A195058 -id:A195058 - 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 *)

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

A195049 Concentric 21-gonal numbers.

Original entry on oeis.org

0, 1, 21, 43, 84, 127, 189, 253, 336, 421, 525, 631, 756, 883, 1029, 1177, 1344, 1513, 1701, 1891, 2100, 2311, 2541, 2773, 3024, 3277, 3549, 3823, 4116, 4411, 4725, 5041, 5376, 5713, 6069, 6427, 6804, 7183, 7581, 7981, 8400, 8821, 9261, 9703, 10164

Offset: 0

Omar E. Pol, Sep 27 2011

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Column 21 of A195040.

Cf. A032527, A032528, A195048, A195148, A195149, A195058.

A195049:=n->21*n^2/4+17*((-1)^n-1)/8: seq(A195049(n), n=0..100); # Wesley Ivan Hurt, Jan 17 2017

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 21, 43}, 50] (* Amiram Eldar, Jan 17 2023 *)

a(n)=21*n^2/4+17*((-1)^n-1)/8 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 21*n^2/4 + 17*((-1)^n-1)/8.
From Colin Barker, Sep 16 2012: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: x*(1+19*x+x^2)/((1-x)^3*(1+x)). (End)
Sum_{n>=1} 1/a(n) = Pi^2/126 + tan(sqrt(17/21)*Pi/2)*Pi/sqrt(357). - Amiram Eldar, Jan 17 2023

A195158 Concentric 24-gonal numbers.

Original entry on oeis.org

0, 1, 24, 49, 96, 145, 216, 289, 384, 481, 600, 721, 864, 1009, 1176, 1345, 1536, 1729, 1944, 2161, 2400, 2641, 2904, 3169, 3456, 3745, 4056, 4369, 4704, 5041, 5400, 5761, 6144, 6529, 6936, 7345, 7776, 8209, 8664, 9121, 9600, 10081, 10584, 11089

Offset: 0

Omar E. Pol, Sep 28 2011

Sequence found by reading the line from 0, in the direction 0, 24, ..., and the same line from 1, in the direction 1, 49, ..., in the square spiral whose vertices are the generalized tetradecagonal numbers A195818. Main axis, perpendicular to A049598 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).

Column 24 of A195040.

Cf. A032527, A032528, A195143, A195149, A195058.

[(12*n^2+5*(-1)^n-5)/2: n in [0..50]]; // Vincenzo Librandi, Sep 30 2011

LinearRecurrence[{2,0,-2,1},{0,1,24,49},50] (* Harvey P. Dale, Jan 28 2021 *)

a(n)=6*n^2+5*((-1)^n-1)/2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 6*n^2 + 5*((-1)^n-1)/2.
a(n) = -a(n-1) + A069190(n). - Vincenzo Librandi, Sep 30 2011
From Colin Barker, Sep 16 2012: (Start)
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: x*(1+22*x+x^2)/((1-x)^3*(1+x)). (End)
Sum_{n>=1} 1/a(n) = Pi^2/144 + tan(sqrt(5/6)*Pi/2)*Pi/(4*sqrt(30)). - Amiram Eldar, Jan 17 2023

A244632 a(n) = 23*n^2.

Original entry on oeis.org

0, 23, 92, 207, 368, 575, 828, 1127, 1472, 1863, 2300, 2783, 3312, 3887, 4508, 5175, 5888, 6647, 7452, 8303, 9200, 10143, 11132, 12167, 13248, 14375, 15548, 16767, 18032, 19343, 20700, 22103, 23552, 25047, 26588, 28175, 29808, 31487, 33212, 34983, 36800, 38663

Offset: 0

Vincenzo Librandi, Jul 03 2014

First bisection of A195058. - Bruno Berselli, Jul 03 2014

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

Cf. A000290, A008605, A195058.

Cf. similar sequences listed in A244630.

Magma
```
[23*n^2: n in [0..40]];
```

Table[23 n^2, {n, 0, 40}]
LinearRecurrence[{3,-3,1},{0,23,92},50] (* Harvey P. Dale, Jul 14 2024 *)

a(n)=23*n^2 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: 23*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) = 23*A000290(n). - Omar E. Pol, Jul 03 2014
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 23*x*(1 + x)*exp(x).
a(n) = n*A008605(n) = A195058(2*n). (End)

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

Ilya Gutkovskiy, Mar 21 2016

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).

OEIS Wiki, Centered polygonal numbers
Eric Weisstein's World of Mathematics, Centered Polygonal Number
Index entries for linear recurrences with constant coefficients, signature (-2,0,2,1).

Cf. A262221 (centered 25-gonal numbers).

Cf. A032527, A032528, A077043, A077221, A195041, A195042, A195045, A195046, A195047, A195048, A195049, A195058, A195142, A195043, A195143, A195145, A195146, A195147, A195148, A195149, A195158.

[((-1)^n*(50*n^2 + 100*n + 29) - 21)/8 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016

A270693:=n->((-1)^n*(50*n^2 + 100*n + 29) - 21)/8: seq(A270693(n), n=0..100); # Wesley Ivan Hurt, Sep 18 2017

LinearRecurrence[{-2, 0, 2, 1}, {1, -25, 51, -100}, 41]
Table[((-1)^n (50 n^2 + 100 n + 29) - 21)/8, {n, 0, 40}]

x='x+O('x^100); Vec((1-23*x+x^2)/((1-x)*(1+x)^3)) \\ Altug Alkan, Mar 21 2016

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.

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.

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A195149 Concentric 22-gonal numbers.

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A195049 Concentric 21-gonal numbers.

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A195158 Concentric 24-gonal numbers.

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A244632 a(n) = 23*n^2.

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Magma

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A270693 Alternating sum of centered 25-gonal numbers.

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Magma

Maple

Mathematica

PARI

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