A195148 -id:A195148 - cp's OEIS frontend

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

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

Omar E. Pol, Sep 27 2011

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.

			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, ...
  ...

Muniru A Asiru, Rows n=0..100, flattened

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).

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

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

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 *)

A124080 10 times triangular numbers: a(n) = 5n(n + 1).

Original entry on oeis.org

0, 10, 30, 60, 100, 150, 210, 280, 360, 450, 550, 660, 780, 910, 1050, 1200, 1360, 1530, 1710, 1900, 2100, 2310, 2530, 2760, 3000, 3250, 3510, 3780, 4060, 4350, 4650, 4960, 5280, 5610, 5950, 6300, 6660, 7030, 7410, 7800, 8200, 8610, 9030, 9460, 9900, 10350

Offset: 0

Zerinvary Lajos, Nov 24 2006

If Y is a 5-subset of an n-set X then, for n >= 5, a(n-4) is equal to the number of 5-subsets of X having exactly three elements in common with Y. Y is a 5-subset of an n-set X then, for n >= 6, a(n-6) is the number of (n-5)-subsets of X having exactly two elements in common with Y. - Milan Janjic, Dec 28 2007

Also sequence found by reading the line from 0, in the direction 0, 10, ... and the same line from 0, in the direction 0, 30, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Axis perpendicular to A195148 in the same spiral. - Omar E. Pol, Sep 18 2011

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A028895, A046092, A045943, A002378, A028896, A024966, A033996, A027468, A049598, A000217.

[ 5*n*(n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

[seq(10*binomial(n,2),n=1..51)];
seq(n*(n+1)*5, n=0..39); # Zerinvary Lajos, Mar 06 2007

10*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,10,30},50] (* Harvey P. Dale, Jul 21 2011 *)

a(n)=5*n*(n+1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = 10*C(n,2), n >= 1.
a(n) = A049598(n) - A002378(n). - Zerinvary Lajos, Mar 06 2007
a(n) = 5*n*(n + 1), n >= 0. - Zerinvary Lajos, Mar 06 2007
a(n) = 5*n^2 + 5*n = 10*A000217(n) = 5*A002378(n) = 2*A028895(n). - Omar E. Pol, Dec 12 2008
a(n) = 10*n + a(n-1) (with a(0) = 0). - Vincenzo Librandi, Nov 12 2009
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0) = 0, a(1) = 10, a(2) = 30. - Harvey P. Dale, Jul 21 2011
a(n) = A062786(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A131242(10*n+9). - Philippe Deléham, Mar 27 2013
From G. C. Greubel, Aug 22 2017: (Start)
G.f.: 10*x/(1 - x)^3.
E.g.f.: 5*x*(x + 2)*exp(x). (End)
From Amiram Eldar, Sep 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2)-1)/5. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(5/Pi)*cos(3*Pi/(2*sqrt(5))).
Product_{n>=1} (1 + 1/a(n)) = (5/Pi)*cos(Pi/(2*sqrt(5))). (End)

A195143 a(n) = n-th concentric 12-gonal number.

Original entry on oeis.org

0, 1, 12, 25, 48, 73, 108, 145, 192, 241, 300, 361, 432, 505, 588, 673, 768, 865, 972, 1081, 1200, 1321, 1452, 1585, 1728, 1873, 2028, 2185, 2352, 2521, 2700, 2881, 3072, 3265, 3468, 3673, 3888, 4105, 4332, 4561, 4800, 5041, 5292, 5545, 5808, 6073, 6348

Offset: 0

Omar E. Pol, Sep 17 2011

Concentric dodecagonal numbers. [corrected by Ivan Panchenko, Nov 09 2013]
Sequence found by reading the line from 0, in the direction 0, 12,..., and the same line from 1, in the direction 1, 25,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Main axis, perpendicular to A028896 in the same spiral.
Partial sums of A091998. - Reinhard Zumkeller, Jan 07 2012
Column 12 of A195040. - Omar E. Pol, Sep 28 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A135453 and A069190 interleaved.
Cf. A001082, A032527, A032528, A077221, A195043, A195045, A195040, A195142, A195145, A195146, A195147, A195148, A195149.
Cf. A016921 (6n+1), A016969 (6n+5), A091998 (positive integers of the form 12*k +- 1), A092242 (positive integers of the form 12*k +- 5).

a195143 n = a195143_list !! n
a195143_list = scanl (+) 0 a091998_list
-- Reinhard Zumkeller, Jan 07 2012

[(3*n^2+(-1)^n-1): n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

Table[Sum[2*(-1)^(n - k + 1) + 6*k - 3, {k, n}], {n, 0, 47}] (* L. Edson Jeffery, Sep 14 2014 *)

From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = 3*n^2+(-1)^n-1.
a(n) = -a(n-1) + 6*n^2 - 6*n + 1. (End)
G.f.: -x*(1+10*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = Sum_{k=1..n} (2*(-1)^(n-k+1) + 3*(2*k-1)), n>0, a(0) = 0. - L. Edson Jeffery, Sep 14 2014
Sum_{n>=1} 1/a(n) = Pi^2/72 + tan(Pi/sqrt(6))*Pi/(4*sqrt(6)). - Amiram Eldar, Jan 16 2023

A195145 Concentric 14-gonal numbers.

Original entry on oeis.org

0, 1, 14, 29, 56, 85, 126, 169, 224, 281, 350, 421, 504, 589, 686, 785, 896, 1009, 1134, 1261, 1400, 1541, 1694, 1849, 2016, 2185, 2366, 2549, 2744, 2941, 3150, 3361, 3584, 3809, 4046, 4285, 4536, 4789, 5054, 5321, 5600, 5881, 6174, 6469, 6776, 7085, 7406

Offset: 0

Omar E. Pol, Sep 17 2011

Also concentric tetradecagonal numbers or concentric tetrakaidecagonal numbers. Also sequence found by reading the line from 0, in the direction 0, 14, ..., and the same line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Main axis, perpendicular to A024966 in the same spiral.

Partial sums of A113801. - Reinhard Zumkeller, Jan 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A144555 and A195314 interleaved.
Cf. A024966, A032527, A032528, A077221, A113801, A195045, A195046, A195142, A195143, A195146, A195147, A195148, A195149.
Column 14 of A195040. - Omar E. Pol, Sep 28 2011

a195145 n = a195145_list !! n
a195145_list = scanl (+) 0 a113801_list
-- Reinhard Zumkeller, Jan 07 2012

[(14*n^2+5*(-1)^n-5)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 14, 29}, 50] (* Amiram Eldar, Jan 16 2023 *)

G.f.: -x*(1+12*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = (14*n^2 + 5*(-1)^n - 5)/4;
a(n) = a(-n) = -a(n-1) + 7*n^2 - 7*n + 1. (End)
Sum_{n>=1} 1/a(n) = Pi^2/84 + tan(sqrt(5/7)*Pi/2)*Pi/(2*sqrt(35)). - Amiram Eldar, Jan 16 2023
E.g.f.: (7*x*(x + 1)*cosh(x) + (7*x^2 + 7*x - 5)*sinh(x))/2. - Stefano Spezia, Nov 30 2024

A195149 Concentric 22-gonal numbers.

Original entry on oeis.org

0, 1, 22, 45, 88, 133, 198, 265, 352, 441, 550, 661, 792, 925, 1078, 1233, 1408, 1585, 1782, 1981, 2200, 2421, 2662, 2905, 3168, 3433, 3718, 4005, 4312, 4621, 4950, 5281, 5632, 5985, 6358, 6733, 7128, 7525, 7942, 8361, 8800, 9241, 9702, 10165, 10648, 11133

Offset: 0

Omar E. Pol, Sep 17 2011

Sequence found by reading the line from 0, in the direction 0, 22,..., and the same line from 1, in the direction 1, 45,..., in the square spiral whose vertices are the generalized tridecagonal numbers A195313. Main axis, perpendicular to A152740 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A195323 and A195318 interleaved.
Cf. A032528, A077221, A195142, A195143, A195145, A195146, A195147, A195148.
Cf. A032527, A195049, A195058. Column 22 of A195040. - Omar E. Pol, Sep 29 2011

[(22*n^2+9*(-1)^n-9)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

A195149:=n->(22*n^2+9*(-1)^n-9)/4: seq(A195149(n), n=0..50); # Wesley Ivan Hurt, Jul 07 2014

Table[(22*n^2 + 9*(-1)^n - 9)/4, {n, 0, 50}] (* Wesley Ivan Hurt, Jul 07 2014 *)

a(n)=(22*n^2+9*(-1)^n-9)/4 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: -x*(1+20*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = (22*n^2+9*(-1)^n-9)/4; a(n) = -a(n-1)+11*n^2-11*n+1. - Vincenzo Librandi, Sep 27 2011
Sum_{n>=1} 1/a(n) = Pi^2/132 + tan(3*Pi/(2*sqrt(11)))*Pi/(6*sqrt(11)). - Amiram Eldar, Jan 17 2023
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Wesley Ivan Hurt, Jun 19 2025

A195142 Concentric 10-gonal numbers.

Original entry on oeis.org

0, 1, 10, 21, 40, 61, 90, 121, 160, 201, 250, 301, 360, 421, 490, 561, 640, 721, 810, 901, 1000, 1101, 1210, 1321, 1440, 1561, 1690, 1821, 1960, 2101, 2250, 2401, 2560, 2721, 2890, 3061, 3240, 3421, 3610, 3801, 4000, 4201, 4410, 4621, 4840, 5061, 5290

Offset: 0

Omar E. Pol, Sep 17 2011

Also concentric decagonal numbers. Also sequence found by reading the line from 0, in the direction 0, 10, ..., and the same line from 1, in the direction 1, 21, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. Main axis, perpendicular to A028895 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A033583 and A069133 interleaved.
Cf. A028895, A032527, A032528, A077221, A085787, A195042, A195143, A195145, A195146, A195147, A195148, A195149.
Cf. A090771 (first differences).
Column 10 of A195040. - Omar E. Pol, Sep 28 2011

a195142 n = a195142_list !! n
a195142_list = scanl (+) 0 a090771_list
-- Reinhard Zumkeller, Jan 07 2012

[(10*n^2+3*(-1)^n-3)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

RecurrenceTable[{a[0]==0,a[1]==1,a[n]==a[n-2]+10(n-1)},a[n],{n,50}] (* or *) LinearRecurrence[{2,0,-2,1},{0,1,10,21},50] (* Harvey P. Dale, Sep 29 2011 *)

G.f.: -x*(1+8*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = -a(n-1) + 5*n^2 - 5*n + 1, a(0)=0. - Vincenzo Librandi, Sep 27 2011
From Bruno Berselli, Sep 27 2011: (Start)
a(n) = a(-n) = (10*n^2 + 3*(-1)^n - 3)/4.
a(n) = a(n-2) + 10*(n-1). (End)
a(n) = 2*a(n-1) + 0*a(n-2) - 2*a(n-3) + a(n-4); a(0)=0, a(1)=1, a(2)=10, a(3)=21. - Harvey P. Dale, Sep 29 2011
Sum_{n>=1} 1/a(n) = Pi^2/60 + tan(sqrt(3/5)*Pi/2)*Pi/(2*sqrt(15)). - Amiram Eldar, Jan 16 2023

A195146 Concentric 16-gonal numbers.

Original entry on oeis.org

0, 1, 16, 33, 64, 97, 144, 193, 256, 321, 400, 481, 576, 673, 784, 897, 1024, 1153, 1296, 1441, 1600, 1761, 1936, 2113, 2304, 2497, 2704, 2913, 3136, 3361, 3600, 3841, 4096, 4353, 4624, 4897, 5184, 5473, 5776, 6081, 6400, 6721, 7056, 7393, 7744, 8097, 8464

Offset: 0

Omar E. Pol, Sep 17 2011

Concentric hexadecagonal numbers or concentric hexakaidecagonal numbers.

Sequence found by reading the line from 0, in the direction 0, 16, ..., and the same line from 1, in the direction 1, 33, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. Main axis, perpendicular to A033996 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A016802 and A195315 interleaved.
Cf. A032528, A033996, A074377, A077221, A195142, A195143, A195145, A195147, A195148, A195149.
Column 16 of A195040.

[(8*n^2+3*(-1)^n-3)/2: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 16, 33}, 50] (* Amiram Eldar, Jan 16 2023 *)

a(n)=(8*n^2+3*(-1)^n-3)/2 \\ Charles R Greathouse IV, Oct 07 2015

From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = (8*n^2 + 3*(-1)^n - 3)/2;
a(n) = -a(n-1) + 8*n^2 - 8*n + 1. (End)
G.f. -x*(1+14*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
Sum_{n>=1} 1/a(n) = Pi^2/96 + tan(sqrt(3)*Pi/4)*Pi/(8*sqrt(3)). - Amiram Eldar, Jan 16 2023

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

Omar E. Pol, Sep 17 2011

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.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A195321 and A195316 interleaved.
Cf. A032528, A077221, A195142, A195143, A195145, A195146, A195148, A195149.
Cf. A032527, A195047, A195048. Column 18 of A195040. - Omar E. Pol, Sep 29 2011

[(18*n^2+7*(-1)^n-7)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 18, 37}, 50] (* Amiram Eldar, Jan 17 2023 *)

a(n)=(18*n^2+7*(-1)^n-7)/4 \\ Charles R Greathouse IV, Sep 28 2015

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

A195317 Centered 40-gonal numbers.

Original entry on oeis.org

1, 41, 121, 241, 401, 601, 841, 1121, 1441, 1801, 2201, 2641, 3121, 3641, 4201, 4801, 5441, 6121, 6841, 7601, 8401, 9241, 10121, 11041, 12001, 13001, 14041, 15121, 16241, 17401, 18601, 19841, 21121, 22441, 23801, 25201, 26641, 28121, 29641, 31201, 32801, 34441, 36121

Offset: 1

Omar E. Pol, Sep 16 2011

Also centered tetracontagonal numbers or centered tetrakaicontagonal numbers. Also sequence found by reading the line from 1, in the direction 1, 41, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semi-axis opposite to A195322 in the same spiral.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195148.
Cf. A003154, A069129, A069133, A069190, A195314, A195315, A195316, A195318.
Cf. A195162, A195322.

[20*n^2 - 20*n + 1: n in [1..50]]; // Vincenzo Librandi, Sep 21 2011

Table[20*n^2 - 20*n + 1, {n, 1, 40}] (* Amiram Eldar, Feb 11 2022 *)

a(n)=20*n^2-20*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 20*n^2 - 20*n + 1.
Sum_{n>=1} 1/a(n) = Pi*tan(Pi/sqrt(5))/(8*sqrt(5)). - Amiram Eldar, Feb 11 2022
G.f.: -x*(1+38*x+x^2)/(x-1)^3. - R. J. Mathar, May 07 2024
From Elmo R. Oliveira, Nov 15 2024: (Start)
E.g.f.: exp(x)*(20*x^2 + 1) - 1.
a(n) = 2*A069133(n) - 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3.  (End)

A195322 a(n) = 20*n^2.

Original entry on oeis.org

0, 20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, 2420, 2880, 3380, 3920, 4500, 5120, 5780, 6480, 7220, 8000, 8820, 9680, 10580, 11520, 12500, 13520, 14580, 15680, 16820, 18000, 19220, 20480, 21780, 23120, 24500, 25920, 27380, 28880, 30420, 32000, 33620, 35280

Offset: 0

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 0, in the direction 0, 20, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semiaxis opposite to A195317 in the same spiral.
a(n) is the sum of all the integers less than 10*n which are not multiple of 2 or 5. a(2) = (1 + 3 + 7 + 9) + (11 + 13 + 17 + 19) = 20 + 60 = 80 = 20 * 2^2. (Link Crux Mathematicorum). - Bernard Schott, May 15 2017
Number of terms less than 10^k (k=0, 1, 2, ...): 1, 1, 3, 8, 23, 71, 224, 708, 2237, 7072, 22361, 70711, ... - Muniru A Asiru, Feb 01 2018

			From _Muniru A Asiru_, Feb 01 2018: (Start)
n=0, a(0) = 20*0^2 = 0.
n=1, a(1) = 20*1^2 = 20.
n=1, a(2) = 20*2^2 = 80.
n=1, a(3) = 20*3^2 = 180.
n=1, a(4) = 20*4^2 = 320.
...
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Léo Sauvé, Problem 53, Crux Mathematicorum, Vol. 1, Nov. 1975, page 88.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195148.
Cf. A016802, A033581, A033583, A135453, A139098, A144555, A195321, A195323.
Cf. A000290, A001105, A005899, A010014, A016742, A033429, A195162, A195317.
Cf. A008602.

List([0..10^3],n->20*n^2); # Muniru A Asiru, Feb 01 2018

[20*n^2: n in [0..40]]; // Vincenzo Librandi, Sep 20 2011

a := n -> 20*n^2; seq(a(n), n=0..10^3); # Muniru A Asiru, Feb 01 2018

20 Range[0, 40]^2 (* or *) LinearRecurrence[{3, -3, 1}, {0, 20, 80}, 50] (* Harvey P. Dale, Jan 18 2013 *)

a(n) = 20*n^2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 20*A000290(n) = 10*A001105(n) = 5*A016742(n) = 4*A033429(n) = 2*A033583(n).
a(0)=0, a(1)=20, a(2)=80; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jan 18 2013
a(n) = A010014(n) - A005899(n) for n > 0. - R. J. Cano, Sep 29 2015
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 20*x*(1 + x)/(1-x)^3.
E.g.f.: 20*x*(1 + x)*exp(x).
a(n) = n*A008602(n) = A195148(2*n). (End)

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

A124080 10 times triangular numbers: a(n) = 5*n*(n + 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A195143 a(n) = n-th concentric 12-gonal number.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Formula

A195145 Concentric 14-gonal numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Formula

A195149 Concentric 22-gonal numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A195142 Concentric 10-gonal numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

Formula

A195146 Concentric 16-gonal numbers.

Views

Author

Keywords

Comments

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

A124080 10 times triangular numbers: a(n) = 5n(n + 1).