A147875 -id:A147875 - cp's OEIS frontend

A000566 Heptagonal numbers (or 7-gonal numbers): n(5n-3)/2.

0, 1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, 616, 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, 1918, 2059, 2205, 2356, 2512, 2673, 2839, 3010, 3186, 3367, 3553, 3744, 3940, 4141, 4347, 4558, 4774, 4995, 5221, 5452, 5688

Offset: 0

N. J. A. Sloane

Binomial transform of (0, 1, 5, 0, 0, 0, ...). Binomial transform is A084899. - Paul Barry, Jun 10 2003
Row sums of triangle A131413. - Gary W. Adamson, Jul 08 2007
Sequence starting (1, 7, 18, 34, ...) = binomial transform of (1, 6, 5, 0, 0, 0, ...). Also row sums of triangle A131896. - Gary W. Adamson, Jul 24 2007
Also the partial sums of A016861, a zero added in front; therefore a(n) = n (mod 5). - R. J. Mathar, Mar 19 2008
Also sequence found by reading the line from 0, in the direction 0, 7, ..., and the line from 1, in the direction 1, 18, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. These parallel lines are the semi-axes perpendicular to the main axis A195015 in the same spiral. - Omar E. Pol, Oct 14 2011
Also sequence found by reading the line from 0, in the direction 0, 7, ... and the parallel line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012
Partial sums give A002413. - Omar E. Pol, Jan 12 2013
The n-th heptagonal number equals the sum of the n consecutive integers starting at 2*n-1; for example, 1, 3+4, 5+6+7, 7+8+9+10, etc. In general, the n-th (2k+1)-gonal number is the sum of the n consecutive integers starting at (k-1)*n - (k-2). When k = 1 and 2, this result generates the triangular numbers, A000217, and the pentagonal numbers, A000326, respectively. - Charlie Marion, Mar 02 2022

			G.f. = x + 7*x^2 + 18*x^3 + 34*x^4 + 55*x^5 + 81*x^6 + 112*x^7 + ... - _Michael Somos_, Jan 25 2019

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
Leonard E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
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).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Daniel Mondot, Table of n, a(n) for n = 0..10000 (first 1000 terms by T. D. Noe)
S. Barbero, U. Cerruti, and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7., section 4.4.
C. K. Cook and M. R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 341.
Bir Kafle, Florian Luca, and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae, pp. 137-145.
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.
Omar E. Pol, Illustration of initial terms of A000217, A000290, A000326, A000384, A000566, A000567.
B. Srinivasa Rao, Heptagonal Numbers in the Pell Sequence and Diophantine Equations 2x^2 = y^2(5y - 3)^2 ± 2, Fib. Quarterly, 43 (2005), 194-201.
B. Srinivasa Rao, Heptagonal numbers in the associated Pell sequence and Diophantine equations x^2(5x - 3)^2 = 8y^2 ± 4, Fib. Quarterly, 43 (2005), 302-306.
Leo Tavares, Illustration.
Eric Weisstein's World of Mathematics, Heptagonal Number.
Index to sequences related to polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014637, A014640, A014773, A014792, A069099, A131413, A131896, A134483, A000384.
a(n)= A093562(n+1, 2), (5, 1)-Pascal column.
Cf. A006564, A147875, A244639.
Cf. sequences listed in A254963.

a000566 n = n * (5 * (n - 1) + 2) `div` 2
a000566_list = scanl (+) 0 a016861_list  -- Reinhard Zumkeller, Jun 16 2013

a000566:=func< n | n*(5*n-3) div 2 >; [ a000566(n): n in [0..50] ];

A000566 := proc(n)
        n*(5*n-3)/2 ;
end proc:
seq(A000566(n),n=0..30); # R. J. Mathar, Oct 02 2020

Table[n (5n - 3)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 7}, 50] (* Harvey P. Dale, Oct 13 2011 *)
Join[{0},Accumulate[Range[1,315,5]]] (* Harvey P. Dale, Mar 26 2016 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[7], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
PolygonalNumber[7,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 23 2021 *)

makelist(n*(5*n-3)/2, n, 0, 20); /* Martin Ettl, Dec 11 2012 */

PARI
```
a(n) = n * (5*n - 3) / 2
```

# Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 5, y + 5
A000566 = aList()
print([next(A000566) for i in range(49)]) # Peter Luschny, Aug 04 2019

[n*(5*n-3)//2 for n in range(50)] # Gennady Eremin, Mar 24 2022

G.f.: x*(1 + 4*x)/(1 - x)^3. Simon Plouffe in his 1992 dissertation.
a(n) = C(n, 1) + 5*C(n, 2). - Paul Barry, Jun 10 2003
a(n) = Sum_{k = 1..n} (4*n - 3*k). - Paul Barry, Sep 06 2005
a(n) = n + 5*A000217(n-1) - Floor van Lamoen, Oct 14 2005
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for a(0) = 0, a(1) = 1, a(2) = 7. -  Jaume Oliver Lafont, Dec 02 2008
a(n+1) = A153126(n) + n mod 2; a(2*n+1) = A033571(n); a(2*(n+1)) = A153127(n) + 1. - Reinhard Zumkeller, Dec 20 2008
40*a(n)+ 9 = A017354(n-1). - Ken Rosenbaum, Dec 02 2009.
a(n) = 2*a(n-1) - a(n-2) + 5, with a(0) = 0 and a(1) = 1. - Mohamed Bouhamida, May 05 2010
a(n) = A000217(n) + 4*A000217(n-1). - Vincenzo Librandi, Nov 20 2010
a(n) = a(n-1) + 5*n - 4, with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = A130520(5*n). - Philippe Deléham, Mar 26 2013
a(5*a(n) + 11*n + 1) = a(5*a(n) + 11*n) + a(5*n + 1). - Vladimir Shevelev, Jan 24 2014
Sum_{n>=1} 1/a(n) = sqrt(1 - 2/sqrt(5))*Pi/3 + 5*log(5)/6 - sqrt(5)*log((1 + sqrt(5))/2)/3 = 1.32277925312238885674944226131... . See A244639. - Vaclav Kotesovec, Apr 27 2016
E.g.f.: x*(2 + 5*x)*exp(x)/2. - Ilya Gutkovskiy, Aug 27 2016
From Charlie Marion, May 02 2017: (Start)
a(n+m) = a(n) + 5*n*m + a(m);
a(n-m) = a(n) - 5*n*m + a(m) + 3*m;
a(n) - a(m) = (5*(n + m) - 3)*(n - m)/2.
In general, let P(k,n) be the n-th k-gonal number. Then
P(k,n+m) = P(k,n) + (k - 2)*n*m + P(k,m);
P(k,n-m) = P(k,n) - (k - 2)*n*m + P(k,m) + (k - 4)*m;
P(k,n) - P(k,m) = ((k-2)*(n + m) + 4 - k)*(n - m)/2.
(End)
a(n) = A147875(-n) for all n in Z. - Michael Somos, Jan 25 2019
a(n) = A000217(n-1) + A000217(2*n-1). - Charlie Marion, Dec 19 2019
Product_{n>=2} (1 - 1/a(n)) = 5/7. - Amiram Eldar, Jan 21 2021
a(n) + a(n+1) = (2*n+1)^2 + n^2 - 2*n. In general, if we let P(k,n) = the n-th k-gonal number, then P(k^2-k+1,n)+ P(k^2-k+1,n+1) = ((k-1)*n+1)^2 + (k-2)*(n^2-2*n) = ((k-1)*n+1)^2 + (k-2)*A005563(n-2). When k = 2, this formula reduces to the well-known triangular number formula: T(n) + T(n+1) = (n+1)^2. - Charlie Marion, Jul 01 2021

Partially edited by Joerg Arndt, Mar 11 2010

A014105 Second hexagonal numbers: a(n) = n(2n + 1).

Original entry on oeis.org

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278

Offset: 0

N. J. A. Sloane, Jun 14 1998

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013

Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Matthew Cho, Anton Dochtermann, Ryota Inagaki, Suho Oh, Dylan Snustad, and Bailee Zacovic, Chip-firing and critical groups of signed graphs, arXiv:2306.09315 [math.CO], 2023. See p. 22.
Robert FERREOL, Illustration: triangular numbers of even order
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Milan Janjic, Two Enumerative Functions, University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Kival Ngaokrajang, Illustration of half circle spiral.
Markus Scheuer, show that; strange sum yields triangular numbers, Mathematics StackExchange.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019).
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
Leo Tavares, Illustration: Squared Hexagons.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A000384, A002378, A002943, A100040, A100041, A081266, A144312.
Cf. A033567, A033585, A059255, A130757, A132440, A027907.
Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.
Cf. A016061, A003215, A000290, A188859.

List([0..50],n->n*(2*n+1)); # Muniru A Asiru, Oct 31 2018

a014105 n = n * (2 * n + 1)
a014105_list = scanl (+) 0 a004767_list  -- Reinhard Zumkeller, Oct 03 2012

[ n*(2*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014

seq(binomial(2*n+1,2), n=0..46); # Zerinvary Lajos, Jan 21 2007

Table[n*(2*n+1), {n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Nov 16 2008 *)
LinearRecurrence[{3,-3,1},{0,3,10},50] (* Harvey P. Dale, Feb 10 2015 *)
CoefficientList[Series[x*(3 + x)/(1 - x)^3,{x, 0, 50}], x] (* Stefano Spezia, Sep 02 2018 *)

PARI
```
a(n)=n*(2*n+1)
```

[n*(2*n+1) for n in range(50)] # G. C. Greubel, Dec 16 2018

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A085787 Generalized heptagonal numbers: m(5m - 3)/2, m = 0, +-1, +-2 +-3, ...

Original entry on oeis.org

0, 1, 4, 7, 13, 18, 27, 34, 46, 55, 70, 81, 99, 112, 133, 148, 172, 189, 216, 235, 265, 286, 319, 342, 378, 403, 442, 469, 511, 540, 585, 616, 664, 697, 748, 783, 837, 874, 931, 970, 1030, 1071, 1134, 1177, 1243, 1288, 1357, 1404, 1476, 1525, 1600, 1651, 1729

Offset: 0

Jon Perry, Jul 23 2003

Zero together with the partial sums of A080512. - Omar E. Pol, Sep 10 2011
Second heptagonal numbers (A147875) and positive terms of A000566 interleaved. - Omar E. Pol, Aug 04 2012
These numbers appear in a theta function identity. See the Hardy-Wright reference, Theorem 355 on p. 284. See the g.f. of A113429. - Wolfdieter Lang, Oct 28 2016
Characteristic function is A133100. - Michael Somos, Jan 30 2017
40*a(n) + 9 is a square. - Bruno Berselli, Apr 18 2018
Numbers k such that the concatenation k225 is a square. - Bruno Berselli, Nov 07 2018
The sequence terms occur as exponents in the expansion of Sum_{n >= 0} q^(n*(n+1)) * Product_{k >= n+1} 1 - q^k = 1 - q - q^4 + q^7 + q^13 - q^18 - q^27 + + - -  ... (see Hardy and Wright, Theorem 363, p. 290). - Peter Bala, Dec 15 2024

			From the first formula: a(5) = A000217(5) + A000217(2) = 15 + 3 = 18.

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

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves, and Millie Jeske, Cyclic permutations avoiding patterns in both one-line and cycle forms, arXiv:2312.05145 [math.CO], 2023. See p. 2.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Column 3 of A195152.
Sequences of generalized k-gonal numbers: A001318 (k=5), A000217 (k=6), this sequence (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).
Cf. A001622, A080512, A113429, A133100.

a085787 n = a085787_list !! n
a085787_list = scanl (+) 0 a080512_list
-- Reinhard Zumkeller, Apr 06 2015

[5*n*(n+1)/8-1/16+(-1)^n*(2*n+1)/16: n in [0..60]]; // Vincenzo Librandi, Sep 11 2011

Select[Table[(n*(n+1)/2-1)/5,{n,500}],IntegerQ] (* Vladimir Joseph Stephan Orlovsky, Feb 06 2012 *)

t(n)=n*(n+1)/2
for(i=0,40,print1(t(i)+t(floor(i/2)), ", "))

{a(n) = (5*(-n\2)^2 - (-n\2)*3*(-1)^n) / 2}; /* Michael Somos, Oct 17 2006 */

a(n) = A000217(n) + A000217(floor(n/2)).
a(2*n-1) = A000566(n).
a(2*n) = A147875(n). - Bruno Berselli, Apr 18 2018
G.f.: x * (1 + 3*x + x^2) / ((1 - x) * (1 - x^2)^2). a(n) = a(-1-n) for all n in Z. - Michael Somos, Oct 17 2006
a(n) = 5*n*(n + 1)/8 - 1/16 + (-1)^n*(2*n + 1)/16. - R. J. Mathar, Jun 29 2009
a(n) = (A000217(n) + A001082(n))/2 = (A001318(n) + A118277(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = A002378(n) - A001318(n). - Omar E. Pol, Oct 23 2013
Sum_{n>=1} 1/a(n) = 10/9 + (2*sqrt(1 - 2/sqrt(5))*Pi)/3. - Vaclav Kotesovec, Oct 05 2016
E.g.f.: (x*(9 + 5*x)*exp(x) - (1 - 2*x)*sinh(x))/8. - Franck Maminirina Ramaharo, Nov 07 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(5)/3 - 10/9 - 2*sqrt(5)*log(phi)/3, where phi is the golden ratio (A001622). - Amiram Eldar, Feb 28 2022

New name from T. D. Noe, Apr 21 2006

Formula in sequence name added by Omar E. Pol, May 28 2012

A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

Original entry on oeis.org

0, 5, 16, 33, 56, 85, 120, 161, 208, 261, 320, 385, 456, 533, 616, 705, 800, 901, 1008, 1121, 1240, 1365, 1496, 1633, 1776, 1925, 2080, 2241, 2408, 2581, 2760, 2945, 3136, 3333, 3536, 3745, 3960, 4181, 4408, 4641, 4880, 5125, 5376, 5633, 5896, 6165, 6440

Offset: 0

R. K. Guy

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the n-th term of the sequence found by reading the line from 0 in the direction 0,5,.... The spiral begins:
                 .
                 85--84--83--82--81--80
                   .                   \
                   56--55--54--53--52  79
                   / .               \   \
                 57  33--32--31--30  51  78
                 /   / .           \   \   \
               58  34  16--15--14  29  50  77
               /   /   / .       \   \   \   \
             59  35  17   5---4  13  28  49  76
             /   /   /   / .   \   \   \   \   \
           60  36  18   6   0   3  12  27  48  75
           /   /   /   /   /   /   /   /   /   /
         61  37  19   7   1---2  11  26  47  74
           \   \   \   \         /   /   /   /
           62  38  20   8---9--10  25  46  73
             \   \   \             /   /   /
             63  39  21--22--23--24  45  72
               \   \                 /   /
               64  40--41--42--43--44  71
                 \                     /
                 65--66--67--68--69--70
(End)
Connection to triangular numbers: a(n) = 4*T_n + S_n where T_n is the n-th triangular number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010
Also, second octagonal numbers. - Bruno Berselli, Jan 13 2011
Sequence found by reading the line from 0, in the direction 0, 16, ... and the line from 5, in the direction 5, 33, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012
Let P denote the points from the n X n grid. A(n-1) also coincides with the minimum number of points Q needed to "block" P, that is, every line segment spanned by two points from P must contain one point from Q. - Manfred Scheucher, Aug 30 2018
Also the number of internal edges of an (n+1)*(n+1) "square" of hexagons; i.e., n+1 rows, each of n+1 edge-adjacent hexagons, stacked with minimal overhang. - Jon Hart, Sep 29 2019
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n+2; {1, 2n-1, 1, 1, 1, 2n-1, 1, 18n+4}]. - Magus K. Chu, Oct 13 2022

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
M. Janjic and B. Petkovic, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Leo Tavares, Illustration: Square Stars
Leo Tavares, Illustration: Split Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A001859. See Comments of A135713.
Cf. A000217, A000567, A001082, A002378, A016969, A049450, A174709.
Cf. second n-gonal numbers: A005449, A014105, A147875, A179986, A033954, A062728, A135705.
Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734, A139273.
Cf. A000290, A022144, A005563, A001105.
Cf. A056109.
Cf. A003154.

[n*(3*n+2) : n in [0..100]]; // Wesley Ivan Hurt, Sep 24 2017

Table[n*(3n+2), {n,0,60}] (* Harvey P. Dale, May 05 2011 *)
LinearRecurrence[{3,-3,1},{0,5,16},60] (* Harvey P. Dale, Jan 19 2016 *)
CoefficientList[Series[x*(5 + x)/(1 - x)^3,{x, 0, 60}], x] (* Stefano Spezia, Sep 01 2018 *)

a(n)=n*(3*n+2) \\ Charles R Greathouse IV, Nov 20 2012

O.g.f.: x*(5+x)/(1-x)^3. - R. J. Mathar, Jan 07 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=0, a(1)=5, a(2)=16. - Harvey P. Dale, May 06 2011
a(n) = a(n-1) + 6*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 18 2010
For n > 0, a(n)^3 + (a(n)+1)^3 + ... + (a(n)+n)^3 + 2*A000217(n)^2 = (a(n) + n + 1)^3 + ... + (a(n) + 2n)^3; see also A033954. - Charlie Marion, Dec 08 2007
a(n) = Sum_{i=0..n-1} A016969(i) for n > 0. - Bruno Berselli, Jan 13 2011
a(n) = A174709(6*n+4). - Philippe Deléham, Mar 26 2013
a(n) = A001082(2*n). - Michael Turniansky, Aug 24 2013
Sum_{n>=1} 1/a(n) = (9 + sqrt(3)*Pi - 9*log(3))/12 = 0.3794906245574721941... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A002378(n) + A014105(n). - J. M. Bergot, Apr 24 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(12) - 3/4. - Amiram Eldar, Jul 03 2020
E.g.f.: exp(x)*x*(5 + 3*x). - Stefano Spezia, Jun 08 2021
From Leo Tavares, Oct 14 2021: (Start)
a(n) = A000290(n) + 4*A000217(n). See Square Stars illustration.
a(n) = A000567(n+2) - A022144(n+1)
a(n) = A005563(n) + A001105(n).
a(n) = A056109(n) - 1. (End)
From Leo Tavares, Oct 06 2022: (Start)
a(n) = A003154(n+1) - A000567(n+1). See Split Stars illustration.
a(n) = A014105(n) + 2*A000217(n). (End)

A033954 Second 10-gonal (or decagonal) numbers: n(4n+3).

Original entry on oeis.org

0, 7, 22, 45, 76, 115, 162, 217, 280, 351, 430, 517, 612, 715, 826, 945, 1072, 1207, 1350, 1501, 1660, 1827, 2002, 2185, 2376, 2575, 2782, 2997, 3220, 3451, 3690, 3937, 4192, 4455, 4726, 5005, 5292, 5587, 5890, 6201, 6520, 6847, 7182, 7525, 7876, 8235

Offset: 0

N. J. A. Sloane

Same as A033951 except start at 0. See example section.

Bisection of A074377. Also sequence found by reading the line from 0, in the direction 0, 22, ... and the line from 7, in the direction 7, 45, ..., in the square spiral whose vertices are the generalized 10-gonal numbers A074377. - Omar E. Pol, Jul 24 2012

			  36--37--38--39--40--41--42
   |                       |
  35  16--17--18--19--20  43
   |   |               |   |
  34  15   4---5---6  21  44
   |   |   |       |   |   |
  33  14   3   0===7==22==45==76=>
   |   |   |   |   |   |
  32  13   2---1   8  23
   |   |           |   |
  31  12--11--10---9  24
   |                   |
  30--29--28--27--26--25

S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Emilio Apricena, A version of the Ulam spiral.
Leo Tavares, Illustration: V numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002620, A033951.
Sequences from spirals: A001107, A002939, A007742, A033951, A033952, A033953, A033954, A033989, A033990, A033991, A002943, A033996, A033988.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, this sequence, A062728, A135705.
Cf. A060544.

List([0..50], n-> n*(4*n+3)) # G. C. Greubel, May 24 2019

[n*(4*n+3): n in [0..50]]; // G. C. Greubel, May 24 2019

Table[n(4n+3),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,22},50] (* Harvey P. Dale, May 06 2018 *)

PARI
```
a(n)=4*n^2+3*n
```

[n*(4*n+3) for n in (0..50)] # G. C. Greubel, May 24 2019

a(n) = A001107(-n) = A074377(2*n).
G.f.: x*(7+x)/(1-x)^3. - Michael Somos, Mar 03 2003
a(n) = a(n-1) + 8*n - 1 with a(0)=0. - Vincenzo Librandi, Jul 20 2010
For n>0, Sum_{j=0..n} (a(n) + j)^4 + (4*A000217(n))^3 = Sum_{j=n+1..2n} (a(n) + j)^4; see also A045944. - Charlie Marion, Dec 08 2007, edited by Michel Marcus, Mar 14 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 7, a(2) = 22. - Philippe Deléham, Mar 26 2013
a(n) = A118729(8n+6). - Philippe Deléham, Mar 26 2013
a(n) = A002943(n) + n = A007742(n) + 2n = A016742(n) + 3n = A033991(n) + 4n = A002939(n) + 5n = A001107(n) + 6n = A033996(n) - n. - Philippe Deléham, Mar 26 2013
Sum_{n>=1} 1/a(n) = 4/9 + Pi/6 - log(2) = 0.2748960394827980081... . - Vaclav Kotesovec, Apr 27 2016
E.g.f.: exp(x)*x*(7 + 4*x). - Stefano Spezia, Jun 08 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(2)) + log(2)/3 - 4/9 - sqrt(2)*arcsinh(1)/3. - Amiram Eldar, Nov 28 2021
For n>0, (a(n)^2 + n)/(a(n) + n) = (4*n + 1)^2/4, a ratio of two squares. - Rick L. Shepherd, Feb 23 2022
a(n) = A060544(n+1) - A000217(n+1). - Leo Tavares, Mar 31 2022

A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

Original entry on oeis.org

0, 5, 15, 30, 50, 75, 105, 140, 180, 225, 275, 330, 390, 455, 525, 600, 680, 765, 855, 950, 1050, 1155, 1265, 1380, 1500, 1625, 1755, 1890, 2030, 2175, 2325, 2480, 2640, 2805, 2975, 3150, 3330, 3515, 3705, 3900, 4100, 4305, 4515, 4730, 4950, 5175, 5405, 5640

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 5, ... and the same line from 0, in the direction 0, 15, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. Axis perpendicular to A195142 in the same spiral. - Omar E. Pol, Sep 18 2011
Bisection of A195014. Sequence found by reading the line from 0, in the direction 0, 5, ..., and the same line from 0, in the direction 0, 15, ..., in the square spiral whose edges have length A195013 and whose vertices are the numbers A195014. This is the main diagonal of the spiral. - Omar E. Pol, Sep 25 2011
a(n) = the Wiener index of the graph obtained by applying Mycielski's construction to the complete graph K(n) (n>=2). - Emeric Deutsch, Aug 29 2013
Sum of the numbers from 2*n to 3*n for n=0,1,2,... - Wesley Ivan Hurt, Nov 27 2015
Numbers k such that the concatenation k625 is a square, where also 625 is a square. - Bruno Berselli, Nov 07 2018
From Paul Curtz, Nov 29 2019: (Start)
Main column of the pentagonal spiral for n (A001477):
                         50
                      49 30 31
                   48 29 15 16 32
                47 28 14  5  6 17 33
             46 27 13  4  0  1  7 18 34
               45 26 12  3  2  8 19 35
                 44 25 11 10  9 20 36
                   43 24 23 22 21 37
                    42 41 40 39 38
(End)

D. B. West, Introduction to Graph Theory, 2nd ed., Prentice-Hall, NJ, 2001, p. 205.

Ivan Panchenko, Table of n, a(n) for n = 0..1000
Rangaswami Balakrishnan and S. Francis Raj, The Wiener number of powers of the Mycielskian, Discussiones Math. Graph Theory, Vol. 30, No. 3 (2010), pp. 489-498 (see Theorem 2.1).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001068, A005891, A028896, A046092, A055998, A085787, A130520, A195013, A195014, A195142.
Cf. index to numbers of the form n*(d*n+10-d)/2 in A140090.
Cf. A000566, A005475, A005476, A033583, A085787, A147875, A192136, A326725 (all in the spiral).

[5*n*(n+1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Jun 09 2014

[seq(5*binomial(n,2), n=1..45)]; # Zerinvary Lajos, Nov 24 2006

Table[Sum[i + 2*n - 1, {i, 2, n}], {n, 45}] (* Zerinvary Lajos, Jul 11 2009 *)
Table[5 n (n + 1)/2, {n, 0, 50}] (* Bruno Berselli, Sep 23 2016 *)
5 Accumulate[Range[0,60]] (* Harvey P. Dale, Jun 17 2025 *)

a(n)=5*n*(n+1)/2 \\ Charles R Greathouse IV, Sep 24 2015

def A028895(n): return 5*n*(n+1)>>1 # Chai Wah Wu, Aug 07 2025

G.f.: 5*x/(1-x)^3.
a(n) = 5*n*(n+1)/2 = 5*A000217(n).
a(n+1) = 5*n+a(n). - Vincenzo Librandi, Aug 05 2010
a(n) = A005891(n) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A130520(5n+4). - Philippe Deléham, Mar 26 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Wesley Ivan Hurt, Nov 27 2015
a(n) = Sum_{i=0..n} A001068(4i). - Wesley Ivan Hurt, May 06 2016
E.g.f.: 5*x*(2 + x)*exp(x)/2. - Ilya Gutkovskiy, May 06 2016
a(n) = A055998(3*n) - A055998(2*n). - Bruno Berselli, Sep 23 2016
From Amiram Eldar, Feb 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/5)*(2*log(2) - 1). (End)
Product_{n>=1} (1 - 1/a(n)) = -(5/(2*Pi))*cos(sqrt(13/5)*Pi/2). - Amiram Eldar, Feb 21 2023

A130520 a(n) = Sum_{k=0..n} floor(k/5). (Partial sums of A002266.)

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 7, 9, 11, 13, 15, 18, 21, 24, 27, 30, 34, 38, 42, 46, 50, 55, 60, 65, 70, 75, 81, 87, 93, 99, 105, 112, 119, 126, 133, 140, 148, 156, 164, 172, 180, 189, 198, 207, 216, 225, 235, 245, 255, 265, 275, 286, 297, 308, 319, 330, 342, 354, 366

Offset: 0

Hieronymus Fischer, Jun 01 2007

Complementary with A130483 regarding triangular numbers, in that A130483(n) + 5*a(n) = n*(n+1)/2 = A000217(n).

Given a sequence b(n) defined by variables b(0) to b(5) and recursion b(n) = -(b(n-6) * a(n-2) * (b(n-4) * b(n-2)^3 - b(n-3)^3 * b(n-1)) - b(n-5) * b(n-3) * b(n-1) * (b(n-5) * b(n-2)^2 - b(n-4)^2 * b(n-1)))/(b(n-4) * (b(n-5) * b(n-3)^3 - b(n-4)^3 * b(n-2))). The denominator of b(n+1) has a factor of (b(1) * b(3)^3 - b(2)^3 * b(4))^a(n+1). For example, if b(0) = 2, b(1) = b(2) = b(3) = 1, b(4) = 1+x, b(5) = 4, then the denominator of b(n+1) is x^a(n+1). - Michael Somos, Nov 15 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,1,-2,1).

Cf. A002264, A002265, A004526, A010872, A010873, A010874, A130481, A130482.

List([0..70], n-> Int((n-1)*(n-2)/10)); # G. C. Greubel, Aug 31 2019

[Round(n*(n-3)/10): n in [0..70]]; // Vincenzo Librandi, Jun 25 2011

seq(floor((n-1)*(n-2)/10), n=0..70); # G. C. Greubel, Aug 31 2019

Accumulate[Floor[Range[0,70]/5]] (* Harvey P. Dale, May 25 2016 *)

a(n) = sum(k=0, n, k\5); \\ Michel Marcus, May 13 2016

[floor((n-1)*(n-2)/10) for n in (0..70)] # G. C. Greubel, Aug 31 2019

a(n) = floor(n/5)*(2*n - 3 - 5*floor(n/5))/2.
a(n) = A002266(n)*(2*n - 3 - 5*A002266(n))/2.
a(n) = A002266(n)*(n -3 +A010874(n))/2.
G.f.: x^5/((1-x^5)*(1-x)^2) =  x^5/( (1+x+x^2+x^3+x^4)*(1-x)^3 ).
a(n) = floor((n-1)*(n-2)/10). - Mitch Harris, Sep 08 2008
a(n) = round(n*(n-3)/10) = ceiling((n+1)*(n-4)/10) = round((n^2 - 3*n - 1)/10). - Mircea Merca, Nov 28 2010
a(n) = A008732(n-5), n > 4. - R. J. Mathar, Nov 22 2008
a(n) = a(n-5) + n - 4, n > 4. - Mircea Merca, Nov 28 2010
a(5n) = A000566(n), a(5n+1) = A005476(n), a(5n+2) = A005475(n), a(5n+3) = A147875(n), a(5n+4) = A028895(n). - Philippe Deléham, Mar 26 2013
From Amiram Eldar, Sep 17 2022: (Start)
Sum_{n>=5} 1/a(n) = 518/45 - 2*sqrt(2*(sqrt(5)+5))*Pi/3.
Sum_{n>=5} (-1)^(n+1)/a(n) = 8*sqrt(5)*arccoth(3/sqrt(5))/3 + 92*log(2)/15 - 418/45. (End)

A131242 Partial sums of A059995: a(n) = sum_{k=0..n} floor(k/10).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 156, 162, 168, 174, 180, 186, 192, 198

Offset: 0

Hieronymus Fischer, Jun 21 2007

Complementary with A130488 regarding triangular numbers, in that A130488(n)+10*a(n)=n(n+1)/2=A000217(n).

			As square array :
    0,   0,   0,   0,   0,   0,   0,   0,   0,    0
    1,   2,   3,   4,   5,   6,   7,   8,   9,   10
   12,  14,  16,  18,  20,  22,  24,  26,  28,   30
   33,  36,  39,  42,  45,  48,  51,  54,  57,   60
   64,  68,  72,  76,  80,  84,  88,  92,  96,  100
  105, 110, 115, 120, 125, 130, 135, 140, 145,  150
  156, 162, 168, 174, 180, 186, 192, 198, 204,  210
... - _Philippe Deléham_, Mar 27 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,0,1,-2,1).

Cf. A008728, A059995, A010879, A002266, A130488, A000217, A002620, A130518, A130519, A130520, A174709, A174738, A118729, A218470.

Table[(1/2)*Floor[n/10]*(2*n - 8 - 10*Floor[n/10]), {n,0,50}] (* G. C. Greubel, Dec 13 2016 *)
Accumulate[Table[FromDigits[Most[IntegerDigits[n]]],{n,0,110}]] (* or *) LinearRecurrence[{2,-1,0,0,0,0,0,0,0,1,-2,1},{0,0,0,0,0,0,0,0,0,0,1,2},120] (* Harvey P. Dale, Apr 06 2017 *)

for(n=0,50, print1((1/2)*floor(n/10)*(2n-8-10*floor(n/10)), ", ")) \\ G. C. Greubel, Dec 13 2016

a(n)=my(k=n\10); k*(n-5*k-4) \\ Charles R Greathouse IV, Dec 13 2016

a(n) = (1/2)*floor(n/10)*(2n-8-10*floor(n/10)).
a(n) = A059995(n)*(2n-8-10*A059995(n))/2.
a(n) = (1/2)*A059995(n)*(n-8+A010879(n)).
a(n) = (n-A010879(n))*(n+A010879(n)-8)/20.
G.f.: x^10/((1-x^10)(1-x)^2).
From Philippe Deléham, Mar 27 2013: (Start)
a(10n)   = A051624(n).
a(10n+1) = A135706(n).
a(10n+2) = A147874(n+1).
a(10n+3) = 2*A005476(n).
a(10n+4) = A033429(n).
a(10n+5) = A202803(n).
a(10n+6) = A168668(n).
a(10n+7) = 2*A147875(n).
a(10n+8) = A135705(n).
a(10n+9) = A124080(n). (End)
a(n) = A008728(n-10) for n>= 10. - Georg Fischer, Nov 03 2018

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

Original entry on oeis.org

0, 6, 19, 39, 66, 100, 141, 189, 244, 306, 375, 451, 534, 624, 721, 825, 936, 1054, 1179, 1311, 1450, 1596, 1749, 1909, 2076, 2250, 2431, 2619, 2814, 3016, 3225, 3441, 3664, 3894, 4131, 4375, 4626, 4884, 5149, 5421, 5700, 5986, 6279, 6579, 6886

Offset: 0

Bruno Berselli, Jan 13 2011

This sequence is a bisection of A118277 (even part).
Sequence found by reading the line from 0, in the direction 0, 19... and the line from 6, in the direction 6, 39,..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 24 2012
The early part of this sequence is a strikingly close approximation to the early part of A100752. - Peter Munn, Nov 14 2019

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

Cf. second k-gonal numbers: A005449 (k=5), A014105 (k=6), A147875 (k=7), A045944 (k=8), this sequence (k=9), A033954 (k=10), A062728 (k=11), A135705 (k=12).

Cf. A001106, A022264, A022265, A024966, A055998, A100752, A174738, A186029, A218471.

[n*(7*n+5)/2: n in [0..50]]; // Bruno Berselli, Sep 23 2016

I:=[0, 6, 19]; [n le 3 select I[n] else 3*Self(n-1) -3*Self(n-2) +Self(n-3): n in [1..60]]; // Vincenzo Librandi, Oct 15 2012

f[n_] := n (7 n + 5)/2; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 6, 19}, 60] (* or *) Array[(#(7# + 5))/2&, 60, 0] (* Harvey P. Dale, Aug 19 2011 *)
CoefficientList[Series[x (6 + x)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2012 *)

a(n)=n*(7*n+5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(6 + x)/(1 - x)^3.
a(n) = Sum_{i=0..(n-1)} A017053(i) for n>0.
a(-n) = A001106(n).
Sum_{i=0..n} (a(n)+i)^2 = ( Sum_{i=(n+1)..2*n} (a(n)+i)^2 ) + 21*A000217(n)^2 for n>0.
a(n) = a(n-1)+7*n-1 for n>0, with a(0)=0. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=6, a(2)=19; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2011
a(n) = A174738(7n+5). - Philippe Deléham, Mar 26 2013
a(n) = A001477(n) + 2*A000290(n) + 3*A000217(n). - J. M. Bergot, Apr 25 2014
a(n) = A055998(4*n) - A055998(3*n). - Bruno Berselli, Sep 23 2016
E.g.f.: (x/2)*(12 + 7*x)*exp(x). - G. C. Greubel, Aug 19 2017

cp's OEIS Frontend

A183298 Complement of A147875.

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Python

Formula

A000566 Heptagonal numbers (or 7-gonal numbers): n*(5*n-3)/2.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Maxima

PARI

Python

Python

Formula

Extensions

A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

A085787 Generalized heptagonal numbers: m*(5*m - 3)/2, m = 0, +-1, +-2 +-3, ...

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

PARI

Formula

Extensions

A045944 Rhombic matchstick numbers: a(n) = n*(3*n+2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A033954 Second 10-gonal (or decagonal) numbers: n*(4*n+3).

Views

A000566 Heptagonal numbers (or 7-gonal numbers): n(5n-3)/2.

A014105 Second hexagonal numbers: a(n) = n(2n + 1).

A085787 Generalized heptagonal numbers: m(5m - 3)/2, m = 0, +-1, +-2 +-3, ...

A045944 Rhombic matchstick numbers: a(n) = n(3n+2).

A033954 Second 10-gonal (or decagonal) numbers: n(4n+3).

A028895 5 times triangular numbers: a(n) = 5n(n+1)/2.

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.