A051624 -id:A051624 - cp's OEIS frontend

A227321 a(n) is the least r>=3 such that the difference between the nearest r-gonal number >= n and n is an r-gonal number.

3, 3, 3, 3, 4, 3, 3, 3, 4, 3, 3, 5, 3, 8, 3, 3, 4, 5, 3, 11, 3, 3, 3, 5, 4, 3, 10, 3, 3, 11, 3, 17, 4, 3, 5, 3, 3, 7, 14, 3, 4, 15, 3, 23, 3, 3, 5, 11, 4, 3, 5, 5, 3, 19, 3, 3, 3, 8, 5, 21, 3, 32, 14, 3, 4, 3, 3, 15, 3, 5, 5, 25, 3, 38, 7, 3, 6, 3, 3, 13, 4, 3

Offset: 0

Vladimir Shevelev, Jul 30 2013

nonn

The n-th r-gonal numbers is n((n-1)r-2(n-2))/2, such that 3-gonal numbers are triangular numbers, 4-gonal numbers are squares, etc.

Peter J. C. Moses, Table of n, a(n) for n = 0..1999

Cf. A000217 (r=3), A000290 (r=4), A000326 (r=5), A000384 (r=6), A000566 (r=7), A000567 (r=8), A001106-7 (r=9,10), A051682 (r=11), A051624 (r=12), A051865-A051876 (r=13-24).

rGonalQ[r_,0]:=True; rGonalQ[r_,n_]:=IntegerQ[(Sqrt[((8r-16)n+(r-4)^2)]+r-4)/(2r-4)]; nthrGonal[r_,n_]:=(n (r-2)(n-1))/2+n; nextrGonal[r_,n_]:=nthrGonal[r,Ceiling[(Sqrt[((8r-16)n+(r-4)^2)]+r-4)/(2r-4)]]; (* next r-gonal number greater than or equal to n *) Table[NestWhile[#+1&,3,!rGonalQ[#,nextrGonal[#,n]-n]&],{n,0,99}] (* Peter J. C. Moses, Aug 03 2013 *)

If n is prime, then n == 1 or 2 mod (a(n)-2). If n >= 13 is the greater of a pair of twin primes (A006512), then a(n) = (n+3)/2. - Vladimir Shevelev, Aug 07 2013

More terms from Peter J. C. Moses, Jul 30 2013

A236257 a(n) = 2n^2 - 7n + 9.

Original entry on oeis.org

9, 4, 3, 6, 13, 24, 39, 58, 81, 108, 139, 174, 213, 256, 303, 354, 409, 468, 531, 598, 669, 744, 823, 906, 993, 1084, 1179, 1278, 1381, 1488, 1599, 1714, 1833, 1956, 2083, 2214, 2349, 2488, 2631, 2778, 2929, 3084, 3243, 3406, 3573, 3744, 3919, 4098, 4281, 4468

Offset: 0

Vladimir Shevelev, Jan 21 2014

If zero polygonal numbers are ignored, then for n >= 3, the a(n)-th n-gonal number is a sum of the (a(n)-1)-th n-gonal number and the (2*n-3)-th n-gonal number.

			a(7)=58. This means that the 58th heptagonal number 8323 (cf. A000566) is a sum of two heptagonal numbers. We have 8323 = 8037 + 286 with indices in A000566 58,57,11.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
W. Burrows, C. Tuffley, Maximising common fixtures in a round robin tournament with two divisions, arXiv preprint arXiv:1502.06664 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A001106, A001107, A051624, A051682, A051865-A051876, A152948.

Table[2 n^2 - 7 n + 9, {n, 0, 48}] (* Michael De Vlieger, Apr 19 2015 *)
LinearRecurrence[{3,-3,1},{9,4,3},50] (* Harvey P. Dale, Nov 24 2017 *)

Vec(-(18*x^2-23*x+9)/(x-1)^3 + O(x^100)) \\ Colin Barker, Jan 21 2014

From Colin Barker, Jan 21 2014: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: -(18*x^2 - 23*x + 9)/(x-1)^3. (End)
E.g.f.: exp(x)*(9 - 5*x + 2*x^2). - Elmo R. Oliveira, Nov 13 2024

A287152 Number T(n,k) of tilings of a 2 X n X n box using k bricks of shape 2 X 1 X 1 and 2*(n^2-k) bricks of shape 1 X 1 X 1; triangle T(n,k), n>=0, 0<=k<=n^2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 12, 42, 44, 9, 1, 33, 436, 2984, 11434, 24766, 29180, 16984, 3993, 229, 1, 64, 1816, 30208, 328214, 2456736, 13022504, 49492032, 135062729, 262610832, 357580896, 331384336, 200032432, 73483328, 14707328, 1308928, 32000, 1, 105, 5112, 153364, 3178256

Offset: 0

Alois P. Heinz, May 20 2017

			Triangle T(n,k) begins:
1;
1,  1;
1, 12,  42,   44,     9;
1, 33, 436, 2984, 11434, 24766, 29180, 16984, 3993, 229;

Alois P. Heinz, Rows n = 0..8, flattened

Columns k=0-1 give: A000012, A051624.
Row sums give A287153.
T(n,n^2) gives A181205.

A214421 Numbers not representable as the sum of three 12-gonal numbers.

Original entry on oeis.org

4, 5, 6, 7, 8, 9, 10, 11, 15, 16, 17, 18, 19, 20, 21, 22, 23, 26, 27, 28, 29, 30, 31, 32, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 68, 69, 70, 71, 72, 73, 74, 75, 79, 80, 81, 82, 83, 84, 85, 86, 87, 89

Offset: 1

T. D. Noe, Jul 17 2012

nonn

There are an infinite number of numbers that are not the sum of three 12-gonal numbers.

R. K. Guy, Unsolved Problems in Number Theory, D3.

T. D. Noe, Table of n, a(n) for n = 1..10000
R. K. Guy, Every number is expressible as the sum of how many polygonal numbers?, Amer. Math. Monthly 101 (1994), 169-172.

Cf. A051624 (12-gonal numbers).

Cf. A118278, A118279.

nn = 100; dod = Table[n*(5n-4), {n, 0, nn}]; t = Table[0, {dod[[-1]]}]; Do[n = dod[[i]] + dod[[j]] + dod[[k]]; If[n <= dod[[-1]], t[[n]] = 1], {i, nn}, {j, i, nn}, {k, j, nn}]; Flatten[Position[t, 0]]

A366827 -a(n)/7! is the coefficient of x^7 in the Taylor expansion of the k-th iteration of sin(x).

Original entry on oeis.org

0, 1, 128, 731, 2160, 4765, 8896, 14903, 23136, 33945, 47680, 64691, 85328, 109941, 138880, 172495, 211136, 255153, 304896, 360715, 422960, 491981, 568128, 651751, 743200, 842825, 950976, 1068003, 1194256, 1330085, 1475840, 1631871, 1798528, 1976161, 2165120, 2365755, 2578416

Offset: 0

Jianing Song, Oct 25 2023

a(n)/7! is the coefficient of x^7 in the Taylor expansion of the k-th iteration of sinh(x). This is most easily seen from the relation -i*sin(...sin(sin(sin(i*x)))...) = -i*sin(...sin(sin(i*sinh(x)))...) = -i*sin(...sin(i*sinh(sinh(x)))...) = ... = sinh(...sinh(sinh(sinh(x)))...).

			sin(sin(x)) = x - 2*x^3/3! + 12*x^5/5! - 128*x^7/7! + ...;
sin(sin(sin(x))) = x - 3*x^3/3! + 33*x^5/5! - 731*x^7/7! + ...;
sin(sin(sin(sin(x)))) = x - 4*x^3/3! + 64*x^5/5! - 2160*x^7/7! + ....

Jianing Song, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A366834 (main sequence), A051624 (coefficient of x^5), A285018, A285019.

a(n) = (175/3)*n^3 - 112*n^2 + (164/3)*n

a(n) = binomial(n,1) + 126*binomial(n,2) + 350*binomial(n,3) = (175*n^2 - 336*n + 164)*n/3. See A366834.

G.f.: x/(1-x)^2 + 126*x^2/(1-x)^3 + 350*x^3/(1-x)^4.

A366834 Square array read by descending antidiagonals: (-1)^nT(n,k)/n! is the coefficient of x^(2n+1) in the Taylor expansion of the k-th iteration of sin(x).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 12, 1, 0, 1, 4, 33, 128, 1, 0, 1, 5, 64, 731, 1872, 1, 0, 1, 6, 105, 2160, 25857, 37600, 1, 0, 1, 7, 156, 4765, 121600, 1311379, 990784, 1, 0, 1, 8, 217, 8896, 368145, 10138880, 89060065, 32333824, 1, 0, 1, 9, 288, 14903, 873936, 42807605, 1162426880, 7778778091, 1272660224, 1, 0

Offset: 0

Jianing Song, Oct 25 2023

T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sinh(x). This is most easily seen from the relation -i*sin(...sin(sin(sin(i*x)))...) = -i*sin(...sin(sin(i*sinh(x)))...) = -i*sin(...sin(i*sinh(sinh(x)))...) = ... = sinh(...sinh(sinh(sinh(x)))...).

			G.f.s of the first few rows:
n = 0: 1/(1-x);
n = 1: x/(1-x)^2;
n = 2: x/(1-x)^2 + 10*x^2/(1-x)^3;
n = 3: x/(1-x)^2 + 126*x^2/(1-x)^3 + 350*x^3/(1-x)^4:
n = 4: x/(1-x)^2 + 1870*x^2/(1-x)^3 + 20244*x^3/(1-x)^4 + 29400*x^4/(1-x)^5;
n = 5: x/(1-x)^2 + 37598*x^2/(1-x)^3 + 1198582*x^3/(1-x)^4 + 5118960*x^4/(1-x)^5 + 4851000*x^5/(1-x)^6.
The explicit formulas for the first few rows:
T(0,k) = binomial(k,0) = 1 for k = 0, 0 for k > 0;
T(1,k) = binomial(k,1) = k;
T(2,k) = binomial(k,1) + 10*binomial(k,2) = 5*k^2 - 4*k;
T(3,k) = binomial(k,1) + 126*binomial(k,2) + 350*binomial(k,3) = (175/3)*k^3 - 112*k^2 + (164/3)*k;
T(4,k) = binomial(k,1) + 1870*binomial(k,2) + 20244*binomial(k,3) + 29400*binomial(k,4) = 1225*k^4 - 3976*k^3 + 4288*k^2 - 1536*k;
T(5,k) = binomial(k,1) + 37598*binomial(k,2) + 1198582*binomial(k,3) + 5118960*binomial(k,4) + 4851000*binomial(k,5) = 40425*k^5 - 190960*k^4 + (1004696/3)*k^3 - 255552*k^2 + (213568/3)*k.
Table of terms:
Row 0: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Row 1: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
Row 2: 0, 1, 12, 33, 64, 105, 156, 217, 288, 369, 460
Row 3: 0, 1, 128, 731, 2160, 4765, 8896, 14903, 23136, 33945, 47680
Row 4: 0, 1, 1872, 25857, 121600, 368145, 873936, 1776817, 3244032, 5472225, 8687440
Row 5: 0, 1, 37600, 1311379, 10138880, 42807605, 130426016, 323774535, 698156544, 1358249385, 2442955360
Row 6: 0, 1, 990784, 89060065, 1162426880, 6937805945, 27344158016, 83303826249, 212901058560, 478937915985, 977877567040
Row 7: 0, 1, 32333824, 7778778091, 174394695680, 1487589904205, 7634965431296, 28668866786679, 87104014381056, 227079171721785, 527214112015360
Row 8: 0, 1, 1272660224, 849264442881, 33044097597440, 406373544070945, 2731282112246016, 12688038285458529, 45949019179646976, 139088689115885505, 367745105831952640
Row 9: 0, 1, 59527313920, 113234181108643, 7701145601933312, 137461463840219237, 1215573962763120128, 7008667055272520967, 30322784763588771840, 106757902382656031049, 321859857651846029824
Row 10: 0, 1, 3252626013184, 18073465545032353, 2162675327569362944, 56311245536706922889, 657730167421332884480, 4719958813316934631353, 24445625744089126797312, 100254353682662263787313, 345053755346367061654528
Demonstration of terms:
sin(sin(x)) = x - 2*x^3/3! + 12*x^5/5! - 128*x^7/7! + 1872*x^9/9! - 37600*x^11/11! + ...;
sin(sin(sin(x))) = x - 3*x^3/3! + 33*x^5/5! - 731*x^7/7! + 25857*x^9/9! - 1311379*x^11/11! + ...;
sin(sin(sin(sin(x)))) = x - 4*x^3/3! + 64*x^5/5! - 2160*x^7/7! + 121600*x^9/9! - 10138880*x^11/11! + ....

Jianing Song, Table of n, a(n) for n = 0..5150 (diagonals 0..100)
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Jianing Song, Formula for A366834.

Cf. A051624 (row n=2), A366827 (row n=3), A003712 (column k=2 signed), A003715 (column k=3 signed).

Cf. A000447, A285018, A285019.

A160562(n,k) = (-1)^k / (2*k+1)! * sum(j=0, k, (-1)^j * binomial(2*k+1, k-j) * (2*j+1)^(2*n+1)) / 2^(2*k)
coeff_of_n_gfs(n) = my(M=matid(1)); for(k=1, n, M = matconcat([concat(M, matrix(k, 1)); concat(0, matrix(1, k, i, j, A160562(k, j-1))*M)])); M \\ The lower triangle matrix (C(i,j))_{0<=j<=i<=n}
T_mat(n,k) = coeff_of_n_gfs(n)*matrix(n+1, k+1, i, j, binomial(j-1,i-1)) \\ gives T(i,j) for i=0..n and j=0..k

T(0,0) = 1, T(n,0) = 0 for n >= 1; T(n,k) = Sum_{i=0..n} A160562(n,i)*T(i,k-1) for k >= 1, where A160562(n,k) = ((-1)^(n-k)*(2*n+1)!/(2*k+1)!) * [x^(2*n+1)]sin(x)^(2*k+1). Note that this is not very efficient to calculate the terms.
A more efficient way would be to calculate the g.f. for each row: the g.f. of the n-th row is C(n,0)/(1-x) + C(n,1)*x/(1-x)^2 + ... + C(n,n)*x^n/(1-x)^(n+1), where C(0,0) = 1, C(n,0) = 0 for n >= 1; C(n,k) = A160562(n,k-1)*C(k-1,k-1) + ... + A160562(n,n-1)*C(n-1,k-1) for n >= k >= 1, so we have T(n,k) = C(n,0)*binomial(k,0) + C(n,1)*binomial(k,1) + ... + C(n,n)*binomial(k,n). See my pdf in the link section for the proof.
From the formula above we see that the n-th row is a degree-n polynomial of k with leading coefficient C(n,n)/n!. We have C(n,n) = A160562(n,n-1)*C(n-1,n-1) = A000447(n)*C(n-1,n-1) for n >= 1, so it can be shown that C(n,n)/n! = n! * A285018(n)/A285019(n).

A380353 a(n) = (n^2 - n + 2) * (5n^2 - 5n + 2) / 4.

Original entry on oeis.org

1, 12, 64, 217, 561, 1216, 2332, 4089, 6697, 10396, 15456, 22177, 30889, 41952, 55756, 72721, 93297, 117964, 147232, 181641, 221761, 268192, 321564, 382537, 451801, 530076, 618112, 716689, 826617, 948736, 1083916, 1233057, 1397089, 1576972, 1773696, 1988281, 2221777

Offset: 1

Kelvin Voskuijl, Jan 22 2025

First differences of A072474 (sum of next n squares).

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A072474 (partial sums), A051624, A000124.

Cf. A005448 (first difference of sum of next n natural numbers).

Table[((n^2 - n + 2)*(5*n^2 - 5*n + 2))/4, {n, 1, 40}]

a(n) = (n^2 - n + 2) * (5*n^2 - 5*n + 2) / 4

a(n) = A051624(A000124(n-1)).
G.f.: x*(1+3*x+x^2)*(1+4*x+x^2)/(1-x)^5. - Jinyuan Wang, Jan 23 2025
E.g.f.: exp(x)*(4 + 22*x^2 + 20*x^3 + 5*x^4)/4 - 1. - Stefano Spezia, Jan 28 2025

A183224 Complement of the 12-gonal numbers.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101

Offset: 1

Clark Kimberling, Jan 01 2011

nonn

Cf. A051624 (12-gonal numbers.)

Table[n+Floor[1/2+(n/5+1/5)^(1/2)], {n,100}]

from math import isqrt
def A183224(n): return n+(isqrt((n<<2|2)//5)+1>>1) # Chai Wah Wu, Oct 06 2024

a(n) = n + floor(1/2+(n/5+1/5)^(1/2)).

A236333 The (n-2)-th (n>=3) triple of terms gives coefficients of double trinomial P_n(x) = ((n-2)^2x^2 + nx + 2)/2 (see comment).

Original entry on oeis.org

1, 3, 2, 4, 4, 2, 9, 5, 2, 16, 6, 2, 25, 7, 2, 36, 8, 2, 49, 9, 2, 64, 10, 2, 81, 11, 2, 100, 12, 2, 121, 13, 2, 144, 14, 2, 169, 15, 2, 196, 16, 2, 225, 17, 2, 256, 18, 2, 289, 19, 2, 324, 20, 2, 361, 21, 2, 400, 22, 2, 441, 23, 2, 484, 24, 2, 529, 25, 2, 576, 26, 2, 625, 27, 2, 676, 28, 2, 729, 29, 2

Offset: 3

Vladimir Shevelev, Jan 22 2014

Let {G_n(k)}_(k>=0) be sequence of n-gonal numbers. Then G_n(P_n(k)) = G_n(P_n(k)-1) + G_n((n-2)*k+1).

			Let n=5, k=4. Then G_5(k)=k*(3*k-1)/2 (Cf. A000326) and the double trinomial 2*P_5(x)= 9*x^2+5*x+2, P_5(4)=(9*4^2+5*4+2)/2=83,
Thus, we have G_5(83)=G_5(82)+G_5(13), or 83*124 = 41*245 + 13*19 = 10292.

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

Cf. A000217, A000290, A000326, A000384, A000566, A000567, A001106, A001107, A051682, A051624, A051865-A051876.

I:=[1,3,2,4,4,2,9,5,2]; [n le 9 select I[n] else 3*Self(n-3)-3*Self(n-6)+Self(n-9): n in [1..90]]; // Vincenzo Librandi, Feb 02 2014

a[n_]:=Which[Mod[n,3]==0,n^2/9,Mod[n,3]==1,(n+5)/3,True,2]; Map[a,Range[3,103]]
CoefficientList[Series[(-1-3 x-2 x^2-x^3+5 x^4+4 x^5-2 x^7-2 x^8)/((-1+x)^3 (1+x+x^2)^3),{x,0,100}],x]

Vec(-x^3*(2*x^8+2*x^7-4*x^5-5*x^4+x^3+2*x^2+3*x+1)/((x-1)^3*(x^2+x+1)^3) + O(x^100)) \\ Colin Barker, Jan 23 2014

If n==0 (mod 3), then a(n) = n^2/9;
if n==1 (mod 3), then a(n) = (n+5)/3;
if n==2 (mod 3), then a(n) = 2.
G.f.: -x^3*(2*x^8+2*x^7-4*x^5-5*x^4+x^3+2*x^2+3*x+1) / ((x-1)^3*(x^2+x+1)^3). - Colin Barker, Jan 23 2014

A266088 Alternating sum of 12-gonal (or dodecagonal) numbers.

Original entry on oeis.org

0, -1, 11, -22, 42, -63, 93, -124, 164, -205, 255, -306, 366, -427, 497, -568, 648, -729, 819, -910, 1010, -1111, 1221, -1332, 1452, -1573, 1703, -1834, 1974, -2115, 2265, -2416, 2576, -2737, 2907, -3078, 3258, -3439, 3629, -3820, 4020, -4221, 4431, -4642

Offset: 0

Ilya Gutkovskiy, Dec 21 2015

More generally, the ordinary generating function for the alternating sum of k-gonal numbers is -x*(1 - (k - 3)*x)/((1 - x)*(1 + x)^3).

G. C. Greubel, Table of n, a(n) for n = 0..5000
OEIS Wiki, Figurate numbers
Index entries for linear recurrences with constant coefficients, signature (-2,0,2,1).

Cf. A006578, A007587, A035608, A051624, A083392, A089594.

[1+(-1)^n*(5*n^2+n-2)/2: n in [0..50]]; // Vincenzo Librandi, Dec 21 2015

Table[1 + (-1)^n (5 n^2 + n - 2)/2, {n, 0, 43}]
CoefficientList[Series[-x (1 - 9 x)/((1 - x) (1 + x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 21 2015 *)

x='x+O('x^100); concat(0, Vec(-x*(1-9*x)/((1-x)*(1+x)^3))) \\ Altug Alkan, Dec 21 2015

G.f.: -x*(1 - 9*x)/((1 - x)*(1 + x)^3).
a(n) = 1 + (-1)^n*(5*n^2 + n - 2)/2.
a(n) = Sum_{k = 0..n} (-1)^k*A051624(k).
Lim_{n -> infinity} a(n + 1)/a(n) = -1.

cp's OEIS Frontend

A227321 a(n) is the least r>=3 such that the difference between the nearest r-gonal number >= n and n is an r-gonal number.

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A236257 a(n) = 2*n^2 - 7*n + 9.

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A287152 Number T(n,k) of tilings of a 2 X n X n box using k bricks of shape 2 X 1 X 1 and 2*(n^2-k) bricks of shape 1 X 1 X 1; triangle T(n,k), n>=0, 0<=k<=n^2, read by rows.

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A214421 Numbers not representable as the sum of three 12-gonal numbers.

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A366827 -a(n)/7! is the coefficient of x^7 in the Taylor expansion of the k-th iteration of sin(x).

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A366834 Square array read by descending antidiagonals: (-1)^n*T(n,k)/n! is the coefficient of x^(2*n+1) in the Taylor expansion of the k-th iteration of sin(x).

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A380353 a(n) = (n^2 - n + 2) * (5*n^2 - 5*n + 2) / 4.

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A183224 Complement of the 12-gonal numbers.

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Programs

A236257 a(n) = 2n^2 - 7n + 9.

A366834 Square array read by descending antidiagonals: (-1)^nT(n,k)/n! is the coefficient of x^(2n+1) in the Taylor expansion of the k-th iteration of sin(x).

A380353 a(n) = (n^2 - n + 2) * (5n^2 - 5n + 2) / 4.

A236333 The (n-2)-th (n>=3) triple of terms gives coefficients of double trinomial P_n(x) = ((n-2)^2x^2 + nx + 2)/2 (see comment).