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.

A195040 Square array read by antidiagonals with T(n,k) = k*n^2/4+(k-4)*((-1)^n-1)/8, n>=0, k>=0.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 3, 2, 1, 0, 1, 4, 5, 3, 1, 0, 0, 7, 8, 7, 4, 1, 0, 1, 9, 13, 12, 9, 5, 1, 0, 0, 13, 18, 19, 16, 11, 6, 1, 0, 1, 16, 25, 27, 25, 20, 13, 7, 1, 0, 0, 21, 32, 37, 36, 31, 24, 15, 8, 1, 0, 1, 25, 41, 48, 49, 45, 37, 28, 17, 9, 1, 0
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Comments

Also, if k >= 2 and m = 2*k, then column k lists the numbers of the form k*n^2 and the centered m-gonal numbers interleaved.
For k >= 3, this is also a table of concentric polygonal numbers. Column k lists the concentric k-gonal numbers.
It appears that the first differences of column k are the numbers that are congruent to {1, k-1} mod k, if k >= 3.

Examples

			Array begins:
  0,   0,   0,   0,   0,   0,   0,   0,   0,   0, ...
  1,   1,   1,   1,   1,   1,   1,   1,   1,   1, ...
  0,   1,   2,   3,   4,   5,   6,   7,   8,   9, ...
  1,   3,   5,   7,   9,  11,  13,  15,  17,  19, ...
  0,   4,   8,  12,  16,  20,  24,  28,  32,  36, ...
  1,   7,  13,  19,  25,  31,  37,  43,  49,  55, ...
  0,   9,  18,  27,  36,  45,  54,  63,  72,  81, ...
  1,  13,  25,  37,  49,  61,  73,  85,  97, 109, ...
  0,  16,  32,  48,  64,  80,  96, 112, 128, 144, ...
  1,  21,  41,  61,  81, 101, 121, 141, 161, 181, ...
  0,  25,  50,  75, 100, 125, 150, 175, 200, 225, ...
  ...
		

Crossrefs

Rows n: A000004 (n=0), A000012 (n=1), A001477 (n=2), A005408 (n=3), A008586 (n=4), A016921 (n=5), A008591 (n=6), A017533 (n=7), A008598 (n=8), A215145 (n=9), A008607 (n=10).
Columns k: A000035 (k=0), A004652 (k=1), A000982 (k=2), A077043 (k=3), A000290 (k=4), A032527 (k=5), A032528 (k=6), A195041 (k=7), A077221 (k=8), A195042 (k=9), A195142 (k=10), A195043 (k=11), A195143 (k=12), A195045 (k=13), A195145 (k=14), A195046 (k=15), A195146 (k=16), A195047 (k=17), A195147 (k=18), A195048 (k=19), A195148 (k=20), A195049 (k=21), A195149 (k=22), A195058 (k=23), A195158 (k=24).

Programs

  • GAP
    nmax:=13;; T:=List([0..nmax],n->List([0..nmax],k->k*n^2/4+(k-4)*((-1)^n-1)/8));; b:=List([2..nmax],n->OrderedPartitions(n,2));;
    a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Jul 19 2018
  • Maple
    A195040 := proc(n,k)
            k*n^2/4+((-1)^n-1)*(k-4)/8 ;
    end proc:
    for d from 0 to 12 do
            for k from 0 to d do
                    printf("%d,",A195040(d-k,k)) ;
            end do:
    end do; # R. J. Mathar, Sep 28 2011
  • Mathematica
    t[n_, k_] := k*n^2/4+(k-4)*((-1)^n-1)/8; Flatten[ Table[ t[n-k, k], {n, 0, 11}, {k, 0, n}]] (* Jean-François Alcover, Dec 14 2011 *)

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988
Offset: 0

Views

Author

Vincenzo Librandi, Jul 03 2014

Keywords

Comments

First bisection of A195047. - Bruno Berselli, Jul 03 2014
Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

Crossrefs

Cf. similar sequences of the type k*n^2: A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12), A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), this sequence (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).

Programs

Formula

G.f.: 17*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A195147 Concentric 18-gonal numbers.

Original entry on oeis.org

0, 1, 18, 37, 72, 109, 162, 217, 288, 361, 450, 541, 648, 757, 882, 1009, 1152, 1297, 1458, 1621, 1800, 1981, 2178, 2377, 2592, 2809, 3042, 3277, 3528, 3781, 4050, 4321, 4608, 4897, 5202, 5509, 5832, 6157, 6498, 6841, 7200, 7561, 7938, 8317, 8712, 9109
Offset: 0

Views

Author

Omar E. Pol, Sep 17 2011

Keywords

Comments

Concentric octadecagonal numbers or concentric octakaidecagonal numbers.
Sequence found by reading the line from 0, in the direction 0, 18, ..., and the same line from 1, in the direction 1, 37, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Main axis, perpendicular to A027468 in the same spiral.

Crossrefs

A195321 and A195316 interleaved.
Cf. A032527, A195047, A195048. Column 18 of A195040. - Omar E. Pol, Sep 29 2011

Programs

Formula

G.f.: -x*(1+16*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = (18*n^2 + 7*(-1)^n - 7)/4;
a(n) = -a(n-1) + 9*n^2 - 9*n + 1. (End)
Sum_{n>=1} 1/a(n) = Pi^2/108 + tan(sqrt(7)*Pi/6)*Pi/(6*sqrt(7)). - Amiram Eldar, Jan 17 2023

A195048 Concentric 19-gonal numbers.

Original entry on oeis.org

0, 1, 19, 39, 76, 115, 171, 229, 304, 381, 475, 571, 684, 799, 931, 1065, 1216, 1369, 1539, 1711, 1900, 2091, 2299, 2509, 2736, 2965, 3211, 3459, 3724, 3991, 4275, 4561, 4864, 5169, 5491, 5815, 6156, 6499, 6859, 7221, 7600, 7981, 8379, 8779, 9196
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Comments

Also concentric enneadecagonal numbers.

Crossrefs

Programs

Formula

a(n) = (19/4)*n^2 + (15/8)*((-1)^n - 1).
From Colin Barker, Sep 16 2012: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: x*(1 + 17*x + x^2)/((1-x)^3*(1+x)). (End)
Sum_{n>=1} 1/a(n) = Pi^2/114 + tan(sqrt(15/19)*Pi/2)*Pi/sqrt(285). - Amiram Eldar, Jan 17 2023

A195046 Concentric 15-gonal numbers.

Original entry on oeis.org

0, 1, 15, 31, 60, 91, 135, 181, 240, 301, 375, 451, 540, 631, 735, 841, 960, 1081, 1215, 1351, 1500, 1651, 1815, 1981, 2160, 2341, 2535, 2731, 2940, 3151, 3375, 3601, 3840, 4081, 4335, 4591, 4860, 5131, 5415, 5701, 6000, 6301, 6615, 6931, 7260, 7591
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[15n^2/4+11((-1)^n-1)/8,{n,0,50}] (* or *) LinearRecurrence[ {2,0,-2,1},{0,1,15,31},50] (* Harvey P. Dale, Feb 23 2012 *)
  • PARI
    a(n)=15*n^2/4+11*((-1)^n-1)/8 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 15*n^2/4+11*((-1)^n-1)/8.
From Harvey P. Dale, Feb 23 2012: (Start)
a(0)=0, a(1)=1, a(2)=15, a(3)=31, a(n)=2*a(n-1)-2*a(n-3)+a(n-4).
G.f.: -((x*(1+x*(13+x)))/((-1+x)^3*(1+x))). (End)
Sum_{n>=1} 1/a(n) = Pi^2/90 + tan(sqrt(11/15)*Pi/2)*Pi/sqrt(165). - Amiram Eldar, Jan 16 2023

Extensions

a(1)=1 added by Harvey P. Dale, Feb 23 2012

A270693 Alternating sum of centered 25-gonal numbers.

Original entry on oeis.org

1, -25, 51, -100, 151, -225, 301, -400, 501, -625, 751, -900, 1051, -1225, 1401, -1600, 1801, -2025, 2251, -2500, 2751, -3025, 3301, -3600, 3901, -4225, 4551, -4900, 5251, -5625, 6001, -6400, 6801, -7225, 7651, -8100, 8551, -9025, 9501, -10000, 10501
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 21 2016

Keywords

Comments

The absolute value alternating sum of centered k-gonal numbers gives concentric k-gonal numbers.
More generally, the ordinary generating function for the alternating sum of centered k-gonal numbers is (1 - (k - 2)*x + x^2)/((1 - x)*(1 + x)^3).

Crossrefs

Programs

  • Magma
    [((-1)^n*(50*n^2 + 100*n + 29) - 21)/8 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016
  • Maple
    A270693:=n->((-1)^n*(50*n^2 + 100*n + 29) - 21)/8: seq(A270693(n), n=0..100); # Wesley Ivan Hurt, Sep 18 2017
  • Mathematica
    LinearRecurrence[{-2, 0, 2, 1}, {1, -25, 51, -100}, 41]
    Table[((-1)^n (50 n^2 + 100 n + 29) - 21)/8, {n, 0, 40}]
  • PARI
    x='x+O('x^100); Vec((1-23*x+x^2)/((1-x)*(1+x)^3)) \\ Altug Alkan, Mar 21 2016
    

Formula

G.f.: (1 - 23*x + x^2)/((1 - x)*(1 + x)^3).
E.g.f.: (1/8)*(-21*exp(x) + (29 - 150*x + 50*x^2)*exp(-x)).
a(n) = -2*a(n-1) + 2*a(n-3) + a(n-4).
a(n) = ((-1)^n*(50*n^2 + 100*n + 29) - 21)/8.
Showing 1-6 of 6 results.