A011865 -id:A011865 - cp's OEIS frontend

A281026 a(n) = floor(3n(n+1)/4).

0, 1, 4, 9, 15, 22, 31, 42, 54, 67, 82, 99, 117, 136, 157, 180, 204, 229, 256, 285, 315, 346, 379, 414, 450, 487, 526, 567, 609, 652, 697, 744, 792, 841, 892, 945, 999, 1054, 1111, 1170, 1230, 1291, 1354, 1419, 1485, 1552, 1621, 1692, 1764, 1837, 1912, 1989, 2067, 2146

Offset: 0

Bruno Berselli, Jan 13 2017

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of the initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Subsequence of A214068.
Partial sums of A047273.
Cf. A011865, A045943, A274757 (subsequence).
Cf. sequences with formula floor(k*n*(n+1)/4): A011848 (k=1), A000217 (k=2), this sequence (k=3), A002378 (k=4).
Cf. sequences with formula floor(k*n*(n+1)/(k+1)): A000217 (k=1), A143978 (k=2), this sequence (k=3), A281151 (k=4), A194275 (k=5).

Magma
```
[3*n*(n+1) div 4: n in [0..60]];
```

A281026:=n->floor(3*n*(n+1)/4): seq(A281026(n), n=0..100); # Wesley Ivan Hurt, Jan 13 2017

Table[Floor[3 n (n + 1)/4], {n, 0, 60}]
LinearRecurrence[{3,-4,4,-3,1},{0,1,4,9,15},60] (* Harvey P. Dale, Jun 04 2023 *)

makelist(floor(3*n*(n+1)/4), n, 0, 60);

PARI
```
vector(60, n, n--; floor(3*n*(n+1)/4))
```
Python
```
[int(3*n*(n+1)/4) for n in range(60)]
```

[floor(3*n*(n+1)/4) for n in range(60)]

O.g.f.: x*(1 + x + x^2)/((1 + x^2)*(1 - x)^3).
E.g.f.: -(1 - 6*x - 3*x^2)*exp(x)/4 - (1 + i)*(i - exp(2*i*x))*exp(-i*x)/8, where i=sqrt(-1).
a(n) = a(-n-1) = 3*a(n-1) - 4*a(n-2) + 4*a(n-3) - 3*a(n-4) + a(n-5) = a(n-4) + 6*n - 9.
a(n) = 3*n*(n+1)/4 + (i^(n*(n+1)) - 1)/4. Therefore:
a(4*k+r) = 12*k^2 + 3*(2*r+1)*k + r^2, where 0 <= r <= 3.
a(n) = n^2 - floor((n-1)*(n-2)/4).
a(n) = A011865(3*n+2).

A084626 a(n) = floor(C(n+6,6)/C(n+2,2)).

Original entry on oeis.org

1, 2, 4, 8, 14, 22, 33, 47, 66, 91, 121, 158, 204, 258, 323, 399, 487, 590, 708, 843, 996, 1170, 1365, 1583, 1827, 2097, 2397, 2728, 3091, 3490, 3927, 4403, 4921, 5483, 6092, 6751, 7462, 8227, 9050, 9933, 10879, 11891, 12972, 14125, 15353, 16660, 18048

Offset: 0

Paul Barry, Jun 01 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A011865, A084624, A084627, A084628, A084630, A084631.

[Floor(n*(n+9)*(n^2+9*n+38)/360)+1: n in [0..50]]; // Vincenzo Librandi, Aug 02 2013

Table[Floor[n*(n+9)*(n^2+9*n+38)/360] +1, {n,0,50}] (* Vincenzo Librandi, Aug 02 2013 *)
Floor[Binomial[Range[6,76],4]/15] (* G. C. Greubel, Mar 24 2023 *)

[binomial(n+6,4)//15 for n in range(71)] # G. C. Greubel, Mar 24 2023

a(n) = 1 + floor( n*(n+9)*(n^2 +9*n +38)/360 ).

a(n) = floor(binomial(n+6,4)/15). - G. C. Greubel, Mar 24 2023

A268539 Numbers k such that 48*k + 25 is a perfect square.

Original entry on oeis.org

0, 2, 3, 7, 17, 25, 28, 38, 58, 72, 77, 93, 123, 143, 150, 172, 212, 238, 247, 275, 325, 357, 368, 402, 462, 500, 513, 553, 623, 667, 682, 728, 808, 858, 875, 927, 1017, 1073, 1092, 1150, 1250, 1312, 1333, 1397, 1507, 1575, 1598, 1668, 1788, 1862, 1887, 1963, 2093, 2173

Offset: 1

N. J. A. Sloane, Feb 24 2016

Equivalently, integers of the form (h+5)*(h-5)/48, where h must be odd, h = 2*m+1, thus also integers of the form (m+3)*(m-2)/12, with m = 2, 5, 6, 9, 14, 17, 18, ... = {2, 5, 6, 9} + 12 N. - M. F. Hasler, Mar 02 2016
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(8*n))*(1 + q^(8*n-1))*(1 + q^(8*n-7))/(1 + q^n) = Sum_{n >= 0} q^(2*n*(n+1)) * Product_{k >= 2*n+2} 1 - q^k = 1 - q^2 - q^3 + q^7 + q^17 - q^25 - q^28 + + - - ... (by the quintuple product identity and Mc Laughlin et al., S.38, p 16). - Peter Bala, Dec 30 2024
Conjecture: the sequence terms are also the exponents in the expansion of Sum_{n >= 0} q^n/(Product_{k = 1..2*n+1} 1 + q^k) = 1 + q^2 - q^3 - q^7 + q^17 + q^25 - - + + .... - Peter Bala, Jan 15 2025

Zak Seidov, Table of n, a(n) for n = 1..10000
J. Mc Laughlin, A. V. Sills and P. Zimmer, Rogers-Ramanujan-Slater Type Identities, Electronic J. Combinatorics, DS15, 1-59, May 31, 2008.
Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157, December 2015, Pages 223-232, see Corollary 4.4.
Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A078405, A154293.

Subsequence of A011865.

[n: n in [0..2200] | IsSquare(48*n+25)]; // Vincenzo Librandi, Feb 25 2016

L := [5, 11, 13, 19, 29, 35, 37, 43]:
seq(seq(((L[i]+48*j)^2-25)/48, i=1..8), j=0..10); # Robert Israel, Feb 29 2016

Select[Range[0, 2500], IntegerQ[Sqrt[48 # + 25]] &] (* Vincenzo Librandi, Feb 25 2016 *)
Table[(3 (n - 1) n + (2 n - 1) (-1)^((n - 2) (n - 1)/2) - 1)/4, {n, 1, 60}] (* Bruno Berselli, Feb 29 2016 *)
LinearRecurrence[{3, -5, 7, -7, 5, -3, 1}, {0, 2, 3, 7, 17, 25, 28}, 48] (* Robert G. Wilson v, Mar 05 2016 *)
CoefficientList[ Series[ x*(2 - 3x + 8x^2 - 3x^3 + 2x^4)/((1 - x)^3*(1 + x^2)^2), {x, 0, 47}], x] (* Robert G. Wilson v, Mar 05 2016 *)

isok(n) = issquare(48*n+25); \\ Michel Marcus, Feb 25 2016

A268539(n)={my(m=n\4*12+[-3,2,5,6][n%4+1]);(3+m)*(m-2)/12} \\ M. F. Hasler, Mar 03 2016

from gmpy2 import is_square
[k for k in range(2200) if is_square(48*k+25)] # Bruno Berselli, Dec 05 2016

[n for n in (0..2200) if is_square(48*n+25)] # Bruno Berselli, Feb 29 2016

For n>25, a(n) = 3*( a(n-8)-a(n-16) ) + a(n-24). - Zak Seidov, Feb 28 2016
From Robert Israel, Feb 29 2016: (Start)
Let L = [5, 11, 13, 19, 29, 35, 37, 43].
Then a(i + 8*j) = ( (L(i) + 48*j)^2 - 25 )/48 for i = 1..8, j >= 0. (End)
From Bruno Berselli, Feb 29 2016: (Start)
G.f.: x^2*(2 - 3*x + 8*x^2 - 3*x^3 + 2*x^4)/((1 - x)^3*(1 + x^2)^2).
a(n) = a(-n+1) = 3*a(n-1) - 5*a(n-2) + 7*a(n-3) - 7*a(n-4) + 5*a(n-5) - 3*a(n-6) + a(n-7) for n>6.
a(n) = (3*(n-1)*n + (2*n-1)*(-1)^((n-2)*(n-1)/2) - 1)/4. Therefore:
a(4*k)   = k*(12*k -5),
a(4*k+1) = k*(12*k +5),
a(4*k+2) = k*(12*k+11)+2 = (3*k+2)*(4*k+1),
a(4*k+3) = k*(12*k+13)+3 = (3*k+1)*(4*k+3).
From the previous formulas follows that 2, 3, 7 and 17 are the only primes of the sequence. (End)
Sum_{n>=2} 1/a(n) = 12/25 + (4/sqrt(3)-1)*Pi/5. - Amiram Eldar, Jul 30 2024

More terms from Michel Marcus, Feb 25 2016

A084624 a(n) = floor(C(n+5,5)/C(n+2,2)).

Original entry on oeis.org

1, 2, 3, 5, 8, 12, 16, 22, 28, 36, 45, 56, 68, 81, 96, 114, 133, 154, 177, 202, 230, 260, 292, 327, 365, 406, 449, 496, 545, 598, 654, 714, 777, 843, 913, 988, 1066, 1148, 1234, 1324, 1419, 1518, 1621, 1729, 1842, 1960, 2082, 2210, 2342, 2480, 2623, 2772, 2926

Offset: 0

Paul Barry, Jun 01 2003

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-3,3,-1).

Cf. A011865, A084625, A084626, A084627, A084628, A084630, A084631.

[Floor(Binomial(n+5,3)/10): n in [0..60]]; // G. C. Greubel, Mar 24 2023

LinearRecurrence[{3,-3,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-3,3,-1},{1,2,3, 5,8,12,16,22,28,36,45,56,68,81,96,114,133,154,177,202,230, 260,292},53] (* Ray Chandler, Jul 17 2015 *)
Table[Floor[Binomial[n+5,5]/Binomial[n+2,2]],{n,0,60}] (* or *) Table[ Floor[((3+n)(4+n)(5+n))/60],{n,0,60}] (* Harvey P. Dale, Sep 04 2017 *)
Floor[Binomial[Range[5,65],3]/10] (* G. C. Greubel, Mar 24 2023 *)

[(binomial(n+5,3)//10) for n in range(61)] # G. C. Greubel, Mar 24 2023

a(n) = 1 + floor( n*(n^2 + 12*n + 47)/60 ).
From G. C. Greubel, Mar 24 2023: (Start)
a(n) = floor( binomial(n+5,3)/10 ).
G.f.: (1 -x +x^3 -x^6 +2*x^7 -2*x^8 +2*x^9 -x^10 +x^11 -x^12 +x^14 +x^15 -2*x^16 +x^17)/((1-x)^3*(1-x^20)). (End)

A255604 Table read by antidiagonals, T(n,k) is the integer part of the area of a regular k-gon with side length n.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 6, 9, 6, 2, 10, 16, 15, 10, 3, 15, 25, 27, 23, 14, 4, 21, 36, 43, 41, 32, 19, 6, 27, 49, 61, 64, 58, 43, 24, 7, 35, 64, 84, 93, 90, 77, 55, 30, 9, 43, 81, 110, 127, 130, 120, 98, 69, 37, 11, 52, 100, 139, 166, 178, 173, 154, 123, 84, 44, 13, 62, 121, 172

Offset: 1

Kival Ngaokrajang, Feb 27 2015

See illustration in the links.

			See table in the links.

Kival Ngaokrajang, Illustration of T(n,k), n = 1..4, k = 3..6, Example of table

Row 1: A011865.

Columns 1,2,3,4,10: A171971, A000290, A255605, A255606, A172526.

t[n_, k_] := Floor[k*n^2/(4 Tan[Pi/k])]; Table[t[n - k + 1, k], {n, 3, 14}, {k, 3 , n}] // Flatten
(* to view table: Table[t[n, k], {k, 3, 6}, {n, 6}] // TableForm *) (* Robert G. Wilson v, Feb 28 2015 *)

{for(i=1,20,for(n=3,i-1,a=floor(n*(i-n)^2/(4*tan(Pi/n)));print1(a,", ")))}

T(n,k) = floor(k*n^2/(4*tan(Pi/k))), n >=1, k >=3.

A231559 a(n) = floor( A000326(n)/2 ).

Original entry on oeis.org

0, 0, 2, 6, 11, 17, 25, 35, 46, 58, 72, 88, 105, 123, 143, 165, 188, 212, 238, 266, 295, 325, 357, 391, 426, 462, 500, 540, 581, 623, 667, 713, 760, 808, 858, 910, 963, 1017, 1073, 1131, 1190, 1250, 1312, 1376, 1441, 1507, 1575, 1645, 1716, 1788, 1862, 1938

Offset: 0

Bruno Berselli, Nov 11 2013

First trisection of A011865.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-4,4,-3,1).

Cf. pentagonal numbers: A000326.
Cf. A011848 for the triangular numbers: floor(A000217/2); A007590 for the squares: floor(A000290/2); A156859 for the hexagonal numbers: floor(A000384/2).
First differences: A047262.

Magma
```
[Floor(n*(3*n-1)/4): n in [0..60]];
```

Table[Floor[n (3 n - 1)/4], {n, 0, 60}]
CoefficientList[Series[x^2(2+x^2)/((1+x^2)(1-x)^3),{x,0,70}],x] (* or *) LinearRecurrence[{3,-4,4,-3,1},{0,0,2,6,11},70] (* Harvey P. Dale, Jan 28 2022 *)

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

a(n) = ( n*(3*n-1) + i^(n*(n+1)) - 1 )/4, where i=sqrt(-1).

A084267 Partial sums of a binomial quotient.

Original entry on oeis.org

1, 2, 4, 7, 11, 17, 24, 33, 44, 57, 72, 89, 109, 131, 156, 184, 215, 250, 288, 330, 376, 426, 480, 538, 601, 668, 740, 817, 899, 987, 1080, 1179, 1284, 1395, 1512, 1635, 1765, 1901, 2044, 2194, 2351, 2516, 2688, 2868, 3056, 3252, 3456, 3668, 3889, 4118, 4356

Offset: 0

Paul Barry, Jun 01 2003

Partial sums of A011865 are a(n)=sum{k=0..n, floor(C(k+2,4)/C(k+2,2))}.

Index entries for linear recurrences with constant coefficients, signature (3,-3,2,-3,3,-2,3,-3,2,-3,3,-1).

Accumulate[Table[Floor[Binomial[n,4]/Binomial[n,2]],{n,6,70}]]  (* Harvey P. Dale, Jul 19 2012 *)

a(n)=sum{k=0..n, floor(C(k+4, 4)/C(k+2, 2))}

G.f.: (x^4-x^3+x^2-x+1)/[(1-x)^4(1+x^2)(1+x+x^2)(1-x^2+x^4)].

cp's OEIS Frontend

A281026 a(n) = floor(3*n*(n+1)/4).

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A084626 a(n) = floor(C(n+6,6)/C(n+2,2)).

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A268539 Numbers k such that 48*k + 25 is a perfect square.

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A084624 a(n) = floor(C(n+5,5)/C(n+2,2)).

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Magma

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A255604 Table read by antidiagonals, T(n,k) is the integer part of the area of a regular k-gon with side length n.

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Mathematica

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Formula

A231559 a(n) = floor( A000326(n)/2 ).

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Magma

Mathematica

Formula

A084267 Partial sums of a binomial quotient.

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A281026 a(n) = floor(3n(n+1)/4).