A014494 -id:A014494 - cp's OEIS frontend

A352115 Partial sums of the even triangular numbers (A014494).

0, 6, 16, 44, 80, 146, 224, 344, 480, 670, 880, 1156, 1456, 1834, 2240, 2736, 3264, 3894, 4560, 5340, 6160, 7106, 8096, 9224, 10400, 11726, 13104, 14644, 16240, 18010, 19840, 21856, 23936, 26214, 28560, 31116, 33744, 36594, 39520, 42680, 45920, 49406, 52976, 56804

Offset: 0

David James Sycamore, Mar 05 2022

The absolute difference between the n-th partial sum of the odd triangular numbers and the (n-1)-th partial sum of the even triangular numbers is equal to n; see formula.

Partial sums of the even generalized hexagonal numbers. - Omar E. Pol, Mar 05 2022

			a(0) = 0 because 0 is the first even term in A000217.
a(1) = 6, the sum of the first two even terms in A000217, and so on.

Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A001477, A000217, A000292, A014493, A014494, A352116, A000447.

LinearRecurrence[{2, 1, -4, 1, 2, -1}, {0, 6, 16, 44, 80, 146}, 50] (* Amiram Eldar, Mar 05 2022 *)

te(n) = (2*n+1)*(2*n+1-(-1)^n)/2; \\ A014494
a(n) = sum(k=0, n, te(k)); \\ Michel Marcus, Mar 06 2022

def A352115(n): return (n + 1)*(2*n*(n+2) + 3*(n%2))//3 # Chai Wah Wu, Mar 11 2022

a(n) = Sum_{k=0..n-1} A014494(k) = Sum_{k=0..n-1} (2*k+1)(2*k+1-(-1)^k)/2.
|A352116(n) - a(n-1)| = n.
A352116(n) + a(n-1) = A000447(n), (n >= 1).
From Stefano Spezia, Mar 05 2022: (Start)
a(n) = (n + 1)*(4*n^2 + 8*n + 3 - 3*(-1)^n)/6.
G.f.: 2*x*(3 + 2*x + 3*x^2)/((1 - x)^4*(1 + x)^2). (End)

A014601 Numbers congruent to 0 or 3 mod 4.

Original entry on oeis.org

0, 3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40, 43, 44, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 67, 68, 71, 72, 75, 76, 79, 80, 83, 84, 87, 88, 91, 92, 95, 96, 99, 100, 103, 104, 107, 108, 111, 112, 115, 116, 119, 120, 123, 124

Offset: 0

Eric Rains (rains(AT)caltech.edu)

Discriminants of orders in imaginary quadratic fields (negated). [Comment corrected by Christopher E. Thompson, Dec 11 2016]
Numbers such that Langford-Skolem problem has a solution - see A014552.
Complement of A042963. - Reinhard Zumkeller, Oct 04 2004
Also called skew amenable numbers; a number k is skew amenable if there exist a set {a(i)} of integers satisfying the relations k = Sum_{i=1..k} a(i) = -Product_{i=1..k} a(i). Thus we have 8 = 1 + 1 + 1 + 1 + 1 + 1 - 2 + 4 = -(1*1*1*1*1*1*(-2)*4). - Lekraj Beedassy, Jan 07 2005
Possible nonpositive discriminants of quadratic equation a*x^2 + b*x + c or discriminants of binary quadratic forms a*x^2 + b*x*y + c*y^2. - Artur Jasinski, Apr 28 2008
Also, disregarding the 0 term, positive integers m such that, equivalently,
  (i) +-1 +-2 +-... +-m is even for all choices of signs,
  (ii) +-1 +-2 +-... +-m = 0 for some choices of signs,
  (iii) for all -m <= k <= m, k = +-1 +-2 +-... +-(k-1) +-(k+1) +-(k+2) +-... +-m for at least one choice of signs. - Rick L. Shepherd, Oct 29 2008
A145768(a(n)) is even. - Reinhard Zumkeller, Jun 05 2012
Multiples of 4 interleaved with 1 less than multiples of 4. - Wesley Ivan Hurt, Nov 08 2013
((2*k+0) + (2*k+1) + ... + (2*k+m-1) + (2*k+m)) is even if and only if m = a(n) for some n where k is any nonnegative integer. - Gionata Neri, Jul 24 2015
Numbers whose binary reflected Gray code (A014550) ends with 0. - Amiram Eldar, May 17 2021

			G.f. = 3*x + 4*x^2 + 7*x^3 + 8*x^4 + 11*x^5 + 12*x^6 + 15*x^7 + 16*x^8 + ...

H. Cohen, Course in Computational Alg. No. Theory, Springer, 1993, pp. 514-5.
A. Scholz and B. Schoeneberg, Einführung in die Zahlentheorie, 5. Aufl., de Gruyter, Berlin, New York, 1973, p. 108.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
S. F. Barger, Solution to problem 10454: Amenable Numbers, Amer. Math. Monthly, Vol. 105, No. 4 (April 1998), p. 368.
Steven R. Finch, Class number theory [Cached copy, with permission of the author]
Heiko Harborth, Solution of Steinhaus's problem with plus and minus signs, Journal of Combinatorial Theory, Series A, Volume 12, Issue 2 (March 1972), Pages 253-259.
Mickaël Launay, Les routes de Numland, L’énigme maths du "Monde" n°2. In French.
Michael Penn, The most beautiful arrangement of numbers, YouTube video, 2024.
Rick L. Shepherd, Binary quadratic forms and genus theory, Master of Arts Thesis, University of North Carolina at Greensboro, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A004676, A014550, A042948, A079896.

Cf. A274406. - Bruno Berselli, Jun 26 2016

a014601 n = a014601_list !! n
a014601_list = [x | x <- [0..], mod x 4 `elem` [0, 3]]
-- Reinhard Zumkeller, Jun 05 2012

[n: n in [0..200]|n mod 4 in {0,3}]; // Vincenzo Librandi, Dec 24 2010

A014601:=n->3*n-2*floor(n/2); seq(A014601(k), k=0..100); # Wesley Ivan Hurt, Nov 08 2013

aa = {}; Do[Do[Do[d = b^2 - 4 a c; If[d <= 0, AppendTo[aa, -d]], {a, 0, 50}], {b, 0, 50}], {c, 0, 50}]; Union[aa] (* Artur Jasinski, Apr 28 2008 *)
Select[Range[0, 124], Or[Mod[#, 4] == 0, Mod[#, 4] == 3] &] (* Ant King, Nov 18 2010 *)
CoefficientList[Series[2 x/(1 - x)^2 + (1/(1 - x) + 1/(1 + x)) x/2, {x, 0, 100}], x] (* Vincenzo Librandi, May 18 2014 *)
a[ n_] := 2 n + Mod[n, 2]; (* Michael Somos, Jul 24 2015 *)

{a(n) = 2*n + n%2}; /* Michael Somos, Dec 27 2010 */

a(n) = (n + 1)*2 + 1 - n mod 2. - Reinhard Zumkeller, Apr 21 2003
A014494(n) = A000217(a(n)). - Reinhard Zumkeller, Oct 04 2004
a(n) = Sum_{k=1..n} (2 - (-1)^k). - William A. Tedeschi, Mar 20 2008
A139131(a(n)) = A078636(a(n)). - Reinhard Zumkeller, Apr 10 2008
From R. J. Mathar, Sep 25 2009: (Start)
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 2.
G.f.: x*(3+x)/((1+x)*(x-1)^2). (End)
a(n) = 2*n + (n mod 2). - Paolo Valzasina (p.valzasina(AT)gmail.com), Nov 24 2009
a(n) = (4*n - (-1)^n + 1)/2. - Bruno Berselli, Oct 06 2010
a(n) = 4*n - a(n-1) - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = -A042948(-n) for all n in Z. - Michael Somos, Dec 27 2010
G.f.: 2*x / (1 - x)^2 + (1 / (1 - x) + 1 / (1 + x)) * x/2. - Michael Somos, Dec 27 2010
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0) = 3 and b(k) = 2^(k+1) for k > 0. - Philippe Deléham, Oct 17 2011
a(n) = ceiling((4/3)*ceiling(3*n/2)). - Clark Kimberling, Jul 04 2012
a(n) = 3n - 2*floor(n/2). - Wesley Ivan Hurt, Nov 08 2013
a(n) = A042948(n+1) - 1 for all n in Z. - Michael Somos, Jul 24 2015
a(n) + a(n+1) = A004767(n) for all n in Z. - Michael Somos, Jul 24 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2)/4 - Pi/8. - Amiram Eldar, Dec 05 2021
E.g.f.: ((4*x + 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 04 2022

A047522 Numbers that are congruent to {1, 7} mod 8.

Original entry on oeis.org

1, 7, 9, 15, 17, 23, 25, 31, 33, 39, 41, 47, 49, 55, 57, 63, 65, 71, 73, 79, 81, 87, 89, 95, 97, 103, 105, 111, 113, 119, 121, 127, 129, 135, 137, 143, 145, 151, 153, 159, 161, 167, 169, 175, 177, 183, 185, 191, 193, 199, 201, 207, 209, 215, 217, 223, 225, 231, 233

Offset: 1

N. J. A. Sloane

Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2)))  = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001132, A014494, A056575, A010709, A074378, A047336, A056020, A005408, A047209, A007310, A090771, A175885, A091998, A175886, A175887, A058529, A047621, A179260, A352125.
Cf. A077221 (partial sums).
Cf. A000217, A089911, A113801.
Cf. A010503, A049310, A053120.

a047522 n = a047522_list !! (n-1)
a047522_list = 1 : 7 : map (+ 8) a047522_list
-- Reinhard Zumkeller, Jan 07 2012

Select[Range[1, 191, 2], JacobiSymbol[2, # ]==1&]

a(n)=4*n-2+(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(8*A014494(n)+1) = sqrt(16*ceiling(n/2)*(2*n+1)+1) = sqrt(8*A056575(n)-8*(2n+1)*(-1)^n+1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
1 - 1/7 + 1/9 - 1/15 + 1/17 - ... = (Pi/8)*(1 + sqrt(2)). [Jolley] - Gary W. Adamson, Dec 16 2006
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = 4n - 2 + (-1)^n = a(n-2) + 8.
G.f.: x(1+6x+x^2)/((1+x)(1-x)^2). (End)
a(n) = 8*n - a(n-1) - 8. - Vincenzo Librandi, Aug 06 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 8*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
E.g.f.: 1 + (4*x - 1)*cosh(x) + (4*x - 3)*sinh(x). - Stefano Spezia, May 13 2021
E.g.f.: 1 + (4*x - 3)*exp(x) + 2*cosh(x). - David Lovler, Jul 16 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2+sqrt(2)) (A179260).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/8)*cosec(Pi/8) (A352125). (End)

A074378 Even triangular numbers halved.

Original entry on oeis.org

0, 3, 5, 14, 18, 33, 39, 60, 68, 95, 105, 138, 150, 189, 203, 248, 264, 315, 333, 390, 410, 473, 495, 564, 588, 663, 689, 770, 798, 885, 915, 1008, 1040, 1139, 1173, 1278, 1314, 1425, 1463, 1580, 1620, 1743, 1785, 1914, 1958, 2093, 2139, 2280, 2328, 2475

Offset: 0

W. Neville Holmes, Sep 04 2002

Set of integers k such that k + (1 + 2 + 3 + 4 + ... + x) = 3*k, where x is sufficiently large. For example, 203 is a term because 203 + (1 + 2 + 3 + 4 + ... +28) = 609 and 609 = 3*203. - Gil Broussard, Sep 01 2008
Set of all m such that 16*m+1 is a perfect square. - Gary Detlefs, Feb 21 2010
Integers of the form Sum_{k=0..n} k/2. - Arkadiusz Wesolowski, Feb 07 2012
Numbers of the form h*(4*h + 1) for h = 0, -1, 1, -2, 2, -3, 3, ... - Bruno Berselli, Feb 26 2018
Numbers whose distance to nearest square equals their distance to nearest oblong; that is, numbers k such that A053188(k) = A053615(k). - Lamine Ngom, Oct 27 2020
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(8*n))*(1 + q^(8*n-3))*(1 + q^(8*n-5)) = 1 + q^3 + q^5 + q^14 + q^18 + .... - Peter Bala, Dec 30 2024

David A. Corneth, Table of n, a(n) for n = 0..9999
Neville Holmes, More Geometric Integer Sequences
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A011848, A014493, A014494, A074377, A035294, A274757.
Cf. A010709, A047522. [Vincenzo Librandi, Feb 14 2009]
Cf. A266883 (numbers n such that 16*n-15 is a square).
Cf. A016687, A153799.
Cf. A053615, A053188.

f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]]; // Bruno Berselli, Nov 13 2012

Maple

a:=n->(2*n+1)*floor((n+1)/2): seq(a(n),n=0..50); # Muniru A Asiru, Feb 01 2019

Mathematica

1/2 * Select[PolygonalNumber@ Range[0, 100], EvenQ] (* Michael De Vlieger, Jun 01 2017, Version 10.4 *) Select[Accumulate[Range[0,100]],EvenQ]/2 (* Harvey P. Dale, Feb 15 2025 *)

PARI

a(n)=(2*n+1)*(n-n\2)

Formula

Sum_{n>=0} q^a(n) = (Prod_{n>0} (1-q^n))*(Sum_{n>=0} A035294(n)*q^n).

a(n) = n*(n + 1)/4 where n*(n + 1)/2 is even.

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

From Benoit Jubin, Feb 05 2009: (Start)

a(n) = (2*n + 1)*floor((n + 1)/2).

a(2*k) = k*(4*k+1); a(2*k+1) = (k+1)*(4*k+3). (End)

a(2*n) = A007742(n), a(2*n-1) = A033991(n). - Arkadiusz Wesolowski, Jul 20 2012

a(n) = (4*n + 1 - (-1)^n)*(4*n + 3 - (-1)^n)/4^2. - Peter Bala, Jan 21 2019

a(n) = (2*n+1)*(n+1)*(1+(-1)^(n+1))/4 + (2*n+1)*(n)*(1+(-1)^n)/4. - Eric Simon Jacob, Jan 16 2020

From Amiram Eldar, Jul 03 2020: (Start)

Sum_{n>=1} 1/a(n) = 4 - Pi (A153799).

Sum_{n>=1} (-1)^(n+1)/a(n) = 6*log(2) - 4 (See A016687). (End)

a(n) = A014494(n)/2 = A274757(n)/3 = A266883(n) - 1. - Hugo Pfoertner, Dec 31 2024

A058377 Number of solutions to 1 +- 2 +- 3 +- ... +- n = 0.

Original entry on oeis.org

0, 0, 1, 1, 0, 0, 4, 7, 0, 0, 35, 62, 0, 0, 361, 657, 0, 0, 4110, 7636, 0, 0, 49910, 93846, 0, 0, 632602, 1199892, 0, 0, 8273610, 15796439, 0, 0, 110826888, 212681976, 0, 0, 1512776590, 2915017360, 0, 0, 20965992017, 40536016030, 0, 0, 294245741167

Offset: 1

Naohiro Nomoto, Dec 19 2000

nonn

Consider the set { 1,2,3,...,n }. Sequence gives number of ways this set can be partitioned into 2 subsets with equal sums. For example, when n = 7, { 1,2,3,4,5,6,7} can be partitioned in 4 ways: {1,6,7} {2,3,4,5}; {2,5,7} {1,3,4,6}; {3,4,7} {1,2,5,6} and {1,2,4,7} {3,5,6}. - sorin (yamba_ro(AT)yahoo.com), Mar 24 2007
The "equal sums" of Sorin's comment are the positive terms of A074378 (Even triangular numbers halved). In the current sequence a(n) <> 0 iff n is the positive index (A014601) of an even triangular number (A014494). - Rick L. Shepherd, Feb 09 2010
a(n) is the number of partitions of n(n-3)/4 into distinct parts not exceeding n-1. - Alon Amit, Oct 18 2017
a(n) is the coefficient of x^(n*(n+1)/4-1) of Product_{k=2..n} (1+x^k). - Jianing Song, Nov 19 2021

			1+2-3=0, so a(3)=1;
1-2-3+4=0, so a(4)=1;
1+2-3+4-5-6+7=0, 1+2-3-4+5+6-7=0, 1-2+3+4-5+6-7=0, 1-2-3-4-5+6+7=0, so a(7)=4.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..3342 (terms < 10^1000, first 1000 terms from Alois P. Heinz)
Larry Glasser, A formula for A058377, Jul 29 2019

Cf. A000217, A014601, A014494, A025591, A063865, A063866, A063867, A069918, A074378, A111133, A161943, A348639.

Column k=2 of A275714.

b:= proc(n, i) option remember; local m; m:= i*(i+1)/2;
      `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i), i-1) +b(n+i, i-1)))
    end:
a:= n-> `if`(irem(n-1, 4)<2, 0, b(n, n-1)):
seq(a(n), n=1..60);  # Alois P. Heinz, Oct 30 2011

f[n_, s_] := f[n, s] = Which[n == 0, If[s == 0, 1, 0], Abs[s] > (n*(n + 1))/2, 0, True, f[n - 1, s - n] + f[n - 1, s + n]]; Table[ f[n, 0]/2, {n, 1, 50}]

list(n) = my(poly=vector(n), v=vector(n)); poly[1]=1; for(k=2, n, poly[k]=poly[k-1]*(1+'x^k)); for(k=1, n, if(k%4==1||k%4==2, v[k]=0, v[k]=polcoeff(poly[k], k*(k+1)/4-1))); v \\ Jianing Song, Nov 19 2021

a(n) is half the coefficient of q^0 in product('(q^(-k)+q^k)', 'k'=1..n) for n >= 1. - Floor van Lamoen, Oct 10 2005
a(4n+1) = a(4n+2) = 0. - Michael Somos, Apr 15 2007
a(n) = [x^n] Product_{k=1..n-1} (x^k + 1/x^k). - Ilya Gutkovskiy, Feb 01 2024

More terms from Sascha Kurz, Mar 25 2002

Edited and extended by Robert G. Wilson v, Oct 24 2002

A014493 Odd triangular numbers.

Original entry on oeis.org

1, 3, 15, 21, 45, 55, 91, 105, 153, 171, 231, 253, 325, 351, 435, 465, 561, 595, 703, 741, 861, 903, 1035, 1081, 1225, 1275, 1431, 1485, 1653, 1711, 1891, 1953, 2145, 2211, 2415, 2485, 2701, 2775, 3003, 3081, 3321, 3403, 3655, 3741, 4005, 4095, 4371, 4465, 4753, 4851

Offset: 1

Mohammad K. Azarian

Odd numbers of the form n*(n+1)/2.
For n such that n(n+1)/2 is odd see A042963 (congruent to 1 or 2 mod 4).
Even central polygonal numbers minus 1. - Omar E. Pol, Aug 17 2011
Odd generalized hexagonal numbers. - Omar E. Pol, Sep 24 2015

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 68.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
D. H. Lehmer, Recurrence formulas for certain divisor functions, Bull. Amer. Math. Soc., Vol. 49, No. 2 (1943), pp. 150-156.
Eric Weisstein's World of Mathematics, Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A000796, A014494, A019669, A042963, A067589, A128880.

List([1..50], n -> (2*n-1)*(2*n-1-(-1)^n)/2); # G. C. Greubel, Feb 09 2019

[(2*n-1)*(2*n-1-(-1)^n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 18 2011

[(2*n-1)*(2*n-1-(-1)^n)/2$n=1..50]; # Muniru A Asiru, Mar 10 2019

Select[ Table[n(n + 1)/2, {n, 93}], OddQ[ # ] &] (* Robert G. Wilson v, Nov 05 2004 *)
LinearRecurrence[{1,2,-2,-1,1},{1,3,15,21,45},50] (* Harvey P. Dale, Jun 19 2011 *)

a(n)=(2*n-1)*(2*n-1-(-1)^n)/2 \\ Charles R Greathouse IV, Sep 24 2015

def A014493(n): return ((n<<1)-1)*(n-(n&1^1)) # Chai Wah Wu, Feb 12 2023

[(2*n-1)*(2*n-1-(-1)^n)/2 for n in (1..50)] # G. C. Greubel, Feb 09 2019

From Ant King, Nov 17 2010: (Start)
a(n) = (2*n-1)*(2*n - 1 - (-1)^n)/2.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). (End)
G.f.: x*(1 + 2*x + 10*x^2 + 2*x^3 + x^4)/((1+x)^2*(1-x)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 10 2009
a(n) = A000217(A042963(n)). - Reinhard Zumkeller, Feb 14 2012, Oct 04 2004
a(n) = A193868(n) - 1. - Omar E. Pol, Aug 17 2011
Let S = Sum_{n>=0} x^n/a(n), then S = Q(0) where Q(k) = 1 + x*(4*k+1)/(4*k + 3 - x*(2*k+1)*(4*k+3)^2/(x*(2*k+1)*(4*k+3) + (4*k+5)*(2*k+3)/Q(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 27 2013
E.g.f.: (2*x^2+x+1)*cosh(x)+x*(2*x-1)*sinh(x)-1. - Ilya Gutkovskiy, Apr 24 2016
Sum_{n>=1} 1/a(n) = Pi/2 (A019669). - Robert Bilinski, Jan 20 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2). - Amiram Eldar, Mar 06 2022

More terms from Erich Friedman

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

A193867 Odd central polygonal numbers.

Original entry on oeis.org

1, 7, 11, 29, 37, 67, 79, 121, 137, 191, 211, 277, 301, 379, 407, 497, 529, 631, 667, 781, 821, 947, 991, 1129, 1177, 1327, 1379, 1541, 1597, 1771, 1831, 2017, 2081, 2279, 2347, 2557, 2629, 2851, 2927, 3161, 3241, 3487, 3571, 3829, 3917, 4187, 4279

Offset: 1

Omar E. Pol, Aug 15 2011

Even triangular numbers plus 1.

Union of A188135 and A185438 without repetitions (A188135 is a bisection of this sequence. Another bisection is A185438 but without its initial term).

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. A000124, A014494, A014601, A185438, A188135, A193868.

Select[Accumulate[Range[0,100]],EvenQ]+1 (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,7,11,29,37},50] (* Harvey P. Dale, Nov 29 2014 *)

Vec(-x*(x^2+1)*(x^2+6*x+1) / ((1+x)^2*(x-1)^3) + O(x^100)) \\ Colin Barker, Jan 27 2016

a(n) = A000124(A014601(n-1)).
a(n) = 1 + A014494(n-1).
G.f.: -x*(x^2+1)*(x^2+6*x+1) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Aug 25 2011
From Colin Barker, Jan 27 2016: (Start)
a(n) = (4*n^2+2*(-1)^n*n-4*n-(-1)^n+3)/2.
a(n) = 2*n^2-n+1 for n even.
a(n) = 2*n^2-3*n+2 for n odd. (End)
Sum_{n>=1} 1/a(n) = 2*Pi*sinh(sqrt(7)*Pi/4)/(sqrt(7)*(2*cosh(sqrt(7)*Pi/4) - sqrt(2))). - Amiram Eldar, May 11 2025

A353136 Primes whose gaps to both neighbor primes are triangular numbers.

Original entry on oeis.org

53, 157, 173, 251, 257, 263, 337, 373, 547, 557, 563, 577, 587, 593, 607, 653, 733, 947, 977, 1039, 1103, 1123, 1181, 1187, 1223, 1367, 1627, 1747, 1753, 1907, 2017, 2063, 2287, 2417, 2677, 2719, 2897, 2903, 2963, 3307, 3313, 3517, 3547, 3637, 3733, 4013, 4211

Offset: 1

Alois P. Heinz, Apr 25 2022

nonn

			Prime 251 is a term, the gap to the previous prime 241 is 10 and the gap to the next prime 257 is 6 and both gaps are triangular numbers.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A000040, A000217, A014494, A353088, A353135, A353137.

t:= proc(n) option remember; issqr(8*n+1) end:
q:= n-> isprime(n) and andmap(t, [n-prevprime(n), nextprime(n)-n]):
select(q, [$3..5000])[];

t[n_] := IntegerQ@Sqrt[8n+1];
q[n_] := PrimeQ[n] && t[n-NextPrime[n, -1]] && t[NextPrime[n]-n];
Select[Range[3, 5000], q] (* Jean-François Alcover, May 14 2022, after Alois P. Heinz *)

A014738 Squares of even triangular numbers.

Original entry on oeis.org

36, 100, 784, 1296, 4356, 6084, 14400, 18496, 36100, 44100, 76176, 90000, 142884, 164836, 246016, 278784, 396900, 443556, 608400, 672400, 894916, 980100, 1272384, 1382976, 1758276, 1898884, 2371600, 2547216, 3132900, 3348900

Offset: 0

Mohammad K. Azarian

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-4,-6,6,4,-4,-1,1).

Cf. A006752, A014494, A014736.

List([1..30], n-> ((2*n+1)*(2*n+1-(-1)^n))^2/4); # G. C. Greubel, Jul 24 2019

[((2*n+1)*(2*n+1-(-1)^n))^2/4: n in [1..30]]; // G. C. Greubel, Jul 24 2019

Select[Accumulate[Range[100]],EvenQ]^2 (* Harvey P. Dale, Oct 09 2012 *)

vector(30, n, ((2*n+1)*(2*n+1-(-1)^n))^2/4) \\ G. C. Greubel, Jul 24 2019

[((2*n+1)*(2*n+1-(-1)^n))^2/4 for n in (1..30)] # G. C. Greubel, Jul 24 2019

a(n) = A014494(n + 1)^2. - Sean A. Irvine, Nov 18 2018
From G. C. Greubel, Jul 24 2019: (Start)
G.f.: 4*x*(9 +16*x +135*x^2 +64*x^3 +135*x^4 +16*x^5 +9*x^6)/((1-x)^5*(1+x)^4).
E.g.f.: x*(35+41*x+36*x^2+4*x^3)*cosh(x) + (1+9*x+77*x^2+28*x^3+4*x^4)* sinh(x). (End)
From Amiram Eldar, Mar 06 2022: (Start)
Sum_{n>=0} 1/a(n) = 7*Pi^2/12 + 2*Pi - 12.
Sum_{n>=0} (-1)^n/a(n) = 12 - 4*G - 12*log(2), where G is Catalan's constant (A006752). (End)

More terms from James Sellers

cp's OEIS Frontend

A352115 Partial sums of the even triangular numbers (A014494).

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A014601 Numbers congruent to 0 or 3 mod 4.

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A047522 Numbers that are congruent to {1, 7} mod 8.

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A074378 Even triangular numbers halved.

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A058377 Number of solutions to 1 +- 2 +- 3 +- ... +- n = 0.

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A014493 Odd triangular numbers.

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