A057570 -id:A057570 - cp's OEIS frontend

A001318 Generalized pentagonal numbers: m(3m - 1)/2, m = 0, +-1, +-2, +-3, ....

0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335

Offset: 0

N. J. A. Sloane

Partial sums of A026741. - Jud McCranie; corrected by Omar E. Pol, Jul 05 2012
From R. K. Guy, Dec 28 2005: (Start)
"Conway's relation twixt the triangular and pentagonal numbers: Divide the triangular numbers by 3 (when you can exactly):
0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ...
0 - 1 2 .- .5 .7 .- 12 15 .- 22 26 .- .35 .40 .- ..51 ...
.....-.-.....+..+.....-..-.....+..+......-...-.......+....
"and you get the pentagonal numbers in pairs, one of positive rank and the other negative.
"Append signs according as the pair have the same (+) or opposite (-) parity.
"Then Euler's pentagonal number theorem is easy to remember:
"p(n-0) - p(n-1) - p(n-2) + p(n-5) + p(n-7) - p(n-12) - p(n-15) ++-- = 0^n
where p(n) is the partition function, the left side terminates before the argument becomes negative and 0^n = 1 if n = 0 and = 0 if n > 0.
"E.g. p(0) = 1, p(7) = p(7-1) + p(7-2) - p(7-5) - p(7-7) + 0^7 = 11 + 7 - 2 - 1 + 0 = 15."
(End)
The sequence may be used in order to compute sigma(n), as described in Euler's article. - Thomas Baruchel, Nov 19 2003
Number of levels in the partitions of n + 1 with parts in {1,2}.
a(n) is the number of 3 X 3 matrices (symmetrical about each diagonal) M = {{a, b, c}, {b, d, b}, {c, b, a}} such that a + b + c = b + d + b = n + 2, a,b,c,d natural numbers; example: a(3) = 5 because (a,b,c,d) = (2,2,1,1), (1,2,2,1), (1,1,3,3), (3,1,1,3), (2,1,2,3). - Philippe Deléham, Apr 11 2007
Also numbers a(n) such that 24*a(n) + 1 = (6*m - 1)^2 are odd squares: 1, 25, 49, 121, 169, 289, 361, ..., m = 0, +-1, +-2, ... . - Zak Seidov, Mar 08 2008
From Matthew Vandermast, Oct 28 2008: (Start)
Numbers n for which A000326(n) is a member of A000332. Cf. A145920.
This sequence contains all members of A000332 and all nonnegative members of A145919. For values of n such that n*(3*n - 1)/2 belongs to A000332, see A145919. (End)
Starting with offset 1 = row sums of triangle A168258. - Gary W. Adamson, Nov 21 2009
Starting with offset 1 = Triangle A101688 * [1, 2, 3, ...]. - Gary W. Adamson, Nov 27 2009
Starting with offset 1 can be considered the first in an infinite set generated from A026741. Refer to the array in A175005. - Gary W. Adamson, Apr 03 2010
Vertex number of a square spiral whose edges have length A026741. The two axes of the spiral forming an "X" are A000326 and A005449. The four semi-axes forming an "X" are A049452, A049453, A033570 and the numbers >= 2 of A033568. - Omar E. Pol, Sep 08 2011
A general formula for the generalized k-gonal numbers is given by n*((k - 2)*n - k + 4)/2, n=0, +-1, +-2, ..., k >= 5. - Omar E. Pol, Sep 15 2011
a(n) is the number of 3-tuples (w,x,y) having all terms in {0,...,n} and 2*w = 2*x + y. - Clark Kimberling, Jun 04 2012
Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. - Omar E. Pol, Aug 04 2012
a(n) is the sum of the largest parts of the partitions of n+1 into exactly 2 parts. - Wesley Ivan Hurt, Jan 26 2013
Conway's relation mentioned by R. K. Guy is a relation between triangular numbers and generalized pentagonal numbers, two sequences from different families, but as triangular numbers are also generalized hexagonal numbers in this case we have a relation between two sequences from the same family. - Omar E. Pol, Feb 01 2013
Start with the sequence of all 0's. Add n to each value of a(n) and the next n - 1 terms. The result is the generalized pentagonal numbers. - Wesley Ivan Hurt, Nov 03 2014
(6k + 1) | a(4k). (3k + 1) | a(4k+1). (3k + 2) | a(4k+2). (6k + 5) | a(4k+3). - Jon Perry, Nov 04 2014
Enge, Hart and Johansson proved: "Every generalised pentagonal number c >= 5 is the sum of a smaller one and twice a smaller one, that is, there are generalised pentagonal numbers a, b < c such that c = 2a + b." (see link theorem 5). - Peter Luschny, Aug 26 2016
The Enge, et al. result for c >= 5 also holds for c >= 2 if 0 is included as a generalized pentagonal number. That is, 2 = 2*1 + 0. - Michael Somos, Jun 02 2018
Suggestion for title, where n actually matches the list and b-file: "Generalized pentagonal numbers: k(n)*(3*k(n) - 1)/2, where k(n) = A001057(n) = [0, 1, -1, 2, -2, 3, -3, ...], n >= 0" - Daniel Forgues, Jun 09 2018 & Jun 12 2018
Generalized k-gonal numbers are the partial sums of the sequence formed by the multiples of (k - 4) and the odd numbers (A005408) interleaved, with k >= 5. - Omar E. Pol, Jul 25 2018
The last digits form a symmetric cycle of length 40 [0, 1, 2, 5, ..., 5, 2, 1, 0], i.e., a(n) == a(n + 40) (mod 10) and a(n) == a(40*k - n - 1) (mod 10), 40*k > n. - Alejandro J. Becerra Jr., Aug 14 2018
Only 2, 5, and 7 are prime. All terms are of the form k*(k+1)/6, where 3 | k or 3 | k+1. For k > 6, the value divisible by 3 must have another factor d > 2, which will remain after the division by 6. - Eric Snyder, Jun 03 2022
8*a(n) is the product of two even numbers one of which is n + n mod 2. - Peter Luschny, Jul 15 2022
a(n) is the dot product of [1, 2, 3, ..., n] and repeat[1, 1/2]. a(5) = 12 = [1, 2, 3, 4, 5] dot [1, 1/2, 1, 1/2, 1] = [1 + 1 + 3 + 2 + 5]. - Gary W. Adamson, Dec 10 2022
Every nonnegative number is the sum of four terms of this sequence [S. Realis]. - N. J. A. Sloane, May 07 2023
From Peter Bala, Jan 06 2025: (Start)
The sequence terms are the exponents in the expansions of the following infinite products:
1) Product_{n >= 1} (1 - s(n)*q^n) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ..., where s(n) = (-1)^(1 + mod(n+1,3)).
2) Product_{n >= 1} (1 - q^(2*n))*(1 - q^(3*n))^2/((1 - q^n)*(1 - q^(6*n))) = 1 + q + q^2 + q^5 + q^7 + q^12 + q^15 + ....
3) Product_{n >= 1} (1 - q^n)*(1 - q^(4*n))*(1 - q^(6*n))^5/((1 - q^(2*n))*(1 - q^(3*n))*(1 - q^(12*n)))^2 = 1 - q + q^2 - q^5 - q^7 + q^12 - q^15 + q^22 + q^26 - q^35 + ....
4) Product_{n >= 1} (1 - q^(2*n))^13/((1 - (-1)^n*q^n)*(1 - q^(4*n)))^5 = 1 - 5*q + 7*q^2 - 11*q^5 + 13*q^7 - 17*q^12 + 19*q^15 - + .... See Oliver, Theorem 1.1. (End)

			G.f. = x + 2*x^2 + 5*x^3 + 7*x^4 + 12*x^5 + 15*x^6 + 22*x^7 + 26*x^8 + 35*x^9 + ...

Enoch Haga, A strange sequence and a brilliant discovery, chapter 5 of Exploring prime numbers on your PC and the Internet, first revised ed., 2007 (and earlier ed.), pp. 53-70.
Ross Honsberger, Ingenuity in Mathematics, Random House, 1970, p. 117.
Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, (to appear), section 7.2.1.4, equation (18).
Ivan Niven and Herbert S. Zuckerman, An Introduction to the Theory of Numbers, 2nd ed., Wiley, NY, 1966, p. 231.
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).

T. D. Noe, Table of n, a(n) for n = 0..1000
G. E. Andrews and J. A. Sellers, Congruences for the Fishburn Numbers, arXiv preprint arXiv:1401.5345 [math.NT], 2014.
Paul Barry, On sequences with {-1, 0, 1} Hankel transforms, arXiv preprint arXiv:1205.2565 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 18 2012
Burkard Polster (Mathologer), The hardest "What comes next?" (Euler's pentagonal formula), Youtube video, Oct 17 2020.
S. Cooper and M. D. Hirschhorn, Results of Hurwitz type for three squares, Discrete Math., Vol. 274, No. 1-3 (2004), pp. 9-24. See P(q).
Stephen Eberhart, Letter to N. J. A. Sloane, Jan 19 1978.
John Elias, Illustration of Initial Terms: Generalized Penthexagrams.
John Elias, Illustration: Star Number Fractal.
John Elias, Illustration: Generalized Pentagonals In Generalized-Octagonal-Hexagrams.
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016.
Leonhard Euler, Découverte d'une loi tout extraordinaire des nombres par rapport à la somme de leurs diviseurs, Opera Omnia, Series I, Vol. 2 (1751), pp. 241-253.
Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, par. 2
Leonhard Euler, Observatio de summis divisorum p. 8.
Leonhard Euler, An observation on the sums of divisors, p. 8, arXiv:math/0411587 [math.HO], 2004.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math., Vol. 3, No. 2 (2008), pp. 76-114.
Silvia Heubach and Toufik Mansour, Counting rises, levels and drops in compositions, arXiv:math/0310197 [math.CO], 2003.
Alfred Hoehn, Illustration of initial terms.
Barbara H. Margolius, Permutations with inversions, J. Integ. Seq., Vol. 4 (2001), Article 01.2.4.
Johannes W. Meijer, Euler's Ship on the Pentagonal Sea, pdf and jpg.
Johannes W. Meijer and Manuel Nepveu, Euler's ship on the Pentagonal Sea, Acta Nova, Vol. 4, No. 1 (December 2008), pp. 176-187.
Mircea Merca, The Lambert series factorization theorem, The Ramanujan Journal, January 2017; DOI: 10.1007/s11139-016-9856-3.
Mircea Merca and Maxie D. Schmidt, New Factor Pairs for Factorizations of Lambert Series Generating Functions, arXiv:1706.02359 [math.CO], 2017. See Remark 2.2.
Mircea Merca, Euler's partition function in terms of 2-adic valuation, Bol. Soc. Mat. Mex. 30, 76 (2024). See p. 3.
Ivan Niven, Formal power series, Amer. Math. Monthly, Vol. 76, No. 8 (1969), pp. 871-889.
Robert J. Lemke Oliver, Eta quotients and theta functions, Advances in Mathematics, Vol. 241, Jul. 2013, pp. 1-17.
Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
Vladimir Pletser, Finding all squared integers expressible as the sum of consecutive squared integers using generalized Pell equation solutions with Chebyshev polynomials, arXiv preprint arXiv:1409.7972 [math.NT], 2014.
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.
S. Realis, Question 271, Nouv. Corresp. Math., 4 (1878) 27-29.
Steven J. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5 (December 2011), pp. 339-350.
André Weil, Two lectures on number theory, past and present, L'Enseign. Math., Vol. XX (1974), pp. 87-110; Oeuvres III, pp. 279-302.
Eric Weisstein's World of Mathematics, Pentagonal numbers, Partition Function P.
Eric Weisstein's World of Mathematics, Pentagonal Number Theorem.
Wikipedia, Pentagonal number theorem.
M. Wohlgemuth, Pentagon, Kartenhaus und Summenzerlegung.
Keke Zhang, Generalized Catalan numbers, arXiv:2011.09593 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A080995 (characteristic function), A026741 (first differences), A034828 (partial sums), A165211 (mod 2).
Cf. A000326 (pentagonal numbers), A005449 (second pentagonal numbers), A000217 (triangular numbers).
Indices of nonzero terms of A010815, i.e., the (zero-based) indices of 1-bits of the infinite binary word to which the terms of A068052 converge.
Union of A036498 and A036499.
Cf. A153384, A168258, A101688, A174739, A175005.
Cf. A074378, A057569, A057570, A007310.
Sequences of generalized k-gonal numbers: this sequence (k=5), A000217 (k=6), A085787 (k=7), A001082 (k=8), A118277 (k=9), A074377 (k=10), A195160 (k=11), A195162 (k=12), A195313 (k=13), A195818 (k=14), A277082 (k=15), A274978 (k=16), A303305 (k=17), A274979 (k=18), A303813 (k=19), A218864 (k=20), A303298 (k=21), A303299 (k=22), A303303 (k=23), A303814 (k=24), A303304 (k=25), A316724 (k=26), A316725 (k=27), A303812 (k=28), A303815 (k=29), A316729 (k=30).
Column 1 of A195152.
Squares in APs: A221671, A221672.
Cf. A054440, A260664, A260672.
Quadrisection: A049453(k), A033570(k), A033568(k+1), A049452(k+1), k >= 0.
Cf. A002620.

a:=[0,1,2,5];; for n in [5..60] do a[n]:=2*a[n-2]-a[n-4]+3; od; a; # Muniru A Asiru, Aug 16 2018

a001318 n = a001318_list !! n
a001318_list = scanl1 (+) a026741_list -- Reinhard Zumkeller, Nov 15 2015

[(6*n^2 + 6*n + 1 - (2*n + 1)*(-1)^n)/16 : n in [0..50]]; // Wesley Ivan Hurt, Nov 03 2014

[(3*n^2 + 2*n + (n mod 2) * (2*n + 1)) div 8: n in [0..70]]; // Vincenzo Librandi, Nov 04 2014

A001318 := -(1+z+z**2)/(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
A001318 := proc(n) (6*n^2+6*n+1)/16-(2*n+1)*(-1)^n/16 ; end proc: # R. J. Mathar, Mar 27 2011

Table[n*(n+1)/6, {n, Select[Range[0, 100], Mod[#, 3] != 1 &]}]
Select[Accumulate[Range[0,200]]/3,IntegerQ] (* Harvey P. Dale, Oct 12 2014 *)
CoefficientList[Series[x (1 + x + x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
LinearRecurrence[{1,2,-2,-1,1},{0,1,2,5,7},70] (* Harvey P. Dale, Jun 05 2017 *)
a[ n_] := With[{m = Quotient[n + 1, 2]}, m (3 m + (-1)^n) / 2]; (* Michael Somos, Jun 02 2018 *)

{a(n) = (3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8}; /* Michael Somos, Mar 24 2011 */

{a(n) = if( n<0, n = -1-n); polcoeff( x * (1 - x^3) / ((1 - x) * (1-x^2))^2 + x * O(x^n), n)}; /* Michael Somos, Mar 24 2011 */

{a(n) = my(m = (n+1) \ 2); m * (3*m + (-1)^n) / 2}; /* Michael Somos, Jun 02 2018 */

def a(n):
    p = n % 2
    return (n + p)*(3*n + 2 - p) >> 3
print([a(n) for n in range(60)])  # Peter Luschny, Jul 15 2022

def A001318(n): return n*(n+1)-(m:=n>>1)*(m+1)>>1 # Chai Wah Wu, Nov 23 2024

@CachedFunction
def A001318(n):
    if n == 0 : return 0
    inc = n//2 if is_even(n) else n
    return inc + A001318(n-1)
[A001318(n) for n in (0..59)] # Peter Luschny, Oct 13 2012

Euler: Product_{n>=1} (1 - x^n) = Sum_{n=-oo..oo} (-1)^n*x^(n*(3*n - 1)/2).
A080995(a(n)) = 1: complement of A090864; A000009(a(n)) = A051044(n). - Reinhard Zumkeller, Apr 22 2006
Euler transform of length-3 sequence [2, 2, -1]. - Michael Somos, Mar 24 2011
a(-1 - n) = a(n) for all n in Z. a(2*n) = A005449(n). a(2*n - 1) = A000326(n). - Michael Somos, Mar 24 2011. [The extension of the recurrence to negative indices satisfies the signature (1,2,-2,-1,1), but not the definition of the sequence m*(3*m -1)/2, because there is no m such that a(-1) = 0. - Klaus Purath, Jul 07 2021]
a(n) = 3 + 2*a(n-2) - a(n-4). - Ant King, Aug 23 2011
Product_{k>0} (1 - x^k) = Sum_{k>=0} (-1)^k * x^a(k). - Michael Somos, Mar 24 2011
G.f.: x*(1 + x + x^2)/((1 + x)^2*(1 - x)^3).
a(n) = n*(n + 1)/6 when n runs through numbers == 0 or 2 mod 3. - Barry E. Williams
a(n) = A008805(n-1) + A008805(n-2) + A008805(n-3), n > 2. - Ralf Stephan, Apr 26 2003
Sequence consists of the pentagonal numbers (A000326), followed by A000326(n) + n and then the next pentagonal number. - Jon Perry, Sep 11 2003
a(n) = (6*n^2 + 6*n + 1)/16 - (2*n + 1)*(-1)^n/16; a(n) = A034828(n+1) - A034828(n). - Paul Barry, May 13 2005
a(n) = Sum_{k=1..floor((n+1)/2)} (n - k + 1). - Paul Barry, Sep 07 2005
a(n) = A000217(n) - A000217(floor(n/2)). - Pierre CAMI, Dec 09 2007
If n even a(n) = a(n-1) + n/2 and if n odd a(n) = a(n-1) + n, n >= 2. - Pierre CAMI, Dec 09 2007
a(n)-a(n-1) = A026741(n) and it follows that the difference between consecutive terms is equal to n if n is odd and to n/2 if n is even. Hence this is a self-generating sequence that can be simply constructed from knowledge of the first term alone. - Ant King, Sep 26 2011
a(n) = (1/2)*ceiling(n/2)*ceiling((3*n + 1)/2). - Mircea Merca, Jul 13 2012
a(n) = (A008794(n+1) + A000217(n))/2 = A002378(n) - A085787(n). - Omar E. Pol, Jan 12 2013
a(n) = floor((n + 1)/2)*((n + 1) - (1/2)*floor((n + 1)/2) - 1/2). - Wesley Ivan Hurt, Jan 26 2013
From Oskar Wieland, Apr 10 2013: (Start)
a(n) = a(n+1) - A026741(n),
a(n) = a(n+2) - A001651(n),
a(n) = a(n+3) - A184418(n),
a(n) = a(n+4) - A007310(n),
a(n) = a(n+6) - A001651(n)*3 = a(n+6) - A016051(n),
a(n) = a(n+8) - A007310(n)*2 = a(n+8) - A091999(n),
a(n) = a(n+10)- A001651(n)*5 = a(n+10)- A072703(n),
a(n) = a(n+12)- A007310(n)*3,
a(n) = a(n+14)- A001651(n)*7. (End)
a(n) = (A007310(n+1)^2 - 1)/24. - Richard R. Forberg, May 27 2013; corrected by Zak Seidov, Mar 14 2015; further corrected by Jianing Song, Oct 24 2018
a(n) = Sum_{i = ceiling((n+1)/2)..n} i. - Wesley Ivan Hurt, Jun 08 2013
G.f.: x*G(0), where G(k) = 1 + x*(3*k + 4)/(3*k + 2 - x*(3*k + 2)*(3*k^2 + 11*k + 10)/(x*(3*k^2 + 11*k + 10) + (k + 1)*(3*k + 4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
Sum_{n>=1} 1/a(n) = 6 - 2*Pi/sqrt(3). - Vaclav Kotesovec, Oct 05 2016
a(n) = Sum_{i=1..n} numerator(i/2) = Sum_{i=1..n} denominator(2/i). - Wesley Ivan Hurt, Feb 26 2017
a(n) = A000292(A001651(n))/A001651(n), for n>0. - Ivan N. Ianakiev, May 08 2018
a(n) = ((-5 + (-1)^n - 6n)*(-1 + (-1)^n - 6n))/96. - José de Jesús Camacho Medina, Jun 12 2018
a(n) = Sum_{k=1..n} k/gcd(k,2). - Pedro Caceres, Apr 23 2019
Quadrisection. For r = 0,1,2,3: a(r + 4*k) = 6*k^2 + sqrt(24*a(r) + 1)*k + a(r), for k >= 1, with inputs (k = 0) {0,1,2,5}. These are the sequences A049453(k), A033570(k), A033568(k+1), A049452(k+1), for k >= 0, respectively. - Wolfdieter Lang, Feb 12 2021
a(n) = a(n-4) + sqrt(24*a(n-2) + 1), n >= 4. - Klaus Purath, Jul 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 6*(log(3)-1). - Amiram Eldar, Feb 28 2022
a(n) = A002620(n) + A008805(n-1). Gary W. Adamson, Dec 10 2022
E.g.f.: (x*(7 + 3*x)*cosh(x) + (1 + 5*x + 3*x^2)*sinh(x))/8. - Stefano Spezia, Aug 01 2024

A154293 Integers of the form t/6, where t is a triangular number (A000217).

Original entry on oeis.org

0, 1, 6, 11, 13, 20, 35, 46, 50, 63, 88, 105, 111, 130, 165, 188, 196, 221, 266, 295, 305, 336, 391, 426, 438, 475, 540, 581, 595, 638, 713, 760, 776, 825, 910, 963, 981, 1036, 1131, 1190, 1210, 1271, 1376, 1441, 1463, 1530, 1645, 1716, 1740, 1813, 1938, 2015

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Old definition was "Integers of the form: 1/6+2/6+3/6+4/6+5/6+...".
1/6 + 2/6 + 3/6 = 1, 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 + 7/6 + 8/6 = 6, ...
a(n) is the set of all integers k such that 48k+1 is a perfect square. The square roots of 48*a(n) + 1 = 1, 7, 17, 23, 25, ... = 8*(n-floor(n/4)) + (-1)^n. - Gary Detlefs, Mar 01 2010
Conjecture: A193828 divided by 2. - Omar E. Pol, Aug 19 2011
The above conjecture is correct. - Charles R Greathouse IV, Jan 02 2012
Quasipolynomial of order 4. - Charles R Greathouse IV, Jan 02 2012
It appears that the sequence terms occur as exponents in the expansion Sum_{n >= 0} x^n/Product_{k = 1..2*n} (1 + x^k) = 1 + x - x^6 - x^11 + x^13 + x^20 - x^35 - x^46  + + - - .... Cf. A218171. [added Jan 21 2025: this is correct  - see Berndt et al., Theorem 3.2.] - Peter Bala, Feb 04 2021
From Peter Bala, Dec 12 2024 (Start)
The sequence terms occur as exponents in the expansion of F(x)*Product_{n >= 1} (1 - x^n) = Product_{n >= 1} (1 - x^n)*(1 + x^(4*n))^2*(1 + x^(4*n-2))*(1 + x^(8*n-3))*(1 + x^(8*n-5)) =  1 - x - x^6 + x^11 + x^13 - x^20 - x^35 + x^46 + x^50 - - + + ..., where F(x) is the g.f. of A069910.
It appears that the sequence terms occur as exponents in the expansion 1/(1 - x) * ( - x^2 + Sum_{n >= 1} x^floor((3*n+1)/2) * 1/Product_{k = 1..n} (1 + x^k) ) = x^6 + x^11 - x^13 - x^20 + x^35 + x^46 - - + + .... (End)
It appears that the sequence terms occur as exponents in the expansion Sum_{n >= 0} x^(n+1)/Product_{k = 1..2*n+2} (1 + x^k) =  x - x^6 - x^11 + x^13 + x^20 - x^35 - x^46  + + - - .... - Peter Bala, Jan 21 2025

			G.f. = x^2 + 6*x^3 + 11*x^4 + 13*x^5 + 20*x^6 + 35*x^7 + 46*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 1..1000
B. C. Berndt, B. Kim, and A. J. Yee, Ramanujan's lost notebook: Combinatorial proofs of identities associated with Heine's transformation or partial theta functions, J. Comb. Thy. Ser. A, 117 (2010), 957-973.
Mircea Merca, The bisectional pentagonal number theorem, Journal of Number Theory, Volume 157 (December 2015), Pages 223-232.
Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Cf. A000217, A001318, A069497, A074378, A057569, A057570, A154292, A193828, A218171.

/* By definition: */ [t/6: n in [0..160] | IsIntegral(t/6) where t is n*(n+1)/2]; // Bruno Berselli, Mar 07 2016

f:=n-> 8*(n-floor(n/4))+(-1)^n:seq((f(n)^2-1)/48,n=0..51); # Gary Detlefs, Mar 01 2010

lst={}; s=0; Do[s+=n/6; If[Floor[s]==s, AppendTo[lst, s]], {n, 0, 7!}]; lst (* Orlovsky *)
Join[{0}, Select[Table[Plus@@Range[n]/6, {n, 200}], IntegerQ]] (* Alonso del Arte, Jan 20 2012 *)
LinearRecurrence[{3,-5,7,-7,5,-3,1},{0,1,6,11,13,20,35},60] (* Charles R Greathouse IV, Jan 20 2012 *)
a[ n_] := (3 n^2 + If[ OddQ[ Quotient[ n + 1, 2]], -5 n + 2, -n]) / 4; (* Michael Somos, Feb 10 2015 *)
a[ n_] := Module[{m = n}, If[ n < 1, m = 1 - n]; SeriesCoefficient[ x^2 (1 + 4 x + x^2) (1 - x^2) (1 - x^6) / ((1 - x)^2 (1 - x^3) (1 - x^4)^2), {x, 0, m}]]; (* Michael Somos, Feb 10 2015 *)

a(n)=n--;(8*(n-n\4)+(-1)^n)^2\48 \\ Charles R Greathouse IV, Jan 02 2012

{a(n) = (3*n^2 + if( (n+1)\2%2, -5*n+2,-n)) / 4}; /* Michael Somos, Feb 10 2015 */

{a(n) = if( n<1, n = 1-n); polcoeff( x^2 * (1 + 4*x + x^2) * (1 - x^2) * (1 - x^6) / ((1 - x)^2 * (1 - x^3) * (1 - x^4)^2) + x * O(x^n), n)}; /* Michael Somos, Feb 10 2015 */

From R. J. Mathar, Jan 07 2009: (Start)
a(n) = A000217(A108752(n))/6.
G.f.: x^2*(x^2-x+1)*(x^2+4*x+1)/((1+x^2)^2*(1-x)^3) (conjectured). (End)
The conjectured g.f. is correct. - Charles R Greathouse IV, Jan 02 2012
a(n) = (f(n)^2-1)/48 where f(n) = 8*(n-floor(n/4))+(-1)^n, with offset 0, a(0)=0. - Gary Detlefs, Mar 01 2010
a(n) = a(1-n) for all n in Z. - Michael Somos, Oct 27 2012
G.f.: x^2 * (1 + 4*x + x^2) * (1 - x^2) * (1 - x^6) / ((1 - x)^2 * (1 - x^3) * (1 - x^4)^2). - Michael Somos, Feb 10 2015
Sum_{n>=2} 1/a(n) = 12 - (1+4/sqrt(3))*Pi. - Amiram Eldar, Mar 18 2022
a(n) = A069497(n)/6. - Hugo Pfoertner, Nov 19 2024
From Peter Bala, Jan 21 2025: (Start)
a(4*n) = 12*n^2 - n; a(4*n+1) = 12*n^2 + n;
a(4*n+2) = (3*n + 1)*(4*n + 1) = A033577(n); a(4*n+3) = (3*n + 2)*(4*n + 3) = A033578(n+1).
Let T(n) = n*(n + 1)/2 denote the n-th triangular number. Then
a(4*n) = (1/6) * T(12*n-1); a(4*n+1) = (1/6) * T(12*n);
a(4*n+2) = (1/6) * T(12*n+3); a(4*n+3) = (1/6) * T(12*n+8). (End)

Definition rewritten by M. F. Hasler, Dec 31 2012

A057569 Numbers of the form k(5k+1)/2 or k(5k-1)/2.

Original entry on oeis.org

0, 2, 3, 9, 11, 21, 24, 38, 42, 60, 65, 87, 93, 119, 126, 156, 164, 198, 207, 245, 255, 297, 308, 354, 366, 416, 429, 483, 497, 555, 570, 632, 648, 714, 731, 801, 819, 893, 912, 990, 1010, 1092, 1113, 1199, 1221, 1311, 1334, 1428, 1452, 1550

Offset: 1

N. J. A. Sloane, Oct 04 2000

a(n) is the set of all m such that 40*m+1 is a perfect square. - Gary Detlefs, Feb 22 2010
Integers of the form (n^2 - n) / 10. Numbers of the form  n * (5*n - 1) / 2 where n is an integer. - Michael Somos, Jan 13 2012
Also integers of the form sum_{k=1..n} k/5. - Alonso del Arte, Jan 20 2012
These numbers appear in a theta function identity. See the Hardy-Wright reference, Theorem 356 on p. 284. See the G.f. of A113428. - Wolfdieter Lang, Oct 28 2016

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fifth ed., Clarendon Press, Oxford, 2003, p. 284.

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

Cf. A074378, A001318, A057570, A154260. - Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Cf. A001622, A113428.

[(10*(n^2-n)+12*(-1)^n*(n div 2))/16: n in [1..60]]; // Vincenzo Librandi, Oct 29 2016

Select[Table[Plus@@Range[n]/5, {n, 0, 199}], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)
LinearRecurrence[{1,2,-2,-1,1},{0,2,3,9,11},50] (* Harvey P. Dale, Jul 05 2021 *)

{a(n) = (10 * (n^2 - n) + 12 * (-1)^n * (n\2)) / 16}; \\ Michael Somos, Jan 13 2012

Vec(x^2*(2*x^2+x+2) / ((1-x)^3*(1+x)^2) + O(x^60)) \\ Colin Barker, Jun 13 2017

A005475 UNION A005476. G.f.: x^2*(2x^2+x+2)/((1-x)^3*(1+x)^2). a(n) = A132356(n+1)/4. - R. J. Mathar, Apr 07 2008
a(n) = (A090771(n)^2 -1)/40. - Gary Detlefs, Feb 22 2010
|A113428(n)| is the characteristic function of the numbers a(n).
a(n) = a(1 - n) for all n in Z. - Michael Somos, Jan 13 2012
From Colin Barker, Jun 13 2017: (Start)
a(n) = n*(5*n - 2)/8 for n even.
a(n) = (5*n - 3)*(n - 1)/8 for n odd.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
(End)
From Amiram Eldar, Mar 17 2022: (Start)
Sum_{n>=2} 1/a(n) = 10 - 2*sqrt(1+2/sqrt(5))*Pi.
Sum_{n>=2} (-1)^n/a(n) = 2*sqrt(5)*log(phi) - 5*(2-log(5)), where phi is the golden ratio (A001622). (End)

A154296 Primes of the form (1+2+3+...+m)/15 = A000217(m)/15, for some m.

Original entry on oeis.org

3, 7, 29, 31

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition: Primes of the form 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=15.
The corresponding m-values are m=9, 14, 29, 30. It is clear that for m > 30, T(m)/15 = m*(m+1)/30 cannot be a prime. - M. F. Hasler, Dec 31 2012
All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013
Also primes p such that 120*p+1 is a perfect square. - Lamine Ngom, Jul 22 2023

Cf. A057570, A154293, A154297, A154298, A154299, A154300, A154301, A154302, A154303, A154304.

lst={};s=0;Do[s+=n/15;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,9!}];lst
Select[(Accumulate[Range[200]])/15,PrimeQ] (* Harvey P. Dale, Oct 30 2011 *)

select(x->denominator(x)==1 & isprime(x), vector(30,m,m^2+m)/30)  \\ M. F. Hasler, Dec 31 2012

Edited by M. F. Hasler, Dec 31 2012

A154304 Primes of the form (1+2+...+m)/210 = A000217(m)/210.

Original entry on oeis.org

3, 17, 47, 419, 421

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition : Primes of the form 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=210.

The corresponding m-values are m=35,84,140,419,420. It is clear that for m>420, T(m)/210 = m(m+1)/420 cannot be a prime, since then each factor in the numerator is larger than the denominator. All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154303.

lst={};s=0;Do[s+=n/210;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,6*9!}];lst

A154304(d=210)={select(x->denominator(x)==1 && isprime(x), vector(d*=2, m, m^2+m)/d)}  \\ - M. F. Hasler, Jan 06 2013

Edited by M. F. Hasler, Jan 06 2013

A154260 Numbers of the form m(4m +- 1)/2.

Original entry on oeis.org

0, 7, 9, 30, 34, 69, 75, 124, 132, 195, 205, 282, 294, 385, 399, 504, 520, 639, 657, 790, 810, 957, 979, 1140, 1164, 1339, 1365, 1554, 1582, 1785, 1815, 2032, 2064, 2295, 2329, 2574, 2610, 2869, 2907, 3180, 3220, 3507, 3549, 3850, 3894, 4209, 4255, 4584

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Also integers of the form Sum_{k = 1..j} k/4 = j*(j + 1)/8. - Alonso del Arte, Jan 20 2012
Numbers h such that 32*h + 1 is a square. - Bruno Berselli, Mar 30 2014
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(16*n))*(1 - q^(16*n-7))*(1 - q^(16*n-9)) = 1 - q^7 - q^9 + q^30 + q^34 - q^69 - q^75 + + - - .... - Peter Bala, Dec 24 2024

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

Cf. A000217, A001318, A074378, A057569, A057570.

Cf. similar sequences listed in A219257 and A299645.

k:=8; f:=func; [0] cat [f(n*m): m in [-1, 1], n in [1..25]]; // Bruno Berselli, Nov 14 2012

Select[Union[Flatten[Table[{n (4n - 1)/2, n (4n + 1)/2}, {n, 0, 199}]]], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)

print1(0);forstep(n=2,1e2,2,print1(", "n*(4*n-1)/2", "n*(4*n+1)/2)) \\ Charles R Greathouse IV, Jan 20 2012

print1(s=0);for(n=1,1e3,s+=n/4;if(denominator(s)==1,print1(s", "))) \\ Charles R Greathouse IV, Jan 20 2012

def A154260(n): return (n>>1)*((n<<2)+(-3 if n&1 else -1)) # Chai Wah Wu, Mar 11 2025

From R. J. Mathar, Jan 07 2009: (Start)
A139274 UNION A139275.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(7 + 2x + 7x^2)/((1+x)^2*(1-x)^3). (End)
From G. C. Greubel, Sep 08 2016: (Start)
a(n) = (1/4)*(8*n^2 + 6*(-1)^n*n - 8*n - 3*(-1)^n + 3).
E.g.f.: (1/4)*( (3 + 8*x^2)*exp(x) - 3*(1 + 2*x)*exp(-x) ). (End)
From Amiram Eldar, Mar 17 2022: (Start)
Sum_{n>=2} 1/a(n) = 8 - (sqrt(2)+1)*Pi.
Sum_{n>=2} (-1)^n/a(n) = 2*sqrt(2)*log(sqrt(2)+1) - 8*(1-log(2)). (End)
a(n) = (n-1)*(4*n-3)/2 if n is odd and a(n) = n*(4*n-1)/2 if n is even. - Chai Wah Wu, Mar 11 2025

A219191 Numbers of the form k(7k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 6, 8, 26, 30, 60, 66, 108, 116, 170, 180, 246, 258, 336, 350, 440, 456, 558, 576, 690, 710, 836, 858, 996, 1020, 1170, 1196, 1358, 1386, 1560, 1590, 1776, 1808, 2006, 2040, 2250, 2286, 2508, 2546, 2780, 2820, 3066, 3108, 3366, 3410, 3680, 3726, 4008

Offset: 1

Bruno Berselli, Nov 14 2012

Equivalently, numbers m such that 28*m+1 is a square.
Also, integer values of h*(h+1)/7.
Let F(r) = Product_{n >= 1} 1 - q^(14*n-r). The sequence terms are the exponents in the expansion of F(0)*F(6)*F(8) = 1 - q^6 - q^8 + q^26 + q^30 - q^60 - q^66 + + - - ... (by the triple product identity).- Peter Bala, Dec 25 2024

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

Cf. numbers of the form k*(i*k+1) with k in A001057: i=0, A001057; i=1, A110660; i=2, A000217; i=3, A152749; i=4, A074378; i=5, A219190; i=6, A036498; i=7, this sequence; i=8, A154260.
Cf. A113801 (square roots of 28*a(n)+1, see the comment).
Cf. similar sequences listed in A219257.
Subsequence of A011860.

k:=7; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];

I:=[0,6,8,26,30]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A := proc (q) local n; for n from 0 to q do if type(sqrt(28*n+1), integer) then print(n) fi; od; end: A(4100); # Peter Bala, Dec 25 2024

Rest[Flatten[{# (7 # - 1), # (7 # + 1)} & /@ Range[0, 25]]]
CoefficientList[Series[2 x (3 + x + 3 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,6,8,26,30},50] (* Harvey P. Dale, Sep 14 2022 *)

G.f.: 2*x^2*(3+x+3*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (14*n*(n-1)+5*(-1)^n*(2*n-1)+5)/8.
a(n) = 2*A057570(n) = (1/7)*A047335(n)*A047274(n+1).
Sum_{n>=2} 1/a(n) = 7 - cot(Pi/7)*Pi. - Amiram Eldar, Mar 17 2022

A274830 Numbers k such that 7*k+1 is a triangular number (A000217).

Original entry on oeis.org

0, 2, 5, 11, 17, 27, 36, 50, 62, 80, 95, 117, 135, 161, 182, 212, 236, 270, 297, 335, 365, 407, 440, 486, 522, 572, 611, 665, 707, 765, 810, 872, 920, 986, 1037, 1107, 1161, 1235, 1292, 1370, 1430, 1512, 1575, 1661, 1727, 1817, 1886, 1980, 2052, 2150, 2225

Offset: 1

Colin Barker, Jul 08 2016

From Peter Bala, Nov 21 2024: (Start)
Numbers of the form n*(7*n + 3)/2 for n in Z. Cf. A057570.
The sequence terms occur as the exponents in the expansion of Product_{n >= 1} (1 - x^(7*n)) * (1 + x^(7*n-2)) * (1 + x^(7*n-5)) = 1 + x^2 + x^5 + x^11 + x^17 + x^27 + x^36 + .... Cf. A363800. (End)

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. A000217, A057570, A363800.

Cf. similar sequences where k*n+1 is a triangular number: A000096 (k=1), A074377 (k=2), A045943 (k=3), A274681 (k=4), A085787 (k=5), A274757 (k=6).

Table[(14 (n - 1) n + (2 n - 1) (-1)^n + 1)/16, {n, 1, 60}] (* Bruno Berselli, Jul 08 2016 *)

select(n->ispolygonal(7*n+1, 3), vector(3000, n, n-1))

concat(0, Vec(x^2*(2+3*x+2*x^2)/((1-x)^3*(1+x)^2) + O(x^100)))

G.f.: x^2*(2 + 3*x + 2*x^2) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
a(n) = (14*(n - 1)*n + (2*n - 1)*(-1)^n + 1)/16. Therefore:
a(n) = n*(7*n - 6)/8 for n even,
a(n) = (n - 1)*(7*n - 1)/8 for n odd.
E.g.f.: (x*(7*x -1)*cosh(x) + (7*x^2 + x + 1)*sinh(x))/8. - Stefano Spezia, Nov 26 2024

Edited by Bruno Berselli, Jul 08 2016

A154303 Primes of the form (1+2+...+m)/90 = A000217(m)/90.

Original entry on oeis.org

7, 11, 179, 181

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Original definition: "Primes of the form : 1/x+2/x+3/x+4/x+5/x+6/x+7/x+..., x=90."

The corresponding m-values are m=35,44,179,180. It is clear that for m>180, T(m)/90 = m(m+1)/180 cannot be a prime, since then each factor in the numerator is larger than the denominator. All of the sequences A154296, ..., A154304 could or should be grouped together in a single ("fuzzy"?) table. It would be more interesting to have the function f(n) which gives the *number* of primes of the form T(k)/n. - M. F. Hasler, Jan 06 2013

Cf. A057570, A154293, A154296 - A154304.

lst={};s=0;Do[s+=n/90;If[Floor[s]==s,If[PrimeQ[s],AppendTo[lst,s]]],{n,0,5*9!}];lst

d=90*2;for(m=1,999,(m^2+m)%d==0&isprime((m^2+m)/d)&print1(m",")) \\ print the m-values(!) - use A154304(90) to get A154303 as a list/vector. \\ - M. F. Hasler, Jan 06 2013

Edited by M. F. Hasler, Jan 06 2013

A154292 Integers of the form m(6m -+ 1)/2.

Original entry on oeis.org

11, 13, 46, 50, 105, 111, 188, 196, 295, 305, 426, 438, 581, 595, 760, 776, 963, 981, 1190, 1210, 1441, 1463, 1716, 1740, 2015, 2041, 2338, 2366, 2685, 2715, 3056, 3088, 3451, 3485, 3870, 3906, 4313, 4351, 4780, 4820, 5271, 5313, 5786, 5830, 6325, 6371

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

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. A001318, A074378, A057569, A057570.

&cat[[n*(6*n-1) div 2, n*(6*n+1) div 2]: n in [2..60 by 2]]; // Vincenzo Librandi, Sep 10 2016

Flatten[Table[{n (6n-1)/2,n (6n+1)/2},{n,2,50,2}]] (* Harvey P. Dale, Jan 19 2013 *)

Vec(x*(11+2*x+11*x^2)/((1-x)^3*(1+x)^2) + O(x^60)) \\ Colin Barker, Feb 26 2016

From Colin Barker, Feb 26 2016: (Start)
a(n) = (12*n^2 - 10*(-1)^n*n + 12*n - 5*(-1)^n + 5)/4.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
G.f.: x*(11 + 2*x + 11*x^2) / ((1-x)^3*(1+x)^2). (End)
E.g.f.: (1/4)*(-5 + 10*x + (5 + 24*x + 12*x^2)*exp(2*x))*exp(-x). - G. C. Greubel, Sep 10 2016
From Amiram Eldar, Mar 18 2022: (Start)
Sum_{n>=1} 1/a(n) = 131/11 - (2+sqrt(3))*Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = 133/11 - 3*log(12) - 2*sqrt(3)*log(2+sqrt(3)). (End)

cp's OEIS Frontend

A001318 Generalized pentagonal numbers: m*(3*m - 1)/2, m = 0, +-1, +-2, +-3, ....

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A154293 Integers of the form t/6, where t is a triangular number (A000217).

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A057569 Numbers of the form k*(5*k+1)/2 or k*(5*k-1)/2.

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A154296 Primes of the form (1+2+3+...+m)/15 = A000217(m)/15, for some m.

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A154304 Primes of the form (1+2+...+m)/210 = A000217(m)/210.

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A154260 Numbers of the form m*(4*m +- 1)/2.

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A001318 Generalized pentagonal numbers: m(3m - 1)/2, m = 0, +-1, +-2, +-3, ....

A057569 Numbers of the form k(5k+1)/2 or k(5k-1)/2.

A154260 Numbers of the form m(4m +- 1)/2.

A219191 Numbers of the form k(7k+1), where k = 0,-1,1,-2,2,-3,3,...

A154292 Integers of the form m(6m -+ 1)/2.