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.

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A287960 Numbers that are the sum of two centered triangular numbers (A005448).

Original entry on oeis.org

2, 5, 8, 11, 14, 20, 23, 29, 32, 35, 38, 41, 47, 50, 56, 62, 65, 68, 74, 77, 83, 86, 89, 92, 95, 104, 110, 113, 116, 119, 128, 131, 137, 140, 146, 149, 155, 167, 170, 173, 176, 182, 185, 194, 197, 200, 203, 209, 212, 218, 221, 230, 236, 239, 245, 251, 254, 263, 266, 272, 275, 278, 281, 284, 293, 299
Offset: 1

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Author

Ilya Gutkovskiy, Jun 03 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 300; f[x_] := Sum[x^(3 k (k - 1)/2 + 1), {k, 1, 20}]^2; Exponent[#, x] & /@ List @@ Normal[Series[f[x], {x, 0, nmax}]]

Formula

8*a(n) = 10+3*A097269(n). - R. J. Mathar, Jul 26 2017

A295288 Binomial transform of the centered triangular numbers A005448.

Original entry on oeis.org

1, 5, 19, 62, 184, 512, 1360, 3488, 8704, 21248, 50944, 120320, 280576, 647168, 1478656, 3350528, 7536640, 16842752, 37421056, 82706432, 181927936, 398458880, 869269504, 1889533952, 4093640704, 8841592832, 19042140160, 40902852608
Offset: 0

Views

Author

Keywords

Comments

The sequence is column 3 of triangle in A207630.
First difference is given by A055818(n+3,3) for n > 0.

Examples

			a(0) = (3*0^2 + 9*0 + 8)*2^(-3) = 8/8 = 1.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, 1994.

Crossrefs

Programs

  • Magma
    I:=[1,5,19]; [n le 3 select I[n] else 6*Self(n-1) -12*Self(n-2) +8*Self(n-3): n in [1..40]]; // G. C. Greubel, Oct 17 2018
  • Maple
    A:=n->(3*n^2+9*n+8)*2^(n-3); seq(A(n), n=0..70);
  • Mathematica
    Table[(3 n^2 + 9 n + 8) 2^(n-3), {n, 0, 70}]
    LinearRecurrence[{6,-12,8}, {1,5,19}, 50] (* G. C. Greubel, Oct 17 2018 *)
  • Maxima
    makelist((3*n^2 + 9*n + 8)*2^(n - 3), n, 0, 70);
    
  • PARI
    a(n) = (3*n^2 + 9*n + 8)*2^(n - 3) \\ Felix Fröhlich, Nov 19 2017
    

Formula

G.f.: (1 - x + x^2)/(1 - 2*x)^3.
a(n+3) = 8*a(n) - 12*a(n+1) + 6*a(n+2).
a(n+1) = 2*a(n) + 3*(n + 2)*2^(n-1).
a(n+1) = 2*a(n) + 3*A001792(n) = 2*a(n) + A001787(n+2) - A001792(n).
a(n) = (3*n^2 + 9*n + 8)*2^(n - 3).
a(n) = (1/8)*A077588(n+2)*A000079(n).

A107118 Numbers that are both centered triangular numbers (A005448) and centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 19, 631, 21421, 727669, 24719311, 839728891, 28526062969, 969046412041, 32919051946411, 1118278719765919, 37988557420094821, 1290492673563457981, 43838762343737476519, 1489227427013510743651, 50589893756115627807601, 1718567160280917834714769
Offset: 1

Views

Author

Richard Choulet, Sep 18 2007

Keywords

Comments

The centered hexagonal numbers are given by 3*p^2 - 3*p + 1 while the centered triangular numbers are given by (3*r^2 + 3*r + 2)/2. A natural number is both of the above numbers if and only if there exist numbers p and r such that 2*(2p-1)^2 = (2*r+1)^2+1. The Diophantine equation X^2 = 2*Y^2 - 1 has the following solutions: X is given by 1, 7, 41, 239, ..., i.e., A002315, and Y is given by A001653. The first equation gives r with 0, 3, 20, 119, 6906, i.e., A001652, and p with 1, 3, 15, 85, 493, ..., i.e., A011900.

Crossrefs

Cf. A003215 (Centered hexagonal numbers), A005448 (Centered triangular numbers).

Programs

  • Mathematica
    a[n_] := 17*n - 7 + Sqrt[288*n^2 - 252*n + 45]; NestList[a, 1, 20] (* Stefan Steinerberger, Sep 18 2007 *)
    LinearRecurrence[{35,-35,1},{1,19,631},30] (* Harvey P. Dale, Jan 16 2016 *)
  • PARI
    Vec(-x*(x^2-16*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100)) \\ Colin Barker, Jan 02 2015

Formula

a(n+2) = 34*a(n+1) - a(n) - 14.
a(n+1) = 17*a(n) - 7 + sqrt(288*a(n)^2 - 252*a(n) + 45).
G.f.: h(z)=(z*(1-16*z+z^2))/((1-z)*(1-34*z+z^2)).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). - Colin Barker, Jan 02 2015
a(n) = (14+(9+6*sqrt(2))*(17+12*sqrt(2))^(-n)+(9-6*sqrt(2))*(17+12*sqrt(2))^n)/32. - Colin Barker, Mar 02 2016

Extensions

More terms from Stefan Steinerberger, Sep 18 2007

A001844 Centered square numbers: a(n) = 2*n*(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

Original entry on oeis.org

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513
Offset: 0

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Author

Keywords

Comments

These are Hogben's central polygonal numbers denoted by
...2...
....P..
...4.n.
Numbers of the form (k^2+1)/2 for k odd.
(y(2x+1))^2 + (y(2x^2+2x))^2 = (y(2x^2+2x+1))^2. E.g., let y = 2, x = 1; (2(2+1))^2 + (2(2+2))^2 = (2(2+2+1))^2, (2(3))^2 + (2(4))^2 = (2(5))^2, 6^2 + 8^2 = 10^2, 36 + 64 = 100. - Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 08 2002
a(n) is also the number of 3 X 3 magic squares with sum 3(n+1). - Sharon Sela (sharonsela(AT)hotmail.com), May 11 2002
For n > 0, a(n) is the smallest k such that zeta(2) - Sum_{i=1..k} 1/i^2 <= zeta(3) - Sum_{i=1..n} 1/i^3. - Benoit Cloitre, May 17 2002
Number of convex polyominoes with a 2 X (n+1) minimal bounding rectangle.
The prime terms are given by A027862. - Lekraj Beedassy, Jul 09 2004
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*(1 + sqrt(2k - 1)). - Lekraj Beedassy, Jun 04 2006
Integers of the form 1 + x + x^2/2 (generating polynomial is Schur's polynomial as in A127876). - Artur Jasinski, Feb 04 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Row sums of triangle A132778. - Gary W. Adamson, Sep 02 2007
Binomial transform of [1, 4, 4, 0, 0, 0, ...]; = inverse binomial transform of A001788: (1, 6, 24, 80, 240, ...). - Gary W. Adamson, Sep 02 2007
Narayana transform (A001263) of [1, 4, 0, 0, 0, ...]. Equals A128064 (unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Dec 29 2007
k such that the Diophantine equation x^3 - y^3 = x*y + k has a solution with y = x-1. If that solution is (x,y) = (m+1,m) then m^2 + (m+1)^2 = k. Note that this Diophantine equation is an elliptic curve and (m+1,m) is an integer point on it. - James R. Buddenhagen, Aug 12 2008
Numbers k such that (k, k, 2*k-2) are the sides of an isosceles triangle with integer area. Also, k such that 2*k-1 is a square. - James R. Buddenhagen, Oct 17 2008
a(n) is also the least weight of self-conjugate partitions having n+1 different odd parts. - Augustine O. Munagi, Dec 18 2008
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) = A153869 * (1, 2, 3, ...). - Gary W. Adamson, Jan 03 2009
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) where a(n) = 2n*(n-1)+1, all tuples of square numbers (X-Y, X, X+Y) are produced by ((m*(a(n)-2n))^2, (m*a(n))^2, (m*(a(n)+2n-2))^2) where m is a whole number. - Doug Bell, Feb 27 2009
Equals (1, 2, 3, ...) convolved with (1, 3, 4, 4, 4, ...). E.g., a(3) = 25 = (1, 2, 3, 4) dot (4, 4, 3, 1) = (4 + 8 + 9 + 4). - Gary W. Adamson, May 01 2009
The running sum of squares taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009
Equals the odd integers convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 25 2009
Equals the triangular numbers convolved with [1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
When the positive integers are written in a square array by diagonals as in A038722, a(n) gives the numbers appearing on the main diagonal. - Joshua Zucker, Jul 07 2009
The finite continued fraction [n,1,1,n] = (2n+1)/(2n^2 + 2n + 1) = (2n+1)/a(n); and the squares of the first two denominators of the convergents = a(n). E.g., the convergents and value of [4,1,1,4] = 1/4, 1/5, 2/9, 9/41 where 4^2 + 5^2 = 41. - Gary W. Adamson, Jul 15 2010
From Keith Tyler, Aug 10 2010: (Start)
Running sum of A008574.
Square open pyramidal number; that is, the number of elements in a square pyramid of height (n) with only surface and no bottom nodes. (End)
For k>0, x^4 + x^2 + k factors over the integers iff sqrt(k) is in this sequence. - James R. Buddenhagen, Aug 15 2010
Create the simple continued fraction from Pythagorean triples to get [2n + 1; 2n^2 + 2n, 2n^2 + 2n + 1]; its value equals the rational number 2n + 1 + a(n) / (4n^4 + 8n^3 + 6n^2 + 2n + 1). - J. M. Bergot, Sep 30 2011
a(n), n >= 1, has in its prime number factorization only primes of the form 4*k+1, i.e., congruent to 1 (mod 4) (see A002144). This follows from the fact that a(n) is a primitive sum of two squares and odd. See Theorem 3.20, p. 164, in the given Niven-Zuckerman-Montgomery reference. E.g., a(3) = 25 = 5^2, a(6) = 85 = 5*17. - Wolfdieter Lang, Mar 08 2012
From Ant King, Jun 15 2012: (Start)
a(n) is congruent to 1 (mod 4) for all n.
The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 5, 4, 7, 5, 7, 4, 5, 1.
The units' digits of the a(n) form a purely periodic palindromic 5-cycle 1, 5, 3, 5, 1.
(End)
Number of integer solutions (x,y) of |x| + |y| <= n. Geometrically: number of lattice points inside a square with vertices (n,0), (0,-n), (-n,0), (0,n). - César Eliud Lozada, Sep 18 2012
(a(n)-1)/a(n) = 2*x / (1+x^2) where x = n/(n+1). Note that in this form, this is the velocity-addition formula according to the special theory of relativity (two objects traveling at 1/(n+1) slower than c relative to each other appear to travel at 1/a(n) less than c to a stationary observer). - Christian N. K. Anderson, May 20 2013 [Corrected by Rémi Guillaume, May 22 2025]
A geometric curiosity: the envelope of the circles x^2 + (y-a(n)/2)^2 = ((2n+1)/2)^2 is the parabola y = x^2, the n=0 circle being the osculating circle at the parabola vertex. - Jean-François Alcover, Dec 02 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; a(n-1) gives the number of internal regions into which the plane is divided (cf. A051890, A046092); a(n-1) = A051890(n) - 1 = A046092(n-1) + 1. - Jaroslav Krizek, Dec 27 2013
a(n) is also, of course, the scalar product of the 2-vector (n, n+1) (or (n+1, n)) with itself. The unique inverse of (n, n+1) as vector in the Clifford algebra over the Euclidean 2-space is (1/a(n))(0, n, n+1, 0) (similarly for the other vector). In general the unique inverse of such a nonzero vector v (odd element in Cl_2) is v^(-1) = (1/|v|^2) v. Note that the inverse with respect to the scalar product is not unique for any nonzero vector. See the P. Lounesto reference, sects. 1.7 - 1.12, pp. 7-14. See also the Oct 15 2014 comment in A147973. - Wolfdieter Lang, Nov 06 2014
Subsequence of A004431, for n >= 1. - Bob Selcoe, Mar 23 2016
Numbers k such that 2k - 1 is a perfect square. - Juri-Stepan Gerasimov, Apr 06 2016
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 574", based on the 5-celled von Neumann neighborhood. - Robert Price, May 13 2016
a(n) is the first integer in a sum of (2*n + 1)^2 consecutive integers that equals (2*n + 1)^4. - Patrick J. McNab, Dec 24 2016
Central elements of odd-length rows of the triangular array of positive integers. a(n) is the mean of the numbers in the (2*n + 1)-th row of this triangle. - David James Sycamore, Aug 01 2018
Intersection of A000982 and A080827. - David James Sycamore, Aug 07 2018
An off-diagonal of the array of Delannoy numbers, A008288, (or a row/column when the array is shown as a square). As such, this is one of the crystal ball sequences. - Jack W Grahl, Feb 15 2021 and Shel Kaphan, Jan 18 2023
a(n) appears as a solution to a "Riddler Express" puzzle on the FiveThirtyEight website. The Jan 21 2022 issue (problem) and the Jan 28 2022 issue (solution) present the following puzzle and include a proof. - Fold a square piece of paper in half, obtaining a rectangle. Fold again to obtain a square with 1/4 the size of the original square. Then make n cuts through the folded paper. a(n) is the greatest number of pieces of the unfolded paper after the cutting. - Manfred Boergens, Feb 22 2022
a(n) is (1/6) times the number of 2 X 2 triangles in the n-th order hexagram with 12*n^2 cells. - Donghwi Park, Feb 06 2024
If k is a centered square number, its index in this sequence is n = (sqrt(2k-1)-1)/2. - Rémi Guillaume, Mar 30 2025.
Row sums of the symmetric triangle of odd numbers [1]; [1, 3, 1]; [1, 3, 5, 3, 1]; [1, 3, 5, 7, 5, 3, 1]; .... - Marco Zárate, Jun 15 2025

Examples

			G.f.: 1 + 5*x + 13*x^2 + 25*x^3 + 41*x^4 + 61*x^5 + 85*x^6 + 113*x^7 + 145*x^8 + ...
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first four such partitions, corresponding to n = 0,1,2,3, i.e., to a(n) = 1,5,13,25, are 1, 3+1+1, 5+3+3+1+1, 7+5+5+3+3+1+1. - _Augustine O. Munagi_, Dec 18 2008
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • Pertti Lounesto, Clifford Algebras and Spinors, second edition, Cambridge University Press, 2001.
  • S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
  • 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).
  • Travers et al., The Mysterious Lost Proof, Using Advanced Algebra, (1976), pp. 27.

Crossrefs

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

Programs

  • Haskell
    a001844 n = 2 * n * (n + 1) + 1
    a001844_list = zipWith (+) a000290_list $ tail a000290_list
    -- Reinhard Zumkeller, Dec 04 2012
    
  • Magma
    [2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013
    
  • Magma
    [n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016
    
  • Maple
    A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[2n(n + 1) + 1, {n, 0, 50}]
    FoldList[#1 + #2 &, 1, 4 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
    maxn := 47; Flatten[Table[SeriesCoefficient[Series[(n + (n - 1)*x)/(1 - x)^2, {x, 0, maxn}], k], {n, maxn}, {k, n - 1, n - 1}]] (* L. Edson Jeffery, Aug 24 2014 *)
    CoefficientList[ Series[-(x^2 + 2x + 1)/(x - 1)^3, {x, 0, 48}], x] (* or *)
    LinearRecurrence[{3, -3, 1}, {1, 5, 13}, 48] (* Robert G. Wilson v, Aug 01 2018 *)
    Total/@Partition[Range[0,50]^2,2,1] (* Harvey P. Dale, Dec 05 2020 *)
    Table[ j! Coefficient[Series[Exp[x]*(1 + 4*x + 2*x^2), {x, 0, 20}], x,
    j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 07 2023 *)
  • PARI
    {a(n) = 2*n*(n+1) + 1};
    
  • PARI
    x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016
    
  • Python
    print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021
  • Sage
    [i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008
    

Formula

a(n) = 2*n^2 + 2*n + 1 = n^2 + (n+1)^2.
a(n) = 1 + 3 + 5 + ... + 2*n-1 + 2*n+1 + 2*n-1 + ... + 3 + 1. - Amarnath Murthy, May 28 2001
a(n) = 1/real(z(n+1)) where z(1)=i, (i^2=-1), z(k+1) = 1/(z(k)+2i). - Benoit Cloitre, Aug 06 2002
Nearest integer to 1/Sum_{k>n} 1/k^3. - Benoit Cloitre, Jun 12 2003
G.f.: (1+x)^2/(1-x)^3.
E.g.f.: exp(x)*(1+4x+2x^2).
a(n) = a(n-1) + 4n.
a(-n) = a(n-1).
a(n) = A064094(n+3, n) (fourth diagonal).
a(n) = 1 + Sum_{j=0..n} 4*j. - Xavier Acloque, Oct 08 2003
a(n) = A046092(n)+1 = (A016754(n)+1)/2. - Lekraj Beedassy, May 25 2004
a(n) = Sum_{k=0..n+1} (-1)^k*binomial(n, k)*Sum_{j=0..n-k+1} binomial(n-k+1, j)*j^2. - Paul Barry, Dec 22 2004
a(n) = ceiling((2n+1)^2/2). - Paul Barry, Jul 16 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=5, a(2)=13. - Jaume Oliver Lafont, Dec 02 2008
a(n)*a(n-1) = 4*n^4 + 1 for n > 0. - Reinhard Zumkeller, Feb 12 2009
Prefaced with a "1" (1, 1, 5, 13, 25, 41, ...): a(n) = 2*n*(n-1)+1. - Doug Bell, Feb 27 2009
a(n) = sqrt((A056220(n)^2 + A056220(n+1)^2) / 2). - Doug Bell, Mar 08 2009
a(n) = floor(2*(n+1)^3/(n+2)). - Gary Detlefs, May 20 2010
a(n) = A000330(n) - A000330(n-2). - Keith Tyler, Aug 10 2010
a(n) = A069894(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Jun 12 2012
Sum_{n>=0} 1/a(n) = (Pi/2)*tanh(Pi/2) = 1.4406595199775... = A228048. - Ant King, Jun 15 2012
a(n) = A209297(2*n+1,n+1). - Reinhard Zumkeller, Jan 19 2013
a(n)^3 = A048395(n)^2 + A048395(-n-1)^2. - Vincenzo Librandi, Jan 19 2013
a(n) = A000217(2n+1) - n. - Ivan N. Ianakiev, Nov 08 2013
a(n) = A251599(3*n+1). - Reinhard Zumkeller, Dec 13 2014
a(n) = A101321(4,n). - R. J. Mathar, Jul 28 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = Sum_{k=0..n} A008574(k).
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = exp(-1) = A068985. (End)
a(n) = 4 * A000217(n) + 1. - Bruce J. Nicholson, Jul 10 2017
a(n) = A002522(n) + A005563(n) = A002522(n+1) + A005563(n-1). - Bruce J. Nicholson, Aug 05 2017
Sum_{n>=0} a(n)/n! = 7*e. Sum_{n>=0} 1/a(n) = A228048. - Amiram Eldar, Jun 20 2020
a(n) = A000326(n+1) + A000217(n-1). - Charlie Marion, Nov 16 2020
a(n) = Integral_{x=0..2n+2} |1-x| dx. - Pedro Caceres, Dec 29 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi)*sinh(Pi/2). (End)
a(n) = A001651(n+1) + 1 - A028242(n). - Charlie Marion, Apr 05 2022
a(n) = A016754(n) - A046092(n). - Leo Tavares, Sep 16 2022
For n>0, a(n) = A101096(n+2) / 30. - Andy Nicol, Feb 06 2025
From Rémi Guillaume, Apr 21 2025: (Start)
a(n) = (2*A003215(n)+1)/3.
a(n) = (4*A005448(n+1)-1)/3.
a(n) + a(n-1) = A001845(n) - A001845(n-1), for n >= 1.
a(n) = (A005917(n+1))/(2n+1). (End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A003215 Hex (or centered hexagonal) numbers: 3*n*(n+1)+1 (crystal ball sequence for hexagonal lattice).

Original entry on oeis.org

1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, 469, 547, 631, 721, 817, 919, 1027, 1141, 1261, 1387, 1519, 1657, 1801, 1951, 2107, 2269, 2437, 2611, 2791, 2977, 3169, 3367, 3571, 3781, 3997, 4219, 4447, 4681, 4921, 5167, 5419, 5677, 5941, 6211, 6487, 6769
Offset: 0

Views

Author

Keywords

Comments

The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
Crystal ball sequence for A_2 lattice. - Michael Somos, Jun 03 2012
Sixth spoke of hexagonal spiral (cf. A056105-A056109).
Number of ordered integer triples (a,b,c), -n <= a,b,c <= n, such that a+b+c=0. - Benoit Cloitre, Jun 14 2003
Also the number of partitions of 6n into at most 3 parts, A001399(6n). - R. K. Guy, Oct 20 2003
Also, a(n) is the number of partitions of 6(n+1) into exactly 3 distinct parts. - William J. Keith, Jul 01 2004
Number of dots in a centered hexagonal figure with n+1 dots on each side.
Values of second Bessel polynomial y_2(n) (see A001498).
First differences of cubes (A000578). - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
Final digits of Hex numbers (hex(n) mod 10) are periodic with palindromic period of length 5 {1, 7, 9, 7, 1}. Last two digits of Hex numbers (hex(n) mod 100) are periodic with palindromic period of length 100. - Alexander Adamchuk, Aug 11 2006
All divisors of a(n) are congruent to 1, modulo 6. Proof: If p is an odd prime different from 3 then 3n^2 + 3n + 1 = 0 (mod p) implies 9(2n + 1)^2 = -3 (mod p), whence p = 1 (mod 6). - Nick Hobson, Nov 13 2006
For n>=1, a(n) is the side of Outer Napoleon Triangle whose reference triangle is a right triangle with legs (3a(n))^(1/2) and 3n(a(n))^(1/2). - Tom Schicker (tschicke(AT)email.smith.edu), Apr 25 2007
Number of triples (a,b,c) where 0<=(a,b)<=n and c=n (at least once the term n). E.g., for n = 1: (0,0,1), (0,1,0), (1,0,0), (0,1,1), (1,0,1), (1,1,0), (1,1,1), so a(1)=7. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Equals the triangular numbers convolved with [1, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009
From Terry Stickels, Dec 07 2009: (Start)
Also the maximum number of viewable cubes from any one static point while viewing a cube stack of identical cubes of varying magnitude.
For example, viewing a 2 X 2 X 2 stack will yield 7 maximum viewable cubes.
If the stack is 3 X 3 X 3, the maximum number of viewable cubes from any one static position is 19, and so on.
The number of cubes in the stack must always be the same number for width, length, height (at true regular cubic stack) and the maximum number of visible cubes can always be found by taking any cubic number and subtracting the number of the cube that is one less.
Examples: 125 - 64 = 61, 64 - 27 = 37, 27 - 8 = 19. (End)
The sequence of digital roots of the a(n) is period 3: repeat [1,7,1]. - Ant King, Jun 17 2012
The average of the first n (n>0) centered hexagonal numbers is the n-th square. - Philippe Deléham, Feb 04 2013
A002024 is the following array A read along antidiagonals:
1, 2, 3, 4, 5, 6, ...
2, 3, 4, 5, 6, 7, ...
3, 4, 5, 6, 7, 8, ...
4, 5, 6, 7, 8, 9, ...
5, 6, 7, 8, 9, 10, ...
6, 7, 8, 9, 10, 11, ...
and a(n) is the hook sum Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
a(n) is the sum of the terms in the n+1 X n+1 matrices minus those in n X n matrices in an array formed by considering A158405 an array (the beginning terms in each row are 1,3,5,7,9,11,...). - J. M. Bergot, Jul 05 2013
The formula also equals the product of the three distinct combinations of two consecutive numbers: n^2, (n+1)^2, and n*(n+1). - J. M. Bergot, Mar 28 2014
The sides of any triangle ABC are divided into 2n + 1 equal segments by 2n points: A_1, A_2, ..., A_2n in side a, and also on the sides b and c cyclically. If A'B'C' is the triangle delimited by AA_n, BB_n and CC_n cevians, we have (ABC)/(A'B'C') = a(n) (see Java applet link). - Ignacio Larrosa Cañestro, Jan 02 2015
a(n) is the maximal number of parts into which (n+1) triangles can intersect one another. - Ivan N. Ianakiev, Feb 18 2015
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = ((2^m-1)(2n+1))^t mod a(n), where m any positive integer, and t = 0(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
((2^m-1)n)^t mod a(n) = ((2^m-1)(n+1))^t mod a(n) = a(n) - (((2^m-1)(2n+1))^t mod a(n)), where m any positive integer, and t = 3(mod 6). - Alzhekeyev Ascar M, Oct 07 2016
(3n+1)^(a(n)-1) mod a(n) = (3n+2)^(a(n)-1) mod a(n) = 1. If a(n) not prime, then always strong pseudoprime. - Alzhekeyev Ascar M, Oct 07 2016
Every positive integer is the sum of 8 hex numbers (zero included), at most 3 of which are greater than 1. - Mauro Fiorentini, Jan 01 2018
Area enclosed by the segment of Archimedean spiral between n*Pi/2 and (n+1)*Pi/2 in Pi^3/48 units. - Carmine Suriano, Apr 10 2018
This sequence contains all numbers k such that 12*k - 3 is a square. - Klaus Purath, Oct 19 2021
The continued fraction expansion of sqrt(3*a(n)) is [3n+1; {1, 1, 2n, 1, 1, 6n+2}]. For n = 0, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 12 2022

Examples

			G.f. = 1 + 7*x + 19*x^2 + 37*x^3 + 61*x^4 + 91*x^5 + 127*x^6 + 169*x^7 + 217*x^8 + ...
From _Omar E. Pol_, Aug 21 2011: (Start)
Illustration of initial terms:
.
.                                 o o o o
.                   o o o        o o o o o
.         o o      o o o o      o o o o o o
.   o    o o o    o o o o o    o o o o o o o
.         o o      o o o o      o o o o o o
.                   o o o        o o o o o
.                                 o o o o
.
.   1      7          19             37
.
(End)
From _Klaus Purath_, Dec 03 2021: (Start)
(1) a(19) is not a prime number, because besides a(19) = a(9) + P(29), a(19) = a(15) + P(20) = a(2) + P(33) is also true.
(2) a(25) is prime, because except for a(25) = a(12) + P(38) there is no other equation of this pattern. (End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 81.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 18.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=3 of A080853, and column k=2 of A047969.
See also A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.
Cf. A287326(A000124(n), 1).
Cf. A008292.
Cf. A154105.

Programs

Formula

a(n) = 3*n*(n+1) + 1, n >= 0 (see the name).
a(n) = (n+1)^3 - n^3 = a(-1-n).
G.f.: (1 + 4*x + x^2) / (1 - x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = 6*A000217(n) + 1.
a(n) = a(n-1) + 6*n = 2a(n-1) - a(n-2) + 6 = 3*a(n-1) - 3*a(n-2) + a(n-3) = A056105(n) + 5n = A056106(n) + 4*n = A056107(n) + 3*n = A056108(n) + 2*n = A056108(n) + n.
n-th partial arithmetic mean is n^2. - Amarnath Murthy, May 27 2003
a(n) = 1 + Sum_{j=0..n} (6*j). E.g., a(2)=19 because 1+ 6*0 + 6*1 + 6*2 = 19. - Xavier Acloque, Oct 06 2003
The sum of the first n hexagonal numbers is n^3. That is, Sum_{n>=1} (3*n*(n-1) + 1) = n^3. - Edward Weed (eweed(AT)gdrs.com), Oct 23 2003
a(n) = right term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 0 0 / 2 1 0 / 3 3 1]. M^n * [1 1 1] = [1 2n+1 a(n)]. E.g., a(4) = 61, right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 9 61] = [1 2n+1 a(4)]. - Gary W. Adamson, Dec 22 2004
Row sums of triangle A130298. - Gary W. Adamson, Jun 07 2007
a(n) = 3*n^2 + 3*n + 1. Proof: 1) If n occurs once, it may be in 3 positions; for the two other ones, n terms are independently possible, then we have 3*n^2 different triples. 2) If the term n occurs twice, the third one may be placed in 3 positions and have n possible values, then we have 3*n more different triples. 3) The term n may occurs 3 times in one way only that gives the formula. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
Binomial transform of [1, 6, 6, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 6, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = (n-1)*A000166(n) + (n-2)*A000166(n-1) = (n-1)floor(n!*e^(-1)+1) + (n-2)*floor((n-1)!*e^(-1)+1) (with offset 0). - Gary Detlefs, Dec 06 2009
a(n) = A028896(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = integral( (sin((n+1/2)x)/sin(x/2))^3, x=0..Pi)/Pi. - Yalcin Aktar, Dec 03 2011
Sum_{n>=0} 1/a(n) = Pi/sqrt(3)*tanh(Pi/(2*sqrt(3))) = 1.305284153013581... - Ant King, Jun 17 2012
a(n) = A000290(n) + A000217(2n+1). - Ivan N. Ianakiev, Sep 24 2013
a(n) = A002378(n+1) + A056220(n) = A005408(n) + 2*A005449(n) = 6*A000217(n) + 1. - Ivan N. Ianakiev, Sep 26 2013
a(n) = 6*A000124(n) - 5. - Ivan N. Ianakiev, Oct 13 2013
a(n) = A239426(n+1) / A239449(n+1) = A215630(2*n+1,n+1). - Reinhard Zumkeller, Mar 19 2014
a(n) = A243201(n) / A002061(n + 1). - Mathew Englander, Jun 03 2014
a(n) = A101321(6,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (1 + 6*x + 3*x^2)*exp(x). - Ilya Gutkovskiy, Jul 28 2016
a(n) = (A001844(n) + A016754(n))/2. - Bruce J. Nicholson, Aug 06 2017
a(n) = A045943(2n+1). - Miquel Cerda, Jan 22 2018
a(n) = 3*Integral_{x=n..n+1} x^2 dx. - Carmine Suriano, Apr 10 2018
a(n) = A287326(A000124(n), 1). - Kolosov Petro, Oct 22 2018
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 10*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 2/e. (End)
G.f.: polylog(-3, x)*(1-x)/x. See the Simon Plouffe formula above, and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 08 2021
a(n) = T(n-1)^2 - 2*T(n)^2 + T(n+1)^2, n >= 1, T = triangular number A000217. - Klaus Purath, Oct 11 2021
a(n) = 1 + 2*Sum_{j=n..2n} j. - Klaus Purath, Oct 19 2021
a(n) = A069099(n+1) - A000217(n). - Klaus Purath, Nov 03 2021
From Leo Tavares, Dec 03 2021: (Start)
a(n) = A005448(n) + A140091(n);
a(n) = A001844(n) + A002378(n);
a(n) = A005891(n) + A000217(n);
a(n) = A000290(n) + A000384(n+1);
a(n) = A060544(n-1) + 3*A000217(n);
a(n) = A060544(n-1) + A045943(n).
a(2*n+1) = A154105(n).
(End)

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A006003 a(n) = n*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775
Offset: 0

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Author

Keywords

Comments

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

Examples

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020
		

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
  • F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

Programs

  • GAP
    a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018
    
  • Haskell
    a006003 n = n * (n ^ 2 + 1) `div` 2
    a006003_list = scanl (+) 0 a005448_list
    -- Reinhard Zumkeller, Jun 20 2013
    
  • MATLAB
    % Also works with FreeMat.
    for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
    % Stefano Spezia, Aug 12 2018
    
  • Magma
    [n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    [Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015
    
  • Mathematica
    Table[ n(n^2 + 1)/2, {n, 0, 45}]
    LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
    CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
    With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)
  • Maxima
    a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */
    
  • PARI
    {a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */
    
  • PARI
    concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016
    
  • Python
    def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

Formula

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Extensions

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

A045943 Triangular matchstick numbers: a(n) = 3*n*(n+1)/2.

Original entry on oeis.org

0, 3, 9, 18, 30, 45, 63, 84, 108, 135, 165, 198, 234, 273, 315, 360, 408, 459, 513, 570, 630, 693, 759, 828, 900, 975, 1053, 1134, 1218, 1305, 1395, 1488, 1584, 1683, 1785, 1890, 1998, 2109, 2223, 2340, 2460, 2583, 2709, 2838, 2970, 3105, 3243, 3384, 3528
Offset: 0

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Comments

Also, 3 times triangular numbers, a(n) = 3*A000217(n).
In the 24-bit RGB color cube, the number of color-lattice-points in r+g+b = n planes at n < 256 equals the triangular numbers. For n = 256, ..., 765 the number of legitimate color partitions is less than A000217(n) because {r,g,b} components cannot exceed 255. For n = 256, ..., 511, the number of non-color partitions are computable with A045943(n-255), while for n = 512, ..., 765, the number of color points in r+g+b planes equals A000217(765-n). - Labos Elemer, Jun 20 2005
If a 3-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
a(n) is also the smallest number that may be written both as the sum of n-1 consecutive positive integers and n consecutive positive integers. - Claudio Meller, Oct 08 2010
For n >= 3, a(n) equals 4^(2+n)*Pi^(1 - n) times the coefficient of zeta(3) in the following integral with upper bound Pi/4 and lower bound 0: int x^(n+1) tan x dx. - John M. Campbell, Jul 17 2011
The difference a(n)-a(n-1) = 3*n, for n >= 1. - Stephen Balaban, Jul 25 2011 [Comment clarified by N. J. A. Sloane, Aug 01 2024]
Sequence found by reading the line from 0, in the direction 0, 3, ..., and the same line from 0, in the direction 0, 9, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. This is one of the orthogonal axes of the spiral; the other is A032528. - Omar E. Pol, Sep 08 2011
A005449(a(n)) = A000332(3n + 3) = C(3n + 3, 4), a second pentagonal number of triangular matchstick number index number. Additionally, a(n) - 2n is a pentagonal number (A000326). - Raphie Frank, Dec 31 2012
Sum of the numbers from n to 2n. - Wesley Ivan Hurt, Nov 24 2015
Number of orbits of Aut(Z^7) as function of the infinity norm (n+1) of the representative integer lattice point of the orbit, when the cardinality of the orbit is equal to 5376 or 17920 or 20160. - Philippe A.J.G. Chevalier, Dec 28 2015
Also the number of 4-cycles in the (n+4)-triangular honeycomb acute knight graph. - Eric W. Weisstein, Jul 27 2017
Number of terms less than 10^k, k=0,1,2,3,...: 1, 3, 8, 26, 82, 258, 816, 2582, 8165, 25820, 81650, 258199, 816497, 2581989, 8164966, ... - Muniru A Asiru, Jan 24 2018
Numbers of the form 3*m*(2*m + 1) for m = 0, -1, 1, -2, 2, -3, 3, ... - Bruno Berselli, Feb 26 2018
Partial sums of A008585. - Omar E. Pol, Jun 20 2018
Column 1 of A273464. (Number of ways to select a unit lozenge inside an isosceles triangle of side length n; all vertices on a hexagonal lattice.) - R. J. Mathar, Jul 10 2019
Total number of pips in the n-th suit of a double-n domino set. - Ivan N. Ianakiev, Aug 23 2020

Examples

			From _Stephen Balaban_, Jul 25 2011: (Start)
T(n), the triangular numbers = number of nodes,
a(n-1) = number of edges in the T(n) graph:
       o    (T(1) = 1, a(0) = 0)
       o
      / \   (T(2) = 3, a(1) = 3)
     o - o
       o
      / \
     o - o  (T(3) = 6, a(2) = 9)
    / \ / \
   o - o - o
... [Corrected by _N. J. A. Sloane_, Aug 01 2024] (End)
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 543.

Crossrefs

The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, A005449, A045943, A115067, A140090, A140091, A059845, A140672, A140673, A140674, A140675, A151542.
A diagonal of A010027.
Orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A115067, A008585, A005843, A001477, A000217.
Cf. A027480 (partial sums).
Cf. A002378 (3-cycles in triangular honeycomb acute knight graph), A028896 (5-cycles), A152773 (6-cycles).
This sequence: Sum_{k = n..2*n} k.
Cf. A304993: Sum_{k = n..2*n} k*(k+1)/2.
Cf. A050409: Sum_{k = n..2*n} k^2.
Similar sequences are listed in A316466.

Programs

Formula

a(n) is the sum of n+1 integers starting from n, i.e., 1+2, 2+3+4, 3+4+5+6, 4+5+6+7+8, etc. - Jon Perry, Jan 15 2004
a(n) = A126890(n+1,n-1) for n>1. - Reinhard Zumkeller, Dec 30 2006
a(n) + A145919(3*n+3) = 0. - Matthew Vandermast, Oct 28 2008
a(n) = A000217(2*n) - A000217(n-1); A179213(n) <= a(n). - Reinhard Zumkeller, Jul 05 2010
a(n) = a(n-1)+3*n, n>0. - Vincenzo Librandi, Nov 18 2010
G.f.: 3*x/(1-x)^3. - Bruno Berselli, Jan 21 2011
a(n) = A005448(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A001477(n)+A000290(n)+A000217(n). - J. M. Bergot, Dec 08 2012
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>2. - Wesley Ivan Hurt, Nov 24 2015
a(n) = A027480(n)-A027480(n-1). - Peter M. Chema, Jan 18 2017.
2*a(n)+1 = A003215(n). - Miquel Cerda, Jan 22 2018
a(n) = T(2*n) - T(n-1), where T(n) = A000217(n). In general, T(k)*T(n) = Sum_{i=0..k-1} (-1)^i*T((k-i)*(n-i)). - Charlie Marion, Dec 06 2020
E.g.f.: 3*exp(x)*x*(2 + x)/2. - Stefano Spezia, May 19 2021
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(2*log(2)-1)/3. (End)
Product_{n>=1} (1 - 1/a(n)) = -(3/(2*Pi))*cos(sqrt(11/3)*Pi/2). - Amiram Eldar, Feb 21 2023

A008486 Expansion of (1 + x + x^2)/(1 - x)^2.

Original entry on oeis.org

1, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177, 180, 183, 186
Offset: 0

Views

Author

Keywords

Comments

Also the Engel expansion of exp^(1/3); cf. A006784 for the Engel expansion definition. - Benoit Cloitre, Mar 03 2002
Coordination sequence for planar net 6^3 (the graphite net, or the graphene crystal) - that is, the number of atoms at graph distance n from any fixed atom. Also for the hcb or honeycomb net. - N. J. A. Sloane, Jan 06 2013, Mar 31 2018
Coordination sequence for 2-dimensional cyclotomic lattice Z[zeta_3].
Conjecture: This is also the maximum number of edges possible in a planar simple graph with n+2 vertices. - Dmitry Kamenetsky, Jun 29 2008
The conjecture is correct. Proof: For n=0 the theorem holds, the maximum planar graph has n+2=2 vertices and 1 edge. Now suppose that we have a connected planar graph with at least 3 vertices. If it contains a face that is not a triangle, we can add an edge that divides this face into two without breaking its planarity. Hence all maximum planar graphs are triangulations. Euler's formula for planar graphs states that in any planar simple graph with V vertices, E edges and F faces we have V+F-E=2. If all faces are triangles, then F=2E/3, which gives us E=3V-6. Hence for n>0 each maximum planar simple graph with n+2 vertices has 3n edges. - Michal Forisek, Apr 23 2009
a(n) = sum of natural numbers m such that n - 1 <= m <= n + 1. Generalization: If a(n,k) = sum of natural numbers m such that n - k <= m <= n + k (k >= 1) then a(n,k) = (k + n)*(k + n + 1)/2 = A000217(k+n) for 0 <= n <= k, a(n,k) = a(n-1,k) +2k + 1 = ((k + n - 1)*(k + n)/2) + 2k + 1 = A000217(k+n-1) +2k +1 for n >= k + 1 (see e.g. A008486). - Jaroslav Krizek, Nov 18 2009
a(n) = partial sums of A158799(n). Partial sums of a(n) = A005448(n). - Jaroslav Krizek, Dec 06 2009
Integers n dividing a(n) = a(n-1) - a(n-2) with initial conditions a(0)=0, a(1)=1 (see A128834 with offset 0). - Thomas M. Bridge, Nov 03 2013
a(n) is conjectured to be the number of polygons added after n iterations of the polygon expansions (type A, B, C, D & E) shown in the Ngaokrajang link. The patterns are supposed to become the planar Archimedean net 3.3.3.3.3.3, 3.6.3.6, 3.12.12, 3.3.3.3.6 and 4.6.12 respectively when n - > infinity. - Kival Ngaokrajang, Dec 28 2014
Number of reduced words of length n in Coxeter group on 3 generators S_i with relations (S_i)^2 = (S_i S_j)^3 = I. - Ray Chandler, Nov 21 2016
Conjecture: let m = n + 2, p is the polyhedron formed by the convex hull of m points, q is the number of quadrilateral faces of p (see the Wikipedia link below), and f(m) = a(n) - q. Then f(m) would be the solution of the Thompson problem for all m in 3-space. - Sergey Pavlov, Feb 03 2017
Also, sequence defined by a(0)=1, a(1)=3, c(0)=2, c(1)=4; and thereafter a(n) = c(n-1) + c(n-2), and c consists of the numbers missing from a (see A001651). - Ivan Neretin, Mar 28 2017

Examples

			G.f. = 1 + 3*x + 6*x^2 + 9*x^3 + 12*x^4 + 15*x^5 + 18*x^6 + 21*x^7 + 24*x^8 + ...
From _Omar E. Pol_, Aug 20 2011: (Start)
Illustration of initial terms as triangles:
.                                              o
.                                 o           o o
.                      o         o o         o   o
.             o       o o       o   o       o     o
.      o     o o     o   o     o     o     o       o
. o   o o   o o o   o o o o   o o o o o   o o o o o o
.
. 1    3      6        9          12           15
(End)
		

References

  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 158.

Crossrefs

Partial sums give A005448.
List of coordination sequences for uniform planar nets: A008458 (the planar net 3.3.3.3.3.3), A008486 (6^3), A008574(4.4.4.4 and 3.4.6.4), A008576 (4.8.8), A008579(3.6.3.6), A008706 (3.3.3.4.4), A072154 (4.6.12), A219529(3.3.4.3.4), A250120 (3.3.3.3.6), A250122 (3.12.12).
List of coordination sequences for Laves tilings (or duals of uniform planar nets): [3,3,3,3,3.3] = A008486; [3.3.3.3.6] = A298014, A298015, A298016; [3.3.3.4.4] = A298022, A298024; [3.3.4.3.4] = A008574, A296368; [3.6.3.6] = A298026, A298028; [3.4.6.4] = A298029, A298031, A298033; [3.12.12] = A019557, A298035; [4.4.4.4] = A008574; [4.6.12] = A298036, A298038, A298040; [4.8.8] = A022144, A234275; [6.6.6] = A008458.

Programs

  • Haskell
    a008486 0 = 1; a008486 n = 3 * n
    a008486_list = 1 : [3, 6 ..]  -- Reinhard Zumkeller, Apr 17 2015
  • Magma
    [0^n+3*n: n in [0..90] ]; // Vincenzo Librandi, Aug 21 2011
    
  • Mathematica
    CoefficientList[Series[(1 + x + x^2) / (1 - x)^2, {x, 0, 80}], x] (* Vincenzo Librandi, Nov 23 2014 *)
    a[ n_] := If[ n == 0, 1, 3 n]; (* Michael Somos, Apr 17 2015 *)
  • PARI
    {a(n) = if( n==0, 1, 3 * n)}; /* Michael Somos, May 05 2015 */
    

Formula

a(0) = 1; a(n) = 3*n = A008585(n), n >= 1.
Euler transform of length 3 sequence [3, 0, -1]. - Michael Somos, Aug 04 2009
a(n) = a(n-1) + 3 for n >= 2. - Jaroslav Krizek, Nov 18 2009
a(n) = 0^n + 3*n. - Vincenzo Librandi, Aug 21 2011
a(n) = -a(-n) unless n = 0. - Michael Somos, May 05 2015
E.g.f.: 1 + 3*exp(x)*x. - Stefano Spezia, Aug 07 2022

A003154 Centered 12-gonal numbers, or centered dodecagonal numbers: numbers of the form 6*k*(k-1) + 1.

Original entry on oeis.org

1, 13, 37, 73, 121, 181, 253, 337, 433, 541, 661, 793, 937, 1093, 1261, 1441, 1633, 1837, 2053, 2281, 2521, 2773, 3037, 3313, 3601, 3901, 4213, 4537, 4873, 5221, 5581, 5953, 6337, 6733, 7141, 7561, 7993, 8437, 8893, 9361, 9841, 10333, 10837, 11353, 11881, 12421
Offset: 1

Views

Author

Keywords

Comments

Binomial transform of [1, 12, 12, 0, 0, 0, ...]. Narayana transform (A001263) of [1, 12, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
Numbers k such that 6*k+3 is a square, these squares are given in A016946. - Gary Detlefs and Vincenzo Librandi, Aug 08 2010
Odd numbers of the form floor(n^2/6). - Juri-Stepan Gerasimov, Jul 27 2011
Bisection of A032528. - Omar E. Pol, Aug 20 2011
Sequence found by reading the line from 1, in the direction 1, 13, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A033581 in the same spiral. - Omar E. Pol, Sep 08 2011
The digital root has period 3 (1, 4, 1) (A146325), the same digital root as the centered triangular numbers A005448(n). - Peter M. Chema, Dec 20 2023

Examples

			From _Omar E. Pol_, Aug 21 2011: (Start)
1. Classic illustration of initial terms of the star numbers:
.
.                                     o
.                                    o o
.                  o            o o o o o o o
.               o o o o          o o o o o o
.     o          o o o            o o o o o
.               o o o o          o o o o o o
.                  o            o o o o o o o
.                                    o o
.                                     o
.
.     1            13                 37
.
2. Alternative illustration of initial terms using n-1 concentric hexagons around a central element:
.
.                                 o o o o o
.                                o         o
.                o o o          o   o o o   o
.               o     o        o   o     o   o
.     o        o   o   o      o   o   o   o   o
.               o     o        o   o     o   o
.                o o o          o   o o o   o
.                                o         o
.                                 o o o o o
(End)
		

References

  • Martin Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 20.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

Formula

G.f.: x*(1+10*x+x^2)/(1-x)^3. Simon Plouffe in his 1992 dissertation
a(n) = 1 + Sum_{j=0..n} (12*j). E.g., a(2)=37 because 1 + 12*0 + 12*1 + 12*2 = 37. - Xavier Acloque, Oct 06 2003
a(n) = numerator in B_2(x) = (1/2)x^2 - (1/2)x + 1/12 = Bernoulli polynomial of degree 2. - Gary W. Adamson, May 30 2005
a(n) = 12*(n-1) + a(n-1), with n>1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = A049598(n-1) + 1. - Omar E. Pol, Oct 03 2011
Sum_{n>=1} 1/a(n) = A306980 = Pi * tan(Pi/(2*sqrt(3))) / (2*sqrt(3)). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} a(n)/n! = 7*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 7/e - 1. (End)
a(n) = 2*A003215(n-1) - 1. - Leo Tavares, Jul 30 2021
E.g.f.: exp(x)*(1 + 6*x^2) - 1. - Stefano Spezia, Aug 19 2022

Extensions

More terms from Michael Somos

A062786 Centered 10-gonal numbers.

Original entry on oeis.org

1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901, 10351, 10811
Offset: 1

Views

Author

Jason Earls, Jul 19 2001

Keywords

Comments

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003
All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006
Centered decagonal numbers. - Omar E. Pol, Oct 03 2011
The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021
The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022
Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022
All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023

Crossrefs

Programs

  • GAP
    List([1..50], n-> 1+5*n*(n-1)); # G. C. Greubel, Mar 30 2019
    
  • Magma
    [1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019
    
  • Mathematica
    FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
    1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)
  • PARI
    j=[]; for(n=1,75,j=concat(j,(5*n*(n-1)+1))); j
    
  • PARI
    for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009
    
  • Python
    def a(n): return(5*n**2-5*n+1) # Torlach Rush, May 10 2024
  • Sage
    [1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019
    

Formula

a(n) = 5*n*(n-1) + 1.
From Gary W. Adamson, Dec 29 2007: (Start)
Binomial transform of [1, 10, 10, 0, 0, 0, ...];
Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)
G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019
E.g.f.: (1+5*x^2)*exp(x) - 1. - G. C. Greubel, Mar 30 2019
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 6*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)
a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021
a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021
From Leo Tavares, Oct 06 2021: (Start)
a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.
a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.
a(n) = A060544(n) + A000217(n-1). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A016754(n-1) + 2*A000217(n-1).
a(n) = A016754(n-1) + A002378(n-1).
a(n) = A069099(n) + 3*A000217(n-1).
a(n) = A069099(n) + A045943(n-1).
a(n) = A003215(n-1) + 4*A000217(n-1).
a(n) = A003215(n-1) + A046092(n-1).
a(n) = A001844(n-1) + 6*A000217(n-1).
a(n) = A001844(n-1) + A028896(n-1).
a(n) = A005448(n) + 7*A000217(n).
a(n) = A005448(n) + A024966(n). (End)
From Klaus Purath, Oct 30 2022: (Start)
a(n) = a(n-2) + 10*(2*n-3).
a(n) = 2*a(n-1) - a(n-2) + 10.
a(n) = A135705(n-1) + n.
a(n) = A190816(n) - n.
a(n) = 2*A005891(n-1) - 1. (End)

Extensions

Better description from Terrel Trotter, Jr., Apr 06 2002
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