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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A022267 a(n) = n*(9*n + 1)/2.

Original entry on oeis.org

0, 5, 19, 42, 74, 115, 165, 224, 292, 369, 455, 550, 654, 767, 889, 1020, 1160, 1309, 1467, 1634, 1810, 1995, 2189, 2392, 2604, 2825, 3055, 3294, 3542, 3799, 4065, 4340, 4624, 4917, 5219, 5530, 5850, 6179
Offset: 0

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Author

Keywords

Comments

From Floor van Lamoen, Jul 21 2001: (Start)
Write 0, 1, 2, 3, 4, ... in a triangular spiral; then a(n) is the sequence found by reading the line from 0 in the direction 0, 5, ... . The spiral begins:
.
15
/ \
16 14
/ \
17 3 13
/ / \ \
18 4 2 12
/ / \ \
19 5 0---1 11
/ / \
20 6---7---8---9--10
.
(End)
a(n) is the sum of n consecutive integers starting from 4*n+1: (5), (9+10), (13+14+15), ... - Klaus Purath, Jul 07 2020
a(n) with n>0 are the numbers with the periodic length 3 in the Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022

Crossrefs

Cf. similar sequences listed in A254963.
Cf. similar sequences listed in A022289.

Programs

  • Maple
    seq(binomial(9*n+1,2)/9, n=0..37); # Zerinvary Lajos, Jan 21 2007
  • Mathematica
    Table[ n (9 n + 1)/2, {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 19}, 40] (* Harvey P. Dale, Jul 01 2013 *)
  • PARI
    vector(100,n,(n-1)*(9*n-8)/2) \\ Derek Orr, Feb 06 2015

Formula

a(n) = A110449(n, 4) for n>3.
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 4*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A000566(n). (End)
a(n) = 9*n + a(n-1) - 4 with n>0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = A218470(9*n+4). - Philippe Deléham, Mar 27 2013
a(n) = A000217(5*n) - A000217(4*n). - Bruno Berselli, Oct 13 2016
E.g.f.: (1/2)*(9*x^2 + 10*x)*exp(x). - G. C. Greubel, Jul 17 2017
a(n) = A060544(n+1) - A016813(n). - Leo Tavares, Mar 20 2022

A038764 a(n) = (9*n^2 + 3*n + 2)/2.

Original entry on oeis.org

1, 7, 22, 46, 79, 121, 172, 232, 301, 379, 466, 562, 667, 781, 904, 1036, 1177, 1327, 1486, 1654, 1831, 2017, 2212, 2416, 2629, 2851, 3082, 3322, 3571, 3829, 4096, 4372, 4657, 4951, 5254, 5566, 5887, 6217, 6556, 6904, 7261, 7627, 8002, 8386, 8779, 9181
Offset: 0

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Author

N. J. A. Sloane, May 03 2000

Keywords

Comments

Coefficients of x^2 of certain rook polynomials (for n>=1; see p. 18 of the Riordan paper). - Emeric Deutsch, Mar 08 2004
a(n) is also the least weight of self-conjugate partitions having n+1 different parts such that each part is congruent to 1 modulo 3. The first such self-conjugate partitions, corresponding to a(n) = 0, 1, 2, 3, are 1, 4+3, 7+4+4+4+3, 10+7+7+7+4+4+4+3. - Augustine O. Munagi, Dec 18 2008

References

  • J. Riordan, Discordant permutations, Scripta Math., 20 (1954), 14-23.

Crossrefs

Reflection of A060544 in A081272.
Second column of A024462. Also = A064641(n+1, 2).
Shallow diagonal of triangular spiral in A051682.
Partial sums of A122709.

Programs

  • Mathematica
    LinearRecurrence[{3, -3, 1}, {1, 7, 22}, 50] (* Paolo Xausa, Jul 03 2025 *)
  • PARI
    a(n)=n*(9*n+3)/2+1 \\ Charles R Greathouse IV, Jun 17 2017
    
  • PARI
    Vec((1 + 2*x)^2 / (1 - x)^3 + O(x^60)) \\ Colin Barker, Jan 22 2018
  • Sage
    a = lambda n: hypergeometric([-n, -2], [1], 3)
    print([simplify(a(n)) for n in range(46)]) # Peter Luschny, Nov 19 2014
    

Formula

a(n) = binomial(n,0) + 6*binomial(n,1) + 9*binomial(n,2).
From Paul Barry, Mar 15 2003: (Start)
G.f.: (1 + 2*x)^2/(1 - x)^3.
Binomial transform of (1, 6, 9, 0, 0, 0, ...). (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - Colin Barker, Jan 22 2018
a(n) = a(n-1) + 3*(3*n-1) for n>0, a(0)=1. - Vincenzo Librandi, Nov 17 2010
a(n) = hypergeometric([-n, -2], [1], 3). - Peter Luschny, Nov 19 2014
E.g.f.: exp(x)*(2 + 12*x + 9*x^2)/2. - Stefano Spezia, Mar 07 2023

Extensions

More terms from James Sellers, May 03 2000
Entry revised by N. J. A. Sloane, Jan 23 2018

A002114 Glaisher's H' numbers.

Original entry on oeis.org

1, 11, 301, 15371, 1261501, 151846331, 25201039501, 5515342166891, 1538993024478301, 533289474412481051, 224671379367784281901, 113091403397683832932811, 67032545884354589043714301, 46211522130188693681603906171
Offset: 1

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Author

Keywords

Comments

a(n) mod 9 = 1,2,4,8,7,5 repeated period 6 (A153130, see also A001370). a(n) mod 10 = 1. - Paul Curtz, Sep 10 2009

References

  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 76.
  • 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).

Programs

  • Maple
    a := n -> (-1)^n*6^(2*n)*(Zeta(0,-n*2,1/3)-Zeta(0,-n*2, 5/6)):
    seq(a(n), n=1..14);
  • Mathematica
    Select[Rest[With[{nn=28},CoefficientList[Series[1/(2 (2Cos[x]-1)), {x,0,nn}], x]Range[0,nn]!]],#!=0&] (* Harvey P. Dale, Jul 27 2011 *)
    FullSimplify[Table[(-1)^(s+1) * BernoulliB[2*s] * (Zeta[2*s + 1, 1/6] - Zeta[2*s + 1, 5/6]) / (4*Pi*Sqrt[3]*Zeta[2*s]), {s, 1, 20}]]  (* Vaclav Kotesovec, May 05 2020 *)
  • Maxima
    a(n) := sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

Formula

H'(n) = H(n)/3, where H(n)=2^(2n+1)*I(n) (see A002112) and e.g.f. for (-1)^n*I(n) is (3/2)/(1+exp(x)+exp(-x)) (see A047788, A047789).
H'(n) = A000436(n)/2^(2n+1). - Philippe Deléham, Jan 17 2004
For n > 0, H'(n) = Sum{k = 0..n, T(n, k)*9^(n-k)*2^(k-1) }; where DELTA is the operator defined in A084938, T(n, k) is the triangle, read by rows, given by :[0, 1, 0, 4, 0, 9, 0, 16, 0, 25, ...] DELTA [1, 0, 10, 0, 28, 0, 55, 0, 90, ..]= {1}; {0, 1}; {0, 1, 1}; {0, 1, 12, 1}; {0, 1, 63, 123, 1}; {0, 1, 274, 2366, 1234, 1}; ... For 1, 10, 28, 55, 90, 136, ... see A060544 or A060544. - Philippe Deléham, Jan 17 2004
E.g.f. 1/2*1/(2*cos(x)-1). a(n)=sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
E.g.f.: E(x)= x^2/(G(0)-x^2) ; G(k)= 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 03 2012
If E(x)=Sum(k=0,1,..., a(k+1)*x^(2k+2)), then A002114(k) = a(k+1)*(2*k+2)!. - Sergei N. Gladkovskii, Jan 09 2012
a(n) ~ (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - Vaclav Kotesovec, Feb 26 2014
a(n) = (-1)^n*6^(2*n)*(zeta(-n*2,1/3)-zeta(-n*2,5/6)), where zeta(a, z) is the generalized Riemann zeta function.
From Vaclav Kotesovec, May 05 2020: (Start)
a(n) = (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (sqrt(3)*(2*Pi)^(2*n+1)).
a(n) = (-1)^(n+1) * Bernoulli(2*n) * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (4*Pi*sqrt(3)*zeta(2*n)). (End)
Conjectural e.g.f.: Sum_{n >= 1} (-1)^n*Product_{k = 1..n} (1 - exp(A007310(k)*z) ) = z + 11*z^2/2! + 301*z^3/3! + .... - Peter Bala, Dec 09 2021

A151542 Generalized pentagonal numbers: a(n) = 12*n + 3*n*(n-1)/2.

Original entry on oeis.org

0, 12, 27, 45, 66, 90, 117, 147, 180, 216, 255, 297, 342, 390, 441, 495, 552, 612, 675, 741, 810, 882, 957, 1035, 1116, 1200, 1287, 1377, 1470, 1566, 1665, 1767, 1872, 1980, 2091, 2205, 2322, 2442, 2565, 2691, 2820, 2952, 3087, 3225, 3366, 3510, 3657, 3807, 3960
Offset: 0

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Author

N. J. A. Sloane, May 15 2009

Keywords

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.

Programs

  • Mathematica
    s=0;lst={};Do[AppendTo[lst,s+=n],{n,12,6!,3}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 05 2010 *)
    LinearRecurrence[{3,-3,1}, {0,12,27}, 50] (* or *) With[{nn = 50}, CoefficientList[Series[(3/2)*(8*x + x^2)*Exp[x], {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 26 2017 *)
  • PARI
    x='x+O('x^50); concat([0], Vec(serlaplace((3/2)*(8*x + x^2)*exp(x)))) \\ G. C. Greubel, May 26 2017
    
  • PARI
    a(n)=(3*n^2+21*n)/2 \\ Charles R Greathouse IV, Jun 16 2017

Formula

a(n) = a(n-1) + 3*n + 9 (with a(0)=0). - Vincenzo Librandi, Nov 26 2010
G.f.: 3*x*(4 - 3*x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 24 2011
From G. C. Greubel, May 26 2017: (Start)
a(n) = 3*n*(n+7)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: (3/2)*(8*x + x^2)*exp(x). (End)
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 121/490.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/21 - 319/4410. (End)
a(n) = A003154(n+1) - A060544(n). - Leo Tavares, Mar 26 2022

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+n*k+k, n*k+k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 5, 1, 1, 35, 28, 7, 1, 1, 126, 165, 55, 9, 1, 1, 462, 1001, 455, 91, 11, 1, 1, 1716, 6188, 3876, 969, 136, 13, 1, 1, 6435, 38760, 33649, 10626, 1771, 190, 15, 1, 1, 24310, 245157, 296010, 118755, 23751, 2925, 253, 17, 1, 1, 92378, 1562275
Offset: 0

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Author

Henry Bottomley, Apr 02 2001

Keywords

Comments

Main diagonal is A108288. Antidiagonal sums is A108289. Inverse binomial transforms of each row give triangle A108290. G.f. of row n multiplied by (1-x)^(n+1) equals g.f. of row n of triangle A108267 (rows sums of A108267 equal (n+1)^n).

Examples

			row 1: (2*n+1)/1!
row 2: (3*n+1)*(3*n+2)/2!
row 3: (4*n+1)*(4*n+2)*(4*n+3)/3!
row 4: (5*n+1)*(5*n+2)*(5*n+3)*(5*n+4)/4!
row 5: (6*n+1)*(6*n+2)*(6*n+3)*(6*n+4)*(6*n+5)/5!.
Table begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,10,28,55,91,136,190,253,325,406,496,...
1,35,165,455,969,1771,2925,4495,6545,...
1,126,1001,3876,10626,23751,46376,82251,...
1,462,6188,33649,118755,324632,749398,...
1,1716,38760,296010,1344904,4496388,...
		

Crossrefs

Cf. A108290, A108267, A108288, A108289, A060544 (row 2), A015219 (row 3).
Rows include A000012, A001700, A025174. Columns include A000012, A005408, A060544. Main diagonal is A060545.

Programs

  • PARI
    T(n,k)=binomial(n+n*k+k,n*k+k)
    
  • PARI
    { i=0; write("b060543.txt", "0 1"); for (m=0, 20, for (k=0, m + 1, n=m - k + 1; write("b060543.txt", i++, " ", binomial(n + n*k + k, n*k + k))); ) } \\ Harry J. Smith, Jul 06 2009

Formula

a(n) = A060539(n, k)/n = A007318(nk, k)/n = A060540(n, k)/A060540(n-1, k).

Extensions

Entry revised by Paul D. Hanna, May 31 2005
Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 17 2007

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

A086089 Decimal expansion of 3*sqrt(3)/(2*Pi).

Original entry on oeis.org

8, 2, 6, 9, 9, 3, 3, 4, 3, 1, 3, 2, 6, 8, 8, 0, 7, 4, 2, 6, 6, 9, 8, 9, 7, 4, 7, 4, 6, 9, 4, 5, 4, 1, 6, 2, 0, 9, 6, 0, 7, 9, 7, 2, 0, 5, 4, 9, 9, 6, 0, 9, 7, 9, 1, 9, 9, 0, 4, 9, 0, 3, 0, 4, 3, 6, 5, 4, 5, 4, 5, 5, 2, 0, 3, 9, 0, 4, 6, 9, 2, 2, 6, 0, 5, 7, 0, 0, 4, 3, 2, 3, 4, 7, 5, 6, 3, 3, 3, 8, 1, 1
Offset: 0

Views

Author

Eric W. Weisstein, Jul 08 2003

Keywords

Comments

Limiting ratio of areas in the disk-covering problem.
From Daniel Forgues, May 26 2010: (Start)
Consider: A060544 (Centered 9-gonal numbers), starting with a(1)=1, P_c(9, n), n >= 1. Every third triangular number, starting with a(1)=1, P(3, 3n-2), n >= 1. Then:
1/(Sum_{n=0..infinity} 1/binomial(3n+2,2)) = 1/(Sum_{n=1..infinity} 1/binomial(3n-1,2)) = 1/(Sum_{n=1..infinity} 1/P_c(9,n)) = 1/(Sum_{n=1..infinity} 1/P(3,3n-2)) = 1/(Sum_{n=1..infinity} 1/A060544(n)) = this constant. (End)
The area of a regular hexagon circumscribed in a unit-area circle. - Amiram Eldar, Nov 05 2020

Examples

			0.8269933431326880742669897474694541620960797205499609791990...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Sections 5.9 p. 325 and 8.2 p. 486.
  • Steven R. Finch, Mathematical Constants II, Cambridge University Press, 2018, p. 196.

Crossrefs

Programs

  • Mathematica
    RealDigits[3 Sqrt[3]/(2 Pi), 10, 110][[1]] (* or, from the third comment: *) RealDigits[N[Product[1 - 1/(3 n)^2, {n, 1, Infinity}], 110]][[1]] (* Bruno Berselli, Apr 02 2013 *)
  • PARI
    3*sqrt(3)/(2*Pi) \\ Michel Marcus, Nov 05 2020

Formula

Equals Product_{n>=1} (1 - 1/(3n)^2). - Bruno Berselli, Apr 02 2013
Equals sinc(Pi/3). - Peter Luschny, Oct 04 2019
Equals Product{k>=1} cos(Pi/(3*2^k)). - Amiram Eldar, Aug 20 2020
Equals Sum_{k>=0} mu(3*k+1)/(3*k+1) (Nevanlinna, 1973). - Amiram Eldar, Dec 21 2020

A195042 Concentric 9-gonal numbers.

Original entry on oeis.org

0, 1, 9, 19, 36, 55, 81, 109, 144, 181, 225, 271, 324, 379, 441, 505, 576, 649, 729, 811, 900, 991, 1089, 1189, 1296, 1405, 1521, 1639, 1764, 1891, 2025, 2161, 2304, 2449, 2601, 2755, 2916, 3079, 3249, 3421, 3600, 3781, 3969, 4159, 4356, 4555, 4761, 4969, 5184, 5401, 5625
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Comments

Also concentric enneagonal numbers or concentric nonagonal numbers.
A016766 and A069131 interleaved.
Partial sums of A056020. - Reinhard Zumkeller, Jan 07 2012

Crossrefs

Programs

  • Haskell
    a195042 n = a195042_list !! n
    a195042_list = scanl (+) 0 a056020_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Magma
    [(9*n^2+5/2*((-1)^n-1))/4: n in [0..50]]; // Vincenzo Librandi, Sep 29 2011
    
  • Mathematica
    LinearRecurrence[{2,0,-2,1},{0,1,9,19},60] (* Harvey P. Dale, Nov 24 2019 *)
  • PARI
    a(n)=(9*n^2+5/2*((-1)^n-1))/4 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = (9*n^2 + 5/2*((-1)^n - 1))/4.
From R. J. Mathar, Sep 28 2011: (Start)
G.f.: -x*(1+7*x+x^2) / ( (1+x)*(x-1)^3 ).
a(n) + a(n+1) = A060544(n+1). (End)
Sum_{n>=1} 1/a(n) = Pi^2/54 + tan(sqrt(5)*Pi/6)*Pi/(3*sqrt(5)). - Amiram Eldar, Jan 16 2023

A086272 Rectangular array T(n,k) of central polygonal numbers, by antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 13, 10, 5, 1, 21, 19, 13, 6, 1, 31, 31, 25, 16, 7, 1, 43, 46, 41, 31, 19, 8, 1, 57, 64, 61, 51, 37, 22, 9, 1, 73, 85, 85, 76, 61, 43, 25, 10, 1, 91, 109, 113, 106, 91, 71, 49, 28, 11, 1, 111, 136, 145, 141, 127, 106, 81, 55, 31, 12, 1, 133, 166, 181, 181, 169
Offset: 1

Views

Author

Clark Kimberling, Jul 14 2003

Keywords

Comments

In the standard notation, the offset is different: the first row are the 2-gonal, the second row the 3-gonal numbers, etc. - R. J. Mathar, Oct 07 2011

Examples

			First rows:
1,3,7,13,21,31,43,57,73,91,111,..   A002061
1,4,10,19,31,46,64,85,109,136,166,...  A005448
1,5,13,25,41,61,85,113,145,181,221,..   A001844
1,6,16,31,51,76,106,141,181,226,276,...  A005891
1,7,19,37,61,91,127,169,217,271,331,...   A003215
1,8,22,43,71,106,148,197,253,316,386,...    A069099
1,9,25,49,81,121,169,225,289,361,441,...    A016754
1,10,28,55,91,136,190,253,325,406,496,...    A060544
		

Crossrefs

Formula

T(k, n)=(k+1)*binomial(n, 2)+1.

A262221 a(n) = 25*n*(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651
Offset: 0

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Author

Bruno Berselli, Sep 15 2015

Keywords

Comments

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Crossrefs

Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Programs

  • Magma
    [25*n*(n+1)/2+1: n in [0..50]];
  • Mathematica
    Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
    25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)
  • PARI
    vector(50, n, n--; 25*n*(n+1)/2+1)
    
  • Sage
    [25*n*(n+1)/2+1 for n in (0..50)]
    

Formula

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024
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