), with A000332(n) = binomial(n, 4). - Mitch Harris, Oct 17 2006 and Bruce J. Nicholson, Oct 22 2017
a(n) = Sum_{i=1..C(n,2)} i = C(C(n,2) + 1, 2) = A000217(A000217(n+1)). - Enrique Pérez Herrero, Jun 11 2012
Euler transform of length 3 sequence [6, 0, -1]. - Michael Somos, Nov 19 2015
E.g.f.: x*(8 + 16*x + 8*x^2 + x^3)*exp(x)/8. - Ilya Gutkovskiy, Apr 26 2016
Sum_{n>=1} 1/a(n) = 6 - 4*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) = 1.25269064911978447... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(n)*A000124(n)/2.
a(n) = ((n-1)^4 + 3*(n-1)^3 + 2*(n-1)^2 + 2*n))/8. - Bruce J. Nicholson, Apr 05 2017
a(n) = (A016754(n)+ A007204(n)- 2) / 32. - Bruce J. Nicholson, Apr 14 2017
a(n) = a(-1-n) for all n in Z. - Michael Somos, Apr 17 2017
a(n) = T(T(n)) where T are the triangular numbers A000217. - Albert Renshaw, Jan 05 2020
a(n) = 2*n^2 - n + 6*binomial(n, 3) + 3*binomial(n, 4). - Ryan Jean, Mar 20 2021
a(n) = (A008514(n) - 1)/16. - Charlie Marion, Dec 20 2024

Extensions

More terms from Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

A064706 Square of permutation defined by A003188.

Original entry on oeis.org

0, 1, 2, 3, 5, 4, 7, 6, 10, 11, 8, 9, 15, 14, 13, 12, 20, 21, 22, 23, 17, 16, 19, 18, 30, 31, 28, 29, 27, 26, 25, 24, 40, 41, 42, 43, 45, 44, 47, 46, 34, 35, 32, 33, 39, 38, 37, 36, 60, 61, 62, 63, 57, 56, 59, 58, 54, 55, 52, 53, 51, 50, 49, 48, 80, 81, 82, 83, 85, 84, 87, 86
Offset: 0

Views

Author

N. J. A. Sloane, Oct 13 2001

Keywords

Comments

Inverse of sequence A064707 considered as a permutation of the nonnegative integers.
Not the same as A100282: a(n) = A100282(n) = A100280(A100280(n)) only for n < 64. - Reinhard Zumkeller, Nov 11 2004

Crossrefs

Cf. A064707 (inverse), A165211 (mod 2).
Cf. also A054238, A163233, A302846.

Programs

  • MATLAB
    A = 1; for i = 1:7 B = A(end:-1:1); A = [A (B + length(A))]; end A(A) - 1
    
  • Mathematica
    Array[BitXor[#, Floor[#/4]] &, 72, 0] (* Michael De Vlieger, Apr 14 2018 *)
  • PARI
    a(n)=bitxor(n,n\4)
    
  • PARI
    { for (n=0, 1000, write("b064706.txt", n, " ", bitxor(n, n\4)) ) } \\ Harry J. Smith, Sep 22 2009
    
  • Python
    def A064706(n): return n^ n>>2 # Chai Wah Wu, Jun 29 2022
  • R
    maxn <- 63 # by choice
    b <- c(1,0,0)
    for(n in 4:maxn) b[n] <- b[n-1] - b[n-2] + b[n-3]
    # c(1,b) is A133872
    a <- 1
    for(n in 1:maxn) {
    a[2*n  ] <- 2*a[n] + 1 - b[n]
    a[2*n+1] <- 2*a[n] +     b[n]
    }
    (a <- c(0,a))
    # Yosu Yurramendi, Oct 25 2020
    

Formula

a(n) = A003188(A003188(n)).
a(n) = n XOR floor(n/4), where XOR is binary exclusive OR. - Paul D. Hanna, Oct 25 2004
a(n) = A233280(A180201(n)), n > 0. - Yosu Yurramendi, Apr 05 2017
a(n) = A000695(A003188(A059905(n))) + 2*A000695(A003188(A059906(n))). - Antti Karttunen, Apr 14 2018

Extensions

More terms from David Wasserman, Aug 02 2002

A130198 Single paradiddle. In percussion, the paradiddle is a four-note drum sticking pattern consisting of two alternating notes followed by two notes on the same hand.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1
Offset: 0

Views

Author

Simone Severini, May 16 2007

Keywords

Comments

Also the binary expansion of the constant 5/17 = 2^(-2) + 2^(-5) + 2^(-7) + ... - R. J. Mathar, Mar 27 2009
Period 8: repeat [0, 1, 0, 0, 1, 0, 1, 1]. - Wesley Ivan Hurt, Aug 23 2015

Crossrefs

Cf. A121262, A131078. - Jaume Oliver Lafont, Mar 19 2009
Cf. A165211.

Programs

Formula

From R. J. Mathar, Mar 27 2009: (Start)
a(n) = a(n-8) = a(n-1) - a(n-4) + a(n-5).
G.f.: -x*(1+x^3-x)/((x-1)*(1+x^4)). (End)
a(n) = (1-(-1)^((n+5)*(n+6)*(n^2+11*n+32)/8))/2. - Wesley Ivan Hurt, Aug 23 2015
a(n) = A165211(n+5). - Wesley Ivan Hurt, Aug 23 2015
Showing 1-4 of 4 results.