A139277 -id:A139277 - cp's OEIS frontend

A139595 A139277(n) followed by A139273(n+1).

0, 5, 13, 26, 42, 63, 87, 116, 148, 185, 225, 270, 318, 371, 427, 488, 552, 621, 693, 770, 850, 935, 1023, 1116, 1212, 1313, 1417, 1526, 1638, 1755, 1875, 2000, 2128, 2261, 2397, 2538, 2682, 2831, 2983, 3140, 3300, 3465, 3633, 3806

Offset: 0

Omar E. Pol, May 03 2008

Sequence found by reading the line from 0, in the direction 0, 5,... and the same line from 0, in the direction 0, 13,..., in the square spiral whose vertices are the triangular numbers A000217.

			Array begins:
0, 5
13, 26
42, 63
87, 116

Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1)

Cf. A000217, A046092, A139273, A139277, A077221, A139591, A139592, A139593, A139596, A139597, A139598.

LinearRecurrence[{2,0,-2,1},{0,5,13,26},50] (* Harvey P. Dale, Jul 31 2021 *)

Array read by rows: row n gives 8*n^2 + 5n, 8*(n+1)^2 - 3(n+1).

G.f.: -x*(5+3*x) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Feb 13 2011

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A139278 a(n) = n(8n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139276, A139277, A139279, A139280, A139281, A139282.

Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)

a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 7*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A139276 a(n) = n(8n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139275, A139277, A139278, A139279, A139280, A139281, A139282.

a[n_]:=n*(8*n+3); a[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,-3,1},{0,11,38},50] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 8*n^2 + 3*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

Original entry on oeis.org

2, 7, 18, 31, 50, 71, 98, 127, 162, 199, 242, 287, 338, 391, 450, 511, 578, 647, 722, 799, 882, 967, 1058, 1151, 1250, 1351, 1458, 1567, 1682, 1799, 1922, 2047, 2178, 2311, 2450, 2591, 2738, 2887, 3042, 3199, 3362, 3527, 3698, 3871, 4050, 4231, 4418, 4607, 4802

Offset: 0

Bruno Berselli, Sep 21 2011 - based on remarks and sequences by Omar E. Pol

Sequence found by reading the numbers in increasing order on the vertical line containing 2 of the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

Also A077591 (without first term) and A157914 interleaved.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A195605.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A077591, A157914.
Cf. A047621 (contains first differences), A016754 (contains the sum of any two consecutive terms).
Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195241 [incomplete list].

[(4*n*(n+2)+(-1)^n+3)/2: n in [0..48]];

CoefficientList[Series[(2 + 3 x + 4 x^2 - x^3) / ((1 + x) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{2,0,-2,1},{2,7,18,31},50] (* Harvey P. Dale, Jan 21 2017 *)

for(n=0, 48, print1((4*n*(n+2)+(-1)^n+3)/2", "));

G.f.: (2+3*x+4*x^2-x^3)/((1+x)*(1-x)^3).
a(n) = a(-n-2) = 2*a(n-1)-2*a(n-3)+a(n-4).
a(n) = A047524(A000982(n+1)).
Sum_{n>=0} 1/a(n) = 1/2 + Pi^2/16 - cot(Pi/(2*sqrt(2)))*Pi/(4*sqrt(2)). - Amiram Eldar, Mar 06 2023

A299645 Numbers of the form m(8m + 5), where m is an integer.

Original entry on oeis.org

0, 3, 13, 22, 42, 57, 87, 108, 148, 175, 225, 258, 318, 357, 427, 472, 552, 603, 693, 750, 850, 913, 1023, 1092, 1212, 1287, 1417, 1498, 1638, 1725, 1875, 1968, 2128, 2227, 2397, 2502, 2682, 2793, 2983, 3100, 3300, 3423, 3633, 3762, 3982, 4117, 4347, 4488, 4728, 4875

Offset: 1

Bruno Berselli, Feb 26 2018

Equivalently, numbers k such that 32*k + 25 is a square. This means that 4*a(n) + 3 is a triangular number.

Interleaving of A139277 and A139272 (without 0).

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

Cf. A139272, A139277.
Subsequence of A011861, A047222.
Cf. numbers of the form m*(8*m + h): A154260 (h=1), A014494 (h=2), A274681 (h=3), A046092 (h=4), this sequence (h=5), 2*A074377 (h=6), A274979 (h=7).

List([1..50], n -> (8*n*(n-1)-(2*n-1)*(-1)^n-1)/4);

[div((8n*(n-1)-(2n-1)*(-1)^n-1), 4) for n in 1:50] # Peter Luschny, Feb 27 2018

[(8*n*(n-1)-(2*n-1)*(-1)^n-1)/4: n in [1..50]];

seq((exp(I*Pi*x)*(1-2*x)+8*(x-1)*x-1)/4, x=1..50); # Peter Luschny, Feb 27 2018

Table[(8 n (n - 1) - (2 n - 1) (-1)^n - 1)/4, {n, 1, 50}]

makelist((8*n*(n-1)-(2*n-1)*(-1)^n-1)/4, n, 1, 50);

vector(50, n, nn; (8*n*(n-1)-(2*n-1)*(-1)^n-1)/4)

concat(0, Vec(x^2*(3 + 10*x + 3*x^2)/((1 - x)^3*(1 + x)^2) + O(x^60))) \\ Colin Barker, Feb 27 2018

[(8*n*(n-1)-(2*n-1)*(-1)**n-1)/4 for n in range(1, 60)]

def A299645(n): return (n>>1)*((n<<2)+(1 if n&1 else -5)) # Chai Wah Wu, Mar 11 2025

[(8*n*(n-1)-(2*n-1)*(-1)^n-1)/4 for n in (1..50)]

O.g.f.: x^2*(3 + 10*x + 3*x^2)/((1 - x)^3*(1 + x)^2).
E.g.f.: (1 + 2*x - (1 - 8*x^2)*exp(2*x))*exp(-x)/4.
a(n) = a(-n+1) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = (8*n*(n - 1) - (2*n - 1)*(-1)^n - 1)/4 = (2*n + (-1)^n - 1)*(4*n - 3*(-1)^n - 2)/4. Therefore, 3 and 13 are the only prime numbers in this sequence.
a(n) + a(n+1) = 4*n^2 for even n, otherwise a(n) + a(n+1) = 4*n^2 - 1.
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=2} 1/a(n) = 8/25 + (sqrt(2)-1)*Pi/5.
Sum_{n>=2} (-1)^n/a(n) = 8*log(2)/5 - sqrt(2)*log(2*sqrt(2)+3)/5 - 8/25. (End)
a(n) = (n-1)*(4*n+1)/2 if n is odd and a(n) = n*(4*n-5)/2 if n is even. - Chai Wah Wu, Mar 11 2025

A195241 Expansion of (1-x+19x^3-3x^4)/(1-x)^3.

Original entry on oeis.org

1, 2, 3, 23, 59, 111, 179, 263, 363, 479, 611, 759, 923, 1103, 1299, 1511, 1739, 1983, 2243, 2519, 2811, 3119, 3443, 3783, 4139, 4511, 4899, 5303, 5723, 6159, 6611, 7079, 7563, 8063, 8579, 9111, 9659, 10223, 10803, 11399, 12011, 12639, 13283, 13943

Offset: 0

Bruno Berselli, Sep 13 2011 - based on remarks and sequences by Omar E. Pol

Sequence found by reading the line 1, 2, 3, 23,.. in the square spiral whose vertices are the triangular numbers (A000217) - see Pol's comments in other sequences visible in this numerical spiral.

This is a subsequence of A110326 (without signs) and A047838 (apart from the second term, 2).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms: An origin of A195241.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217.

Cf. A033585, A069129, A077221, A102083, A139098, A139271-A139277, A139592, A139593, A188135, A194268, A194431, A195605 [incomplete list].

m:=44; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x+19*x^3-3*x^4)/(1-x)^3));

CoefficientList[Series[(1 - x + 19 x^3 - 3 x^4)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{3,-3,1},{1,2,3,23,59},50] (* Harvey P. Dale, Dec 04 2022 *)

makelist(coeff(taylor((1-x+19*x^3-3*x^4)/(1-x)^3, x, 0, n), x, n), n, 0, 43);

Vec((1-x+19*x^3-3*x^4)/(1-x)^3+O(x^44))

G.f.: (1-x+19*x^3-3*x^4)/(1-x)^3.

a(n) = 8*n^2-20*n+11 for n>1; a(0)=1, a(1)=2.

cp's OEIS Frontend

A139595 A139277(n) followed by A139273(n+1).

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A139273 a(n) = n*(8*n - 3).

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A139278 a(n) = n*(8*n+7).

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A139276 a(n) = n*(8*n+3).

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A195605 a(n) = (4*n*(n+2)+(-1)^n+1)/2 + 1.

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A299645 Numbers of the form m*(8*m + 5), where m is an integer.

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A195241 Expansion of (1-x+19*x^3-3*x^4)/(1-x)^3.

A139273 a(n) = n(8n - 3).

A139278 a(n) = n(8n+7).

A139276 a(n) = n(8n+3).

A195605 a(n) = (4n(n+2)+(-1)^n+1)/2 + 1.

A299645 Numbers of the form m(8m + 5), where m is an integer.

A195241 Expansion of (1-x+19x^3-3x^4)/(1-x)^3.