cp's OEIS Frontend

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

Showing 1-6 of 6 results.

A001105 a(n) = 2*n^2.

Original entry on oeis.org

0, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418
Offset: 0

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Author

Bernd.Walter(AT)frankfurt.netsurf.de

Keywords

Comments

Number of edges of the complete bipartite graph of order 3n, K_{n,2n}. - Roberto E. Martinez II, Jan 07 2002
"If each period in the periodic system ends in a rare gas ..., the number of elements in a period can be found from the ordinal number n of the period by the formula: L = ((2n+3+(-1)^n)^2)/8..." - Nature, Jun 09 1951; Nature 411 (Jun 07 2001), p. 648. This produces the present sequence doubled up.
Let z(1) = i = sqrt(-1), z(k+1) = 1/(z(k)+2i); then a(n) = (-1)*Imag(z(n+1))/Real(z(n+1)). - Benoit Cloitre, Aug 06 2002
Maximum number of electrons in an atomic shell with total quantum number n. Partial sums of A016825. - Jeremy Gardiner, Dec 19 2004
Arithmetic mean of triangular numbers in pairs: (1+3)/2, (6+10)/2, (15+21)/2, ... . - Amarnath Murthy, Aug 05 2005
These numbers form a pattern on the Ulam spiral similar to that of the triangular numbers. - G. Roda, Oct 20 2010
Integral areas of isosceles right triangles with rational legs (legs are 2n and triangles are nondegenerate for n > 0). - Rick L. Shepherd, Sep 29 2009
Even squares divided by 2. - Omar E. Pol, Aug 18 2011
Number of stars when distributed as in the U.S.A. flag: n rows with n+1 stars and, between each pair of these, one row with n stars (i.e., n-1 of these), i.e., n*(n+1)+(n-1)*n = 2*n^2 = A001105(n). - César Eliud Lozada, Sep 17 2012
Apparently the number of Dyck paths with semilength n+3 and an odd number of peaks and the central peak having height n-3. - David Scambler, Apr 29 2013
Sum of the partition parts of 2n into exactly two parts. - Wesley Ivan Hurt, Jun 01 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective odd leg a (A180620); sequence gives values c-a, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
Number of roots in the root systems of type B_n and C_n (for n > 1). - Tom Edgar, Nov 05 2013
Area of a square with diagonal 2n. - Wesley Ivan Hurt, Jun 18 2014
This sequence appears also as the first and second member of the quartet [a(n), a(n), p(n), p(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p(n) = A046092(n). See an Oct 15 2014 comment on A147973 where also a reference is given. - Wolfdieter Lang, Oct 16 2014
a(n) are the only integers m where (A000005(m) + A000203(m)) = (number of divisors of m + sum of divisors of m) is an odd number. - Richard R. Forberg, Jan 09 2015
a(n) represents the first term in a sum of consecutive integers running to a(n+1)-1 that equals (2n+1)^3. - Patrick J. McNab, Dec 24 2016
Also the number of 3-cycles in the (n+4)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 29 2017
Also the Wiener index of the n-cocktail party graph for n > 1. - Eric W. Weisstein, Sep 07 2017
Numbers represented as the palindrome 242 in number base B including B=2 (binary), 3 (ternary) and 4: 242(2)=18, 242(3)=32, 242(4)=50, ... 242(9)=200, 242(10)=242, ... - Ron Knott, Nov 14 2017
a(n) is the square of the hypotenuse of an isosceles right triangle whose sides are equal to n. - Thomas M. Green, Aug 20 2019
The sequence contains all odd powers of 2 (A004171) but no even power of 2 (A000302). - Torlach Rush, Oct 10 2019
From Bernard Schott, Aug 31 2021 and Sep 16 2021: (Start)
Apart from 0, integers such that the number of even divisors (A183063) is odd.
Proof: every n = 2^q * (2k+1), q, k >= 0, then 2*n^2 = 2^(2q+1) * (2k+1)^2; now, gcd(2, 2k+1) = 1, tau(2^(2q+1)) = 2q+2 and tau((2k+1)^2) = 2u+1 because (2k+1)^2 is square, so, tau(2*n^2) = (2q+2) * (2u+1).
The 2q+2 divisors of 2^(2q+1) are {1, 2, 2^2, 2^3, ..., 2^(2q+1)}, so 2^(2q+1) has 2q+1 even divisors {2^1, 2^2, 2^3, ..., 2^(2q+1)}.
Conclusion: these 2q+1 even divisors create with the 2u+1 odd divisors of (2k+1)^2 exactly (2q+1)*(2u+1) even divisors of 2*n^2, and (2q+1)*(2u+1) is odd. (End)
a(n) with n>0 are the numbers with period length 2 for Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022
Number of points at L1 distance = 2 from any given point in Z^n. - Shel Kaphan, Feb 25 2023
Integer that multiplies (h^2)/(m*L^2) to give the energy of a 1-D quantum mechanical particle in a box whenever it is an integer multiple of (h^2)/(m*L^2), where h = Planck's constant, m = mass of particle, and L = length of box. - A. Timothy Royappa, Mar 14 2025

Examples

			a(3) = 18; since 2(3) = 6 has 3 partitions with exactly two parts: (5,1), (4,2), (3,3).  Adding all the parts, we get: 1 + 2 + 3 + 3 + 4 + 5 = 18. - _Wesley Ivan Hurt_, Jun 01 2013
		

References

  • Peter Atkins, Julio De Paula, and James Keeler, "Atkins' Physical Chemistry," Oxford University Press, 2023, p. 31.
  • Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
  • Martin Gardner, The Colossal Book of Mathematics, Classic Puzzles, Paradoxes and Problems, Chapter 2 entitled "The Calculus of Finite Differences," W. W. Norton and Company, New York, 2001, pages 12-13.
  • L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 44.
  • Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000, p. 213.

Crossrefs

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. A058331 and A247375. - Bruno Berselli, Sep 16 2014
Cf. A194715 (4-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017
Integers such that: this sequence (the number of even divisors is odd), A028982 (the number of odd divisors is odd), A028983 (the number of odd divisors is even), A183300 (the number of even divisors is even).

Programs

Formula

a(n) = (-1)^(n+1) * A053120(2*n, 2).
G.f.: 2*x*(1+x)/(1-x)^3.
a(n) = A100345(n, n).
Sum_{n>=1} 1/a(n) = Pi^2/12 =A072691. [Jolley eq. 319]. - Gary W. Adamson, Dec 21 2006
a(n) = A049452(n) - A033991(n). - Zerinvary Lajos, Jun 12 2007
a(n) = A016742(n)/2. - Zerinvary Lajos, Jun 20 2008
a(n) = 2 * A000290(n). - Omar E. Pol, May 14 2008
a(n) = 4*n + a(n-1) - 2, n > 0. - Vincenzo Librandi
a(n) = A002378(n-1) + A002378(n). - Joerg M. Schuetze (joerg(AT)cyberheim.de), Mar 08 2010 [Corrected by Klaus Purath, Jun 18 2020]
a(n) = A176271(n,k) + A176271(n,n-k+1), 1 <= k <= n. - Reinhard Zumkeller, Apr 13 2010
a(n) = A007607(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
For n > 0, a(n) = 1/coefficient of x^2 in the Maclaurin expansion of 1/(cos(x)+n-1). - Francesco Daddi, Aug 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Artur Jasinski, Nov 24 2011
a(n) = A070216(n,n) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) = A014132(2*n-1,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n) = A000217(n) + A000326(n). - Omar E. Pol, Jan 11 2013
(a(n) - A000217(k))^2 = A000217(2*n-1-k)*A000217(2*n+k) + n^2, for all k. - Charlie Marion, May 04 2013
a(n) = floor(1/(1-cos(1/n))), n > 0. - Clark Kimberling, Oct 08 2014
a(n) = A251599(3*n-1) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+4)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = A002061(n+1) + A165900(n). - Torlach Rush, Feb 21 2019
E.g.f.: 2*exp(x)*x*(1 + x). - Stefano Spezia, Oct 12 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Jul 03 2020
From Amiram Eldar, Feb 03 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sqrt(2)*sinh(Pi/sqrt(2))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(2)*sin(Pi/sqrt(2))/Pi. (End)

A303305 Generalized 17-gonal (or heptadecagonal) numbers: m*(15*m - 13)/2 with m = 0, +1, -1, +2, -2, +3, -3, ...

Original entry on oeis.org

0, 1, 14, 17, 43, 48, 87, 94, 146, 155, 220, 231, 309, 322, 413, 428, 532, 549, 666, 685, 815, 836, 979, 1002, 1158, 1183, 1352, 1379, 1561, 1590, 1785, 1816, 2024, 2057, 2278, 2313, 2547, 2584, 2831, 2870, 3130, 3171, 3444, 3487, 3773, 3818, 4117, 4164, 4476, 4525, 4850
Offset: 0

Views

Author

Omar E. Pol, Jun 06 2018

Keywords

Comments

120*a(n) + 169 is a square. - Bruno Berselli, Jun 08 2018
Partial sums of A317313. - Omar E. Pol, Jul 28 2018
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. They are also the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, k >= 5. In this case k = 17. - Omar E. Pol, Apr 25 2021

Examples

			From _Omar E. Pol_, Apr 24 2021: (Start)
Illustration of initial terms as vertices of a rectangular spiral:
        43_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _17
         |                                                   |
         |                         0                         |
         |                         |_ _ _ _ _ _ _ _ _ _ _ _ _|
         |                         1                         14
         |
        48
More generally, all generalized k-gonal numbers can be represented with this kind of spirals, k >= 5". (End)
		

Crossrefs

Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), this sequence (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).

Programs

  • Mathematica
    With[{pp = 17, nn = 55}, {0}~Join~Riffle[Array[PolygonalNumber[pp, #] &, Ceiling[nn/2]], Array[PolygonalNumber[pp, -#] &, Ceiling[nn/2]]]] (* Michael De Vlieger, Jun 06 2018 *)
    Table[(30 n (n + 1) + 11 (2 n + 1) (-1)^n - 11)/16, {n, 0, 60}] (* Bruno Berselli, Jun 08 2018 *)
    CoefficientList[ Series[-x (x^2 + 13x + 1)/((x - 1)^3 (x + 1)^2), {x, 0, 50}], x] (* or *)
    LinearRecurrence[{1, 2, -2, -1, 1}, {0, 1, 14, 17, 43}, 51] (* Robert G. Wilson v, Jul 28 2018 *)
  • PARI
    concat(0, Vec(x*(1 + 13*x + x^2)/((1 + x)^2*(1 - x)^3) + O(x^40))) \\ Colin Barker, Jun 12 2018

Formula

From Bruno Berselli, Jun 08 2018: (Start)
G.f.: x*(1 + 13*x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n-1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = (30*n*(n + 1) + 11*(2*n + 1)*(-1)^n - 11)/16. Therefore:
a(n) = n*(15*n + 26)/8, if n is even, or (n + 1)*(15*n - 11)/8 otherwise.
2*(2*n - 1)*a(n) + 2*(2*n + 1)*a(n-1) - n*(15*n^2 - 13) = 0. (End)

A069131 Centered 18-gonal numbers.

Original entry on oeis.org

1, 19, 55, 109, 181, 271, 379, 505, 649, 811, 991, 1189, 1405, 1639, 1891, 2161, 2449, 2755, 3079, 3421, 3781, 4159, 4555, 4969, 5401, 5851, 6319, 6805, 7309, 7831, 8371, 8929, 9505, 10099, 10711, 11341, 11989, 12655, 13339, 14041, 14761, 15499, 16255, 17029, 17821
Offset: 1

Views

Author

Terrel Trotter, Jr., Apr 07 2002

Keywords

Comments

Equals binomial transform of [1, 18, 18, 0, 0, 0, ...]. Example: a(3) = 55 = (1, 2, 1) dot (1, 18, 18) = (1 + 36 + 18). - Gary W. Adamson, Aug 24 2010
Narayana transform (A001263) of [1, 18, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
From Lamine Ngom, Aug 19 2021: (Start)
Sequence is a spoke of the hexagonal spiral built from the terms of A016777 (see illustration in links section).
a(n) is a bisection of A195042.
a(n) is a trisection of A028387.
a(n) + 1 is promic (A002378).
a(n) + 2 is a trisection of A002061.
a(n) + 9 is the arithmetic mean of its neighbors.
4*a(n) + 5 is a square: A016945(n)^2. (End)

Examples

			a(5) = 181 because 9*5^2 - 9*5 + 1 = 225 - 45 + 1 = 181.
		

Crossrefs

Programs

Formula

a(n) = 9*n^2 - 9*n + 1.
a(n) = 18*n + a(n-1) - 18 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: ( x*(1+16*x+x^2) ) / ( (1-x)^3 ). - R. J. Mathar, Feb 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=19, a(3)=55. - Harvey P. Dale, Jan 20 2014
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(5)*Pi/6)/(3*sqrt(5)).
Sum_{n>=1} a(n)/n! = 10*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 10/e - 1. (End)
From Lamine Ngom, Aug 19 2021: (Start)
a(n) = 18*A000217(n) + 1 = 9*A002378(n) + 1.
a(n) = 3*A003215(n) - 2.
a(n) = A247792(n) - 9*n.
a(n) = A082040(n) + A304163(n) - a(n-1) = A016778(n) + A016790(n) - a(n-1), n > 0.
a(n) + a(n+1) = 2*A247792(n) = A010008(n), n > 0.
a(n+1) - a(n) = 18*n = A008600(n). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n)= A000290(n) + A139278(n-1)
a(n) = A069129(n) + A002378(n-1)
a(n) = A062786(n) + 8*A000217(n-1)
a(n) = A062786(n) + A033996(n-1)
a(n) = A060544(n) + 9*A000217(n-1)
a(n) = A060544(n) + A027468(n-1)
a(n) = A016754(n-1) + 10*A000217(n-1)
a(n) = A016754(n-1) + A124080
a(n) = A069099(n) + 11*A000217(n-1)
a(n) = A069099(n) + A152740(n-1)
a(n) = A003215(n-1) + 12*A000217(n-1)
a(n) = A003215(n-1) + A049598(n-1)
a(n) = A005891(n-1) + 13*A000217(n-1)
a(n) = A005891(n-1) + A152741(n-1)
a(n) = A001844(n) + 14*A000217(n-1)
a(n) = A001844(n) + A163756(n-1)
a(n) = A005448(n) + 15*A000217(n-1)
a(n) = A005448(n) + A194715(n-1). (End)
E.g.f.: exp(x)*(1 + 9*x^2) - 1. - Nikolaos Pantelidis, Feb 06 2023

A069128 Centered 15-gonal numbers: a(n) = (15*n^2 - 15*n + 2)/2.

Original entry on oeis.org

1, 16, 46, 91, 151, 226, 316, 421, 541, 676, 826, 991, 1171, 1366, 1576, 1801, 2041, 2296, 2566, 2851, 3151, 3466, 3796, 4141, 4501, 4876, 5266, 5671, 6091, 6526, 6976, 7441, 7921, 8416, 8926, 9451, 9991, 10546, 11116, 11701, 12301, 12916, 13546, 14191, 14851, 15526
Offset: 1

Views

Author

Terrel Trotter, Jr., Apr 07 2002

Keywords

Comments

Centered pentadecagonal numbers or centered quindecagonal numbers or centered pentakaidecagonal numbers. - Omar E. Pol, Oct 03 2011

Examples

			a(5) = 151 because (15*5^2 - 15*5 + 2)/2 = 151.
		

Crossrefs

Programs

Formula

a(n) = (15*n^2 - 15*n + 2)/2.
a(n) = 15*n+a(n-1)-15 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+13*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
Binomial transform of [1, 15, 15, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 15, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
a(n) = A194715(n-1) + 1. - Omar E. Pol, Oct 03 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(7/15)*Pi/2)/sqrt(105).
Sum_{n>=1} a(n)/n! = 17*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 17/(2*e) - 1. (End)
E.g.f.: exp(x)*(1 + 15*x^2/2) - 1. - Nikolaos Pantelidis, Feb 07 2023

A290391 Number of 5-cycles in the n-triangular honeycomb obtuse knight graph.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 30, 120, 294, 552, 894, 1320, 1830, 2424, 3102, 3864, 4710, 5640, 6654, 7752, 8934, 10200, 11550, 12984, 14502, 16104, 17790, 19560, 21414, 23352, 25374, 27480, 29670, 31944, 34302, 36744, 39270, 41880, 44574, 47352, 50214, 53160, 56190, 59304, 62502
Offset: 1

Views

Author

Eric W. Weisstein, Jul 29 2017

Keywords

Crossrefs

Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A194715 (4-cycles), A290392 (6-cycles).

Programs

  • Mathematica
    Table[If[n < 7, 0, 6 (292 - 90 n + 7 n^2)], {n, 20}]
    Join[{0, 0, 0, 0, 0, 0}, LinearRecurrence[{3, -3, 1}, {30, 120, 294}, 14]]
    CoefficientList[Series[-((6 x^6 (5 + 5 x + 4 x^2))/(-1 + x)^3), {x, 0, 20}], x]

Formula

For n >= 7, a(n) = 6*(292 - 90*n + 7*n^2).
For n >= 10, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
G.f.: -6*x^7*(5 + 5*x + 4*x^2)/(-1 + x)^3.

A290392 Number of 6-cycles in the n-triangular honeycomb obtuse knight graph.

Original entry on oeis.org

0, 0, 0, 0, 4, 13, 98, 415, 1151, 2471, 4385, 6893, 9995, 13691, 17981, 22865, 28343, 34415, 41081, 48341, 56195, 64643, 73685, 83321, 93551, 104375, 115793, 127805, 140411, 153611, 167405, 181793, 196775, 212351, 228521, 245285, 262643, 280595
Offset: 1

Views

Author

Eric W. Weisstein, Jul 29 2017

Keywords

Crossrefs

Cf. A001105 (3-cycles in the triangular honeycomb obtuse knight graph), A194715 (4-cycles), A290391 (5-cycles).

Programs

  • Mathematica
    Table[Piecewise[{{0, n <= 4}, {4, n == 5}, {13, n == 6}, {98, n == 7}, {415, n == 8}, {16001 - 4323 n + 297 n^2, n > 8}}, 0], {n, 20}]
    Join[{0, 0, 0, 0, 4, 13, 98, 415}, LinearRecurrence[{3, -3, 1}, {1151, 2471, 4385}, 12]]
    CoefficientList[Series[(x^4 (-4 - x - 71 x^2 - 156 x^3 - 187 x^4 - 165 x^5 - 10 x^6))/(-1 + x)^3, {x, 0, 20}], x]

Formula

For n >= 9, a(n) = 16001 - 4323*n + 297*n^2.
For n >= 12, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3).
G.f.: x^5*(-4 - x - 71*x^2 - 156*x^3 - 187*x^4 - 165*x^5 - 10*x^6)/(-1 +
x)^3.
Showing 1-6 of 6 results.