A001105 -id:A001105 - cp's OEIS frontend

A144555 a(n) = 14*n^2.

0, 14, 56, 126, 224, 350, 504, 686, 896, 1134, 1400, 1694, 2016, 2366, 2744, 3150, 3584, 4046, 4536, 5054, 5600, 6174, 6776, 7406, 8064, 8750, 9464, 10206, 10976, 11774, 12600, 13454, 14336, 15246, 16184, 17150, 18144, 19166, 20216, 21294, 22400, 23534, 24696

Offset: 0

N. J. A. Sloane, Jan 01 2009

Sequence found by reading the line from 0, in the direction 0, 14, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Also sequence found by reading the same line and direction in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 10 2011

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

See also A033428, A033429, A033581, A033582, A033583, A033584, ... and A249327 for the whole table.
Cf. A000290, A001105, A064761, A118277, A152742, A195019, A195020.
Cf. A008596, A195145.

Table[14*n^2, {n, 0, 45}] (* Amiram Eldar, Feb 03 2021 *)

A144555(n)=14*n^2 \\ M. F. Hasler, Oct 31 2014

a(n) = 14*A000290(n) = 7*A001105(n) = 2*A033582(n). - Omar E. Pol, Jan 01 2009
a(n) = a(n-1) + 14*(2*n-1), with a(0) = 0. - Vincenzo Librandi, Nov 25 2010
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/84.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/168.
Product_{n>=1} (1 + 1/a(n)) = sqrt(14)*sinh(Pi/sqrt(14))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(14)*sin(Pi/sqrt(14))/Pi. (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 14*x*(1 + x)/(1-x)^3.
E.g.f.: 14*x*(1 + x)*exp(x).
a(n) = n*A008596(n) = A195145(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A226488 a(n) = n(13n - 9)/2.

Original entry on oeis.org

0, 2, 17, 45, 86, 140, 207, 287, 380, 486, 605, 737, 882, 1040, 1211, 1395, 1592, 1802, 2025, 2261, 2510, 2772, 3047, 3335, 3636, 3950, 4277, 4617, 4970, 5336, 5715, 6107, 6512, 6930, 7361, 7805, 8262, 8732, 9215, 9711, 10220, 10742, 11277, 11825, 12386, 12960

Offset: 0

Bruno Berselli, Jun 09 2013

Sum of n-th octagonal number and n-th 9-gonal (nonagonal) number.

Sum of reciprocals of a(n), for n>0: 0.629618994194109711163742089971688...

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

Cf. A000567, A001106, A153080 (first differences).

Cf. numbers of the form n*(n*k-k+4)/2 listed in A005843 (k=0), A000096 (k=1), A002378 (k=2), A005449 (k=3), A001105 (k=4), A005476 (k=5), A049450 (k=6), A218471 (k=7), A002939 (k=8), A062708 (k=9), A135706 (k=10), A180223 (k=11), A139267 (n=12), this sequence (k=13), A139268 (k=14), A226489 (k=15), A139271 (k=16), A180232 (k=17), A152995 (k=18), A226490 (k=19), A152965 (k=20), A226491 (k=21), A152997 (k=22).

List([0..50], n-> n*(13*n-9)/2); # G. C. Greubel, Aug 30 2019

Magma
```
[n*(13*n-9)/2: n in [0..50]];
```

I:=[0,2,17]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2) +Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A226488:=n->n*(13*n - 9)/2; seq(A226488(n), n=0..50); # Wesley Ivan Hurt, Feb 25 2014

Table[n(13n-9)/2, {n, 0, 50}]
LinearRecurrence[{3, -3, 1}, {0, 2, 17}, 50] (* Harvey P. Dale, Jun 19 2013 *)
CoefficientList[Series[x(2+11x)/(1-x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)

a(n)=n*(13*n-9)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(13*n-9)/2 for n in (0..50)] # G. C. Greubel, Aug 30 2019

G.f.: x*(2+11*x)/(1-x)^3.
a(n) + a(-n) = A152742(n).
a(0)=0, a(1)=2, a(2)=17; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jun 19 2013
E.g.f.: x*(4 + 13*x)*exp(x)/2. - G. C. Greubel, Aug 30 2019
a(n) = A000567(n) + A001106(n). - Michel Marcus, Aug 31 2019

A347457 Heinz numbers of integer partitions with integer alternating product.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 71, 72, 73, 74, 75, 76, 78

Offset: 1

Gus Wiseman, Sep 26 2021

nonn

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.
We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)).
Also numbers whose multiset of prime indices has integer reverse-alternating product.

			The prime indices of 525 are {2,3,3,4}, with reverse-alternating product 2, so 525 is in the sequence
The prime indices of 135 are {2,2,2,3}, with reverse-alternating product 3/2, so 135 is not in the sequence.

The reciprocal version is A028982.
Allowing any alternating product > 1 gives A028983, reverse A347465.
Factorizations of this type are counted by A347437.
These partitions are counted by A347446.
The reverse reciprocal version A347451.
The odd-length case is A347453.
The reverse version is A347454.
The complement is A347455.
A056239 adds up prime indices, row sums of A112798.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A335433 lists numbers whose prime indices are separable, complement A335448.
A347461 counts possible alternating products of partitions, reverse A347462.
Cf. A001105, A001222, A028260, A119620, A119899, A316523, A344606, A344617, A346703, A346704, A347448, A347450.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}];
Select[Range[100],IntegerQ[altprod[Reverse[primeMS[#]]]]&]

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

A135453 a(n) = 12*n^2.

Original entry on oeis.org

0, 12, 48, 108, 192, 300, 432, 588, 768, 972, 1200, 1452, 1728, 2028, 2352, 2700, 3072, 3468, 3888, 4332, 4800, 5292, 5808, 6348, 6912, 7500, 8112, 8748, 9408, 10092, 10800, 11532, 12288, 13068, 13872, 14700, 15552, 16428, 17328, 18252, 19200, 20172, 21168, 22188

Offset: 0

Ben Paul Thurston, Dec 14 2007

Areas of perfect 4:3 rectangles (for n > 0).
Sequence found by reading the line from 0, in the direction 0, 12, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Semi-axis opposite to A069190 in the same spiral. - Omar E. Pol, Sep 16 2011
(x,y,z) = (-a(n), 1 + n*a(n), 1 - n*a(n)) are solutions of the Diophantine equation x^3 + 2*y^3 + 2*z^3 = 4. - XU Pingya, Apr 30 2022

			192 is on the list since 16*12 is a 4:3 rectangle with integer sides and an area of 192.

Ivan Panchenko, Table of n, a(n) for n = 0..10000
John Elias, Illustration: Even Ordered Star Perimeters.
Leo Tavares, Illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A001105, A016742, A033428, A033581, A069190, A086729.

Cf. A008594, A195143.

List([0..100],n->12*n^2); # Muniru A Asiru, Jan 29 2018

seq(12*h^2,n=0..100); # Muniru A Asiru, Jan 29 2018

Table[12*n^2, {n, 0, 60}] (* Stefan Steinerberger, Dec 17 2007 *)
LinearRecurrence[{3,-3,1},{0,12,48},50] (* Harvey P. Dale, Jan 19 2020 *)

a(n)=12*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 12*A000290(n) = 6*A001105(n) = 4*A033428(n) = 3*A016742(n) = 2*A033581(n). - Omar E. Pol, Dec 13 2008
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/72 (A086729).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/144.
Product_{n>=1} (1 + 1/a(n)) = 2*sqrt(3)*sinh(Pi/(2*sqrt(3)))/Pi.
Product_{n>=1} (1 - 1/a(n)) = 2*sqrt(3)*sin(Pi/(2*sqrt(3)))/Pi. (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 12*x*(1 + x)/(1-x)^3.
E.g.f.: 12*x*(1 + x)*exp(x).
a(n) = n*A008594(n) = A195143(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

More terms from Stefan Steinerberger, Dec 17 2007

Minor edits from Omar E. Pol, Dec 15 2008

A139570 a(n) = 2n(n+3).

Original entry on oeis.org

0, 8, 20, 36, 56, 80, 108, 140, 176, 216, 260, 308, 360, 416, 476, 540, 608, 680, 756, 836, 920, 1008, 1100, 1196, 1296, 1400, 1508, 1620, 1736, 1856, 1980, 2108, 2240, 2376, 2516, 2660, 2808, 2960, 3116, 3276, 3440, 3608, 3780, 3956, 4136, 4320, 4508, 4700, 4896

Offset: 0

Omar E. Pol, May 19 2008

Numbers n such that 2*n + 9 is a square. - Vincenzo Librandi, Nov 24 2010

a(n) appears also as the fourth member of the quartet [p0(n), p1(n), p2(n), a(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n), and p2(n) = A054000(n+1). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 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. A000096, A001105, A020739, A022998, A028552, A046092, A054000, A067728, A147973.

CoefficientList[Series[4 x (2 - x)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, May 23 2014 *)

a(n)=2*n*(n+3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2*A028552(n) = 2*n^2 + 6*n = n*(2*n+6).
a(n) = a(n-1) + 4*n + 4 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
From Paul Curtz, Mar 27 2011: (Start)
a(n) = A022998(n)*A022998(n+3).
a(n) = 4*A000096(n). (End)
G.f.: 4*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 31 2011
From Amiram Eldar, Dec 23 2022: (Start)
Sum_{n>=1} 1/a(n) = 11/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 5/36. (End)
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: 2*exp(x)*x*(4 + x).
a(n) = n*A020739(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A347450 Numbers whose multiset of prime indices has alternating product <= 1.

Original entry on oeis.org

1, 2, 4, 6, 8, 9, 10, 14, 15, 16, 18, 21, 22, 24, 25, 26, 32, 33, 34, 35, 36, 38, 39, 40, 46, 49, 50, 51, 54, 55, 56, 57, 58, 60, 62, 64, 65, 69, 72, 74, 77, 81, 82, 84, 85, 86, 87, 88, 90, 91, 93, 94, 95, 96, 98, 100, 104, 106, 111, 115, 118, 119, 121, 122

Offset: 1

Gus Wiseman, Sep 24 2021

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)).
Also Heinz numbers integer partitions with reverse-alternating product <= 1, where the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
Also numbers whose multiset of prime indices has alternating sum <= 1.

			The initial terms and their prime indices:
      1: {}            26: {1,6}           56: {1,1,1,4}
      2: {1}           32: {1,1,1,1,1}     57: {2,8}
      4: {1,1}         33: {2,5}           58: {1,10}
      6: {1,2}         34: {1,7}           60: {1,1,2,3}
      8: {1,1,1}       35: {3,4}           62: {1,11}
      9: {2,2}         36: {1,1,2,2}       64: {1,1,1,1,1,1}
     10: {1,3}         38: {1,8}           65: {3,6}
     14: {1,4}         39: {2,6}           69: {2,9}
     15: {2,3}         40: {1,1,1,3}       72: {1,1,1,2,2}
     16: {1,1,1,1}     46: {1,9}           74: {1,12}
     18: {1,2,2}       49: {4,4}           77: {4,5}
     21: {2,4}         50: {1,3,3}         81: {2,2,2,2}
     22: {1,5}         51: {2,7}           82: {1,13}
     24: {1,1,1,2}     54: {1,2,2,2}       84: {1,1,2,4}
     25: {3,3}         55: {3,5}           85: {3,7}

The additive version (alternating sum <= 0) is A028260.
The reverse version is A028982, counted by A119620.
Allowing any alternating product < 1 gives A119899.
Factorizations of this type are counted by A339846, complement A339890.
Allowing any alternating product >= 1 gives A344609, multiplicative A347456.
Partitions of this type are counted by A347443.
Allowing any integer alternating product gives A347454, reciprocal A347451.
The complement is A347465, reverse A028983, counted by A347448.
A056239 adds up prime indices, row sums of A112798.
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A335433 lists numbers whose prime indices are separable, complement A335448.
A344606 counts alternating permutations of prime indices.
A347457 lists Heinz numbers of partitions with integer alternating product.
Cf. A001105, A001222, A027193, A344617, A345958, A346703, A346704, A347449, A347461, A347462.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}];
Select[Range[100],altprod[primeMS[#]]<=1&]

Union of A028982 and A119899.

Union of A028260 and A001105.

A347454 Numbers whose multiset of prime indices has integer alternating product.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 25, 27, 28, 29, 31, 32, 36, 37, 41, 42, 43, 44, 45, 47, 48, 49, 50, 52, 53, 59, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 78, 79, 80, 81, 83, 89, 92, 97, 98, 99, 100, 101, 103, 107, 108, 109, 112, 113

Offset: 1

Gus Wiseman, Sep 26 2021

nonn

First differs from A265640 in having 42.
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
We define the alternating product of a sequence (y_1,...,y_k) to be Product_i y_i^((-1)^(i-1)).
Also Heinz numbers of partitions with integer reverse-alternating product, where the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

			The terms and their prime indices begin:
      1: {}            20: {1,1,3}         47: {15}
      2: {1}           23: {9}             48: {1,1,1,1,2}
      3: {2}           25: {3,3}           49: {4,4}
      4: {1,1}         27: {2,2,2}         50: {1,3,3}
      5: {3}           28: {1,1,4}         52: {1,1,6}
      7: {4}           29: {10}            53: {16}
      8: {1,1,1}       31: {11}            59: {17}
      9: {2,2}         32: {1,1,1,1,1}     61: {18}
     11: {5}           36: {1,1,2,2}       63: {2,2,4}
     12: {1,1,2}       37: {12}            64: {1,1,1,1,1,1}
     13: {6}           41: {13}            67: {19}
     16: {1,1,1,1}     42: {1,2,4}         68: {1,1,7}
     17: {7}           43: {14}            71: {20}
     18: {1,2,2}       44: {1,1,5}         72: {1,1,1,2,2}
     19: {8}           45: {2,2,3}         73: {21}

The even-length case is A000290.
The additive version is A026424.
Allowing any alternating product < 1 gives A119899, strict A028260.
Allowing any alternating product >= 1 gives A344609, multiplicative A347456.
Factorizations of this type are counted by A347437.
These partitions are counted by A347445, reverse A347446.
Allowing any alternating product <= 1 gives A347450.
The reciprocal version is A347451.
The odd-length case is A347453.
The version for reversed prime indices is A347457, complement A347455.
Allowing any alternating product > 1 gives A347465, reverse A028983.
A056239 adds up prime indices, row sums of A112798.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A335433 lists numbers whose prime indices are separable, complement A335448.
A344606 counts alternating permutations of prime indices.
A347461 counts possible alternating products of partitions.
A347462 counts possible reverse-alternating products of partitions.
Cf. A001105, A001222, A028982, A119620, A236913, A316523, A344653, A346703, A346704, A347443, A347439.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
altprod[q_]:=Product[q[[i]]^(-1)^(i-1),{i,Length[q]}];
Select[Range[100],IntegerQ[altprod[primeMS[#]]]&]

A093328 a(n) = 2*n^2 + 3.

Original entry on oeis.org

3, 5, 11, 21, 35, 53, 75, 101, 131, 165, 203, 245, 291, 341, 395, 453, 515, 581, 651, 725, 803, 885, 971, 1061, 1155, 1253, 1355, 1461, 1571, 1685, 1803, 1925, 2051, 2181, 2315, 2453, 2595, 2741, 2891, 3045, 3203, 3365, 3531, 3701, 3875, 4053, 4235, 4421, 4611

Offset: 0

Ralf Stephan, Apr 25 2004

Number of 132-avoiding two-stack sortable permutations which also avoid 4321.
Conjecture: no perfect powers. - Zak Seidov, Sep 27 2015
Numbers k such that 2*k - 6 is a square. - Bruno Berselli, Nov 08 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Eric S. Egge and Toufik Mansour, 132-avoiding two-stack sortable permutations, Fibonacci numbers, and Pell numbers, Discrete Applied Mathematics, Vol. 143, No. 1-3 (2004), pp. 72-83; arXiv preprint, arXiv:math/0205206 [math.CO], 2002.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001105, A005893, A058331, A154685.

[2*n^2+3: n in [0..50]]; // Vincenzo Librandi, Jul 08 2012

Table[2 n^2 + 3, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011*)
CoefficientList[Series[(3 - 4 x + 5 x^2)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 08 2012 *)
LinearRecurrence[{3, -3, 1}, {3, 5, 11}, 50] (* Harvey P. Dale, Apr 03 2016 *)

a(n)=2*n^2+3; \\ Zak Seidov, Sep 27 2015

a(n) = A005893(n)+1 = A058331(n)+2 = A001105(n)+3.
a(n+2) = A154685(n+1,n+2).
From Vincenzo Librandi, Jul 08 2012: (Start)
G.f.: (3 - 4*x + 5*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
Sum_{n>=0} 1/a(n) = (1 + sqrt(3/2)*Pi*coth(sqrt(3/2)*Pi))/6. - Amiram Eldar, Nov 25 2020
E.g.f.: exp(x)*(3 + 2*x + 2*x^2). - Elmo R. Oliveira, Jan 17 2025

Simpler definition and new offset from Paul F. Brewbaker, Jun 23 2009

Edited by N. J. A. Sloane, Jun 27 2009

A002593 a(n) = n^2(2n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.

Original entry on oeis.org

0, 1, 28, 153, 496, 1225, 2556, 4753, 8128, 13041, 19900, 29161, 41328, 56953, 76636, 101025, 130816, 166753, 209628, 260281, 319600, 388521, 468028, 559153, 662976, 780625, 913276, 1062153, 1228528, 1413721, 1619100, 1846081

Offset: 0

N. J. A. Sloane

The m-th term, for m = A065549(n), is perfect (A000396). - Lekraj Beedassy, Jun 04 2002
Partial sums of A016755. - Lekraj Beedassy, Jan 06 2004
Also, the k-th triangular number, where k = 2n^2 - 1 = A056220(n), i.e., a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
Also, the j-th hexagonal number, where j = n^2 = A000290(n), i.e., a(n) = A000384(A000290(n)) and a(n) = A056220(n) * A000290(n) or j * k. This sequence is a subsequence of the hexagonal number sequence and retains the aspect intrinsic to the hexagonal number sequence that each number in this sequence can be found by multiplying its triangular number by its hexagonal number. - Bruce J. Nicholson, Aug 22 2017
Odd numbers and their squares both having the form 2x-+1, we may write (2r+1)^3 = (2r+1)*(2s-1), where s = centered squares = (r+1)^2 + r^2. Since 2r+1 = (r+1)^2 - r^2, it follows immediately from summing telescopingly over n-1, the product 2*{(r+1)^4 - r^4} - {(r+1)^2 - r^2}, that Sum_{r=0..n-1} (2r+1)^3 = 2*n^4 - n^2 = n^2*(2n^2 - 1). - Lekraj Beedassy, Jun 16 2004
a(n) is also the starting term in the sum of a number M(n) of consecutive cubed integers equaling a squared integer (A253724) for M(n) equal to twice a squared integer (A001105). Numbers a(n) such that a^3 + (a+1)^3 + ... + (a+M-1)^3 = c^2 has nontrivial solutions over the integers for M equal to twice a squared integer (A001105). If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting from a^3 and equaling a squared integer c^2. For n >= 1, M(n) = 2n^2 (A001105), a(n) = M(M-1)/2 = n^2(2n^2 - 1), and c(n) = sqrt(M/2) (M(M^2-1)/2) = n^3(4n^4 - 1). The trivial solutions with M < 1 and a < 2 are not considered. - Vladimir Pletser, Jan 10 2015
Binomial transform of the sequence with offset 1 is (1, 27, 98, 120, 48, 0, 0, 0, ...). - Gary W. Adamson, Jul 23 2015

Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 169, #31.
F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 47.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
F. E. Croxton and D. J. Cowden, Applied General Statistics, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955. [Annotated scans of just pages 742-743]
Neslihan Kilar, Abdelmejid Bayad, and Yilmaz Simsek, Finite sums involving trigonometric functions and special polynomials: analysis of generating functions and p-adic integrals, Appl. Anal. Disc. Math., hal-04535748, 2024. See p. 22.
Vladimir Pletser, File Triplets (M,a,c) for M=2n^2
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
G. Xiao, Sigma Server, Operate on "(2*n-1)^3".
M. J. Zerger, Proof without words: The sum of consecutive odd cubes is a triangular number, Math. Mag., 68 (1995), 371.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000290, A000384, A000447, A000583, A002309, A253724, A253725, A260810.

[n^2*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

A002593:=-z*(z+1)*(z**2+22*z+1)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation
a:= n-> n^2*(2*n^2-1): seq(a(n), n=0..50);  # Vladimir Pletser, Jan 10 2015

CoefficientList[Series[(-x^4-23x^3-23x^2-x)/(x-1)^5,{x,0, 80}],x]  (* or *)
Table[ n^2 (2n^2-1),{n,0,80}]  (* Harvey P. Dale, Mar 28 2011 *)
Join[{0},Accumulate[Range[1,91,2]^3]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,28,153,496},40] (* Harvey P. Dale, Mar 22 2017 *)

a(n) = n^2*(2*n^2 - 1) \\ Charles R Greathouse IV, Feb 07 2017

a(n) = A000217(A056220(n)). - Lekraj Beedassy, Jun 11 2004
G.f.: (-x^4 - 23*x^3 - 23*x^2 - x)/(x - 1)^5. - Harvey P. Dale, Mar 28 2011
a(n) = n^2*(2n^2 - 1). - Vladimir Pletser, Jan 10 2015
E.g.f.: exp(x)*x*(1 + 13*x + 24*x^2/2! + 12*x^3/3!). - Wolfdieter Lang, Mar 11 2017
a(n) = A000384(A000290(n)) = A056220(n) * A000290(n). - Bruce J. Nicholson, Aug 22 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 1 - Pi^2/6 - cot(Pi/sqrt(2))*Pi/sqrt(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = cosec(Pi/sqrt(2))*Pi/sqrt(2) - Pi^2/12 - 1. (End)

cp's OEIS Frontend

A144555 a(n) = 14*n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A226488 a(n) = n*(13*n - 9)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Magma

Magma

Maple

Mathematica

PARI

Sage

Formula

A347457 Heinz numbers of integer partitions with integer alternating product.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

A135453 a(n) = 12*n^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Formula

Extensions

A139570 a(n) = 2*n*(n+3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A347450 Numbers whose multiset of prime indices has alternating product <= 1.

Views

Author

Keywords

Comments

A226488 a(n) = n(13n - 9)/2.

A139570 a(n) = 2n(n+3).

A002593 a(n) = n^2(2n^2 - 1); also Sum_{k=0..n-1} (2k+1)^3.