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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A254962 Indices of hexagonal numbers (A000384) that are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 2, 12, 31, 211, 552, 3782, 9901, 67861, 177662, 1217712, 3188011, 21850951, 57206532, 392099402, 1026529561, 7035938281, 18420325562, 126254789652, 330539330551, 2265550275451, 5931287624352, 40653650168462, 106432637907781, 729500152756861
Offset: 1

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Author

Colin Barker, Feb 11 2015

Keywords

Comments

Also positive integers x in the solutions to 4*x^2 - 5*y^2 - 2*x + 5*y - 2 = 0, the corresponding values of y being A254627.

Examples

			12 is in the sequence because the 12th hexagonal number is 276, which is also the 11th centered pentagonal number.
		

Crossrefs

Cf. A000032 (Lucas numbers), A000384, A005891, A254627, A254628.

Programs

  • PARI
    Vec(-x*(x^4+x^3-8*x^2+x+1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^100))

Formula

a(n) = a(n-1)+18*a(n-2)-18*a(n-3)-a(n-4)+a(n-5).
G.f.: -x*(x^4+x^3-8*x^2+x+1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)).
a(n) = (2 + (2-r)^n - (-2-r)^n*(-2+r) + 2*(-2+r)^n + r*(-2+r)^n + (2+r)^n)/8 where r = sqrt(5). - Colin Barker, Nov 25 2016
a(n+2) - a(n) = A000032(3*n + 2) if n is odd, A000032(3*n + 1) if n is even. - Diego Rattaggi, May 11 2020

A322801 Number of compositions (ordered partitions) of n into centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 28, 36, 46, 59, 76, 98, 128, 167, 217, 281, 363, 468, 605, 784, 1017, 1320, 1712, 2217, 2869, 3713, 4807, 6227, 8070, 10458, 13549, 17549, 22726, 29430, 38117, 49375, 63962, 82859, 107333, 139026, 180071
Offset: 0

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Author

Ilya Gutkovskiy, Dec 26 2018

Keywords

Crossrefs

Programs

  • Maple
    h:= proc(n) option remember; `if`(n<0, 0, (t->
          `if`(((t+1)*5*t+2)/2>n, t-1, t))(1+h(n-1)))
        end:
    a:= proc(n) option remember; `if`(n=0, 1,
          add(a(n-((i+1)*5*i+2)/2), i=0..h(n)))
        end:
    seq(a(n), n=0..60);  # Alois P. Heinz, Dec 28 2018
  • Mathematica
    nmax = 50; CoefficientList[Series[1/(1 - Sum[x^(5 k (k + 1)/2 + 1), {k, 0, nmax}]), {x, 0, nmax}], x]

Formula

G.f.: 1/(1 - Sum_{k>=0} x^(5*k*(k+1)/2+1)).

A128917 Pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 51, 3151, 195301, 12105501, 750345751, 46509331051, 2882828179401, 178688837791801, 11075825114912251, 686522468286767751, 42553317208664688301, 2637619144468923906901, 163489833639864617539551, 10133732066527137363545251, 628127898291042651922266001
Offset: 1

Views

Author

Steven Schlicker, Apr 24 2007

Keywords

Examples

			a(1)=51 because 51 is the fifth centered pentagonal number and the sixth pentagonal number.
		

Crossrefs

Programs

  • Maple
    CP := n -> 1+1/2*5*(n^2-n): N:=10: u:=4: v:=1: x:=5: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+15*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp;
  • Mathematica
    LinearRecurrence[{63,-63,1},{1,51,3151},20] (* Harvey P. Dale, Nov 26 2022 *)
  • PARI
    Vec(-x*(x^2-12*x+1)/((x-1)*(x^2-62*x+1)) + O(x^100)) \\ Colin Barker, Jan 07 2015

Formula

Define x(n) + y(n)*sqrt(15) = (5+sqrt(15))*(4+sqrt(15))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+5*(s(n)^2-s(n))).
From Richard Choulet, Sep 19 2007: (Start)
We must solve 3*p^2-p=5*r^2+5*r+2, which gives X^2=15*Y^2+10 where X=6*p-1 and Y=2*r+1.
Four other sequences are obtained at the same time:
X is given by 5,35,275,2165,... with the recurrence a(n+2)=8*a(n+1)-a(n) and also a(n+1)=4*a(n)+(15*a(n)^2-150)^(1/2) (numbers such that 15*X^2-150 is a square).
Y is given by 1,9,71,559,... with the recurrence a(n+2)=8*a(n+1)-a(n) and also a(n+1)=4*a(n)+(15*a(n)^2+10)^(1/2) (numbers such that 15*Y^2+10 is a square).
p is given by 1,6,46,361,... with the recurrence a(n+2)=8*a(n+1)-a(n)-1 and also a(n+1)=4*a(n)-0.5+0.5*(60*a(n)^2-20*a(n)-15)^(1/2) (numbers such that 15*(6*p-1)^2-150 is a square).
r is given by 0,4,35,279,... with the recurrence a(n+2)=8*a(n+1)-a(n)+3 and also a(n+1)=4*a(n)+1.5+0.5*(60*a(n)^2+60*a(n)+25)^(1/2) (numbers such that 15*(2*r+1)^2+10 is a square).
a(n+2) = 62*a(n+1)-a(n)-10, a(n+1)=31*a(n)-5+(960*a(n)^2-320*a(n)-45)^(1/2).
G.f.: z*(1-12*z+z^2)/((1-z)*(1-62*z+z^2)). (End)
a(n) = 63*a(n-1)-63*a(n-2)+a(n-3). - Colin Barker, Jan 07 2015

Extensions

Edited by N. J. A. Sloane, Sep 25 2007
More terms from R. J. Mathar, Oct 31 2007

A253470 Indices of centered triangular numbers (A005448) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 5, 36, 280, 2201, 17325, 136396, 1073840, 8454321, 66560725, 524031476, 4125691080, 32481497161, 255726286205, 2013328792476, 15850904053600, 124793903636321, 982500325036965, 7735208696659396, 60899169248238200, 479458145289246201, 3774765993065731405
Offset: 1

Views

Author

Colin Barker, Jan 01 2015

Keywords

Comments

Also indices of pentagonal numbers (A000326) which are also centered pentagonal numbers (A005891).
Also positive integers x in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0, the corresponding values of y being A182432.

Examples

			5 is in the sequence because the 5th centered triangular number is 31, which is also the 4th centered pentagonal number.
		

Crossrefs

Programs

  • PARI
    Vec(x*(4*x-1)/((x-1)*(x^2-8*x+1)) + O(x^100))

Formula

a(n) = 9*a(n-1)-9*a(n-2)+a(n-3).
G.f.: x*(4*x-1) / ((x-1)*(x^2-8*x+1)).
a(n) = (6-(4-sqrt(15))^n*(3+sqrt(15))+(-3+sqrt(15))*(4+sqrt(15))^n)/12. - Colin Barker, Mar 03 2016

A213604 Cumulative sums of digital roots of A005891(n).

Original entry on oeis.org

1, 7, 14, 18, 24, 28, 35, 41, 42, 48, 55, 59, 65, 69, 76, 82, 83, 89, 96, 100, 106, 110, 117, 123, 124, 130, 137, 141, 147, 151, 158, 164, 165, 171, 178, 182, 188, 192, 199, 205, 206, 212, 219, 223, 229, 233, 240, 246, 247, 253, 260, 264, 270, 274, 281, 287
Offset: 0

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Author

Keywords

Crossrefs

Cf. A005891.

Programs

  • Mathematica
    CoefficientList[Series[(1 + 6*x + 7*x^2 + 4 x^3 + 6*x^4 + 4*x^5 + 7*x^6 +
    6*x^7)/((x - 1)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7)), {x, 0, 50}], x] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,1,-1}, {1,7,14,18,24, 28,35,41,42}, 50](* G. C. Greubel, Feb 26 2017 *)
  • PARI
    x='x+O('x^50); Vec((1+6*x+7*x^2+4x^3+6*x^4+4*x^5+7*x^6+6*x^7) / ((x-1)^2 * (1+x+x^2+x^3+x^4+x^5+x^6+x^7))) \\ G. C. Greubel, Feb 26 2017

Formula

a(n+9) = -a(n) + a(n+1) + a(n+8), a(0)=1, a(1)=7, a(2)=14, a(3)=18, a(4)=24, a(5)=28, a(6)=35, a(7)=41, a(8)=42.
G.f.: (1+6*x+7*x^2+4x^3+6*x^4+4*x^5+7*x^6+6*x^7) / ((x-1)^2 * (1+x+x^2+x^3+x^4+x^5+x^6+x^7)).

A254374 Digital roots of centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6, 4, 7, 6, 1, 1, 6, 7, 4, 6
Offset: 1

Views

Author

Colin Barker, Jan 29 2015

Keywords

Comments

The sequence is periodic with period 9.

Examples

			a(3) = 7 because the 3rd centered pentagonal number is 16, the digital root of which is 7.
		

Crossrefs

Programs

  • Mathematica
    FixedPoint[Plus @@ IntegerDigits[#] &, #] & /@ Table[(5 n^2 + 5 n + 2)/2, {n, 0, 80}] (* Michael De Vlieger, Feb 01 2015 *)
    LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 6, 7, 4, 6, 4, 7, 6, 1},86] (* Ray Chandler, Aug 26 2015 *)
    PadRight[{},120,{1,6,7,4,6,4,7,6,1}] (* Harvey P. Dale, Aug 23 2017 *)
  • PARI
    m=5; vector(200, n, (m*n*(n-1)/2)%9+1)

Formula

a(n) = A010888(A005891(n)).
a(n) = a(n-9).
G.f.: -x*(x^8+6*x^7+7*x^6+4*x^5+6*x^4+4*x^3+7*x^2+6*x+1) / ((x-1)*(x^2+x+1)*(x^6+x^3+1)).

A145837 Indices of primes in A005891(n).

Original entry on oeis.org

3, 8, 11, 15, 20, 23, 36, 44, 47, 48, 60, 68, 71, 75, 83, 84, 87, 92, 111, 116, 128, 132, 143, 144, 156, 159, 164, 167, 168, 183, 192, 200, 204, 207, 215, 224, 228, 231, 236, 239, 264, 272, 287, 299, 300, 303, 312, 315, 320, 323, 356, 359, 360, 363, 372, 387
Offset: 1

Views

Author

Alexander Adamchuk, Mar 21 2009

Keywords

Comments

Corresponding centered pentagonal primes are listed in A145838 = {31, 181, 331, 601, 1051, 1381, 3331, ...}.

Crossrefs

Cf. A145838 = Primes in A005891. Cf. A005891 = Centered pentagonal numbers: (5n^2+5n+2)/2.

Programs

Formula

A005891(a(n)) = A145838(n)

Extensions

Extended by R. J. Mathar, Mar 26 2009
Edited by N. J. A. Sloane, Apr 06 2009

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

A005902 Centered icosahedral (or cuboctahedral) numbers, also crystal ball sequence for f.c.c. lattice.

Original entry on oeis.org

1, 13, 55, 147, 309, 561, 923, 1415, 2057, 2869, 3871, 5083, 6525, 8217, 10179, 12431, 14993, 17885, 21127, 24739, 28741, 33153, 37995, 43287, 49049, 55301, 62063, 69355, 77197, 85609, 94611, 104223, 114465, 125357, 136919, 149171, 162133, 175825, 190267, 205479
Offset: 0

Views

Author

Keywords

Comments

Called "magic numbers" in some chemical contexts.
Partial sums of A005901(n). - Lekraj Beedassy, Oct 30 2003
Equals binomial transform of [1, 12, 30, 20, 0, 0, 0, ...]. - Gary W. Adamson, Aug 01 2008
Crystal ball sequence for A_3 lattice. - Michael Somos, Jun 03 2012

Examples

			a(4) = 147 = (1, 3, 3, 1) dot (1, 12, 30, 20) = (1 + 36 + 90 + 20). - _Gary W. Adamson_, Aug 01 2008
G.f. = 1 + 13*x + 55*x^2 + 147*x^3 + 309*x^4 + 561*x^5 + 923*x^6 + 1415*x^7 + ...
		

References

  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(2*n^3-3*n^2+n)+2*n-1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

  • Magma
    [(2*n+1)*(5*n^2+5*n+3)/3: n in [0..30]]; // G. C. Greubel, Dec 01 2017
    
  • Maple
    A005902 := n -> (2*n+1)*(5*n^2+5*n+3)/3;
    A005902:=(z+1)*(z**2+8*z+1)/(z-1)**4; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_] := (2n + 1)(5n^2 + 5n + 3)/3; Array[f, 36, 0] (* Robert G. Wilson v, Feb 02 2011 *)
    LinearRecurrence[{4,-6,4,-1},{1,13,55,147},50] (* Harvey P. Dale, Oct 08 2015 *)
    CoefficientList[Series[(x^3 + 9*x^2 + 9*x + 1)/(x - 1)^4, {x, 0, 50}], x] (* Indranil Ghosh, Apr 08 2017 *)
  • PARI
    {a(n) = (2*n + 1) * (5*n^2 + 5*n + 3) / 3}; /* Michael Somos, Jun 03 2012 */
    
  • PARI
    x='x+O('x^50); Vec((x^3 + 9*x^2 + 9*x + 1)/(x - 1)^4) \\ Indranil Ghosh, Apr 08 2017
    
  • Python
    def a(n): return (2*n+1)*(5*n**2+5*n+3)//3
    print([a(n) for n in range(40)]) # Michael S. Branicky, Jan 13 2021

Formula

a(n) = (2*n+1)*(5*n^2+5*n+3)/3.
For n > 0, n*a(n) = (Sum_{i=0..n-1} a(i)) + 2*A005891(n)*A000217(n). - Bruno Berselli, Feb 02 2011
a(-1 - n) = -a(n). - Michael Somos, Jun 03 2012
From Indranil Ghosh, Apr 08 2017: (Start)
G.f.: (x^3 + 9x^2 + 9x + 1)/(x - 1)^4.
E.g.f.: (1/3)*exp(x)*(10x^3 + 45x^2 + 36x + 3).
(End)
a(n) = A100171(n+1) - A008778(n-1) = A100174(n+1) - A000290(n) = A005917(n+1) - A006331(n) = A051673(n+1) + A000578(n). - Bruce J. Nicholson, Jul 05 2018
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