cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A005408 The odd numbers: a(n) = 2*n + 1.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Leibniz's series: Pi/4 = Sum_{n>=0} (-1)^n/(2n+1) (cf. A072172).
Beginning of the ordering of the natural numbers used in Sharkovski's theorem - see the Cielsielski-Pogoda paper.
The Sharkovski ordering begins with the odd numbers >= 3, then twice these numbers, then 4 times them, then 8 times them, etc., ending with the powers of 2 in decreasing order, ending with 2^0 = 1.
Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0(6).
Also continued fraction for coth(1) (A073747 is decimal expansion). - Rick L. Shepherd, Aug 07 2002
a(1) = 1; a(n) is the smallest number such that a(n) + a(i) is composite for all i = 1 to n-1. - Amarnath Murthy, Jul 14 2003
Smallest number greater than n, not a multiple of n, but containing it in binary representation. - Reinhard Zumkeller, Oct 06 2003
Numbers n such that phi(2n) = phi(n), where phi is Euler's totient (A000010). - Lekraj Beedassy, Aug 27 2004
Pi*sqrt(2)/4 = Sum_{n>=0} (-1)^floor(n/2)/(2n+1) = 1 + 1/3 - 1/5 - 1/7 + 1/9 + 1/11 ... [since periodic f(x)=x over -Pi < x < Pi = 2(sin(x)/1 - sin(2x)/2 + sin(3x)/3 - ...) using x = Pi/4 (Maor)]. - Gerald McGarvey, Feb 04 2005
For n > 1, numbers having 2 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005
a(n) = shortest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1.
First differences of squares (A000290). - Lekraj Beedassy, Jul 15 2006
The odd numbers are the solution to the simplest recursion arising when assuming that the algorithm "merge sort" could merge in constant unit time, i.e., T(1):= 1, T(n):= T(floor(n/2)) + T(ceiling(n/2)) + 1. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 14 2006
2n-5 counts the permutations in S_n which have zero occurrences of the pattern 312 and one occurrence of the pattern 123. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A001248(k)^(n-1)); a(n) = A000005(A000302(n-1)) = A000005(A001019(n-1)) = A000005(A009969(n-1)) = A000005(A087752(n-1)). - Reinhard Zumkeller, Mar 04 2007
For n > 2, a(n-1) is the least integer not the sum of < n n-gonal numbers (0 allowed). - Jonathan Sondow, Jul 01 2007
A134451(a(n)) = abs(A134452(a(n))) = 1; union of A134453 and A134454. - Reinhard Zumkeller, Oct 27 2007
Numbers n such that sigma(2n) = 3*sigma(n). - Farideh Firoozbakht, Feb 26 2008
a(n) = A139391(A016825(n)) = A006370(A016825(n)). - Reinhard Zumkeller, Apr 17 2008
Number of divisors of 4^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Equals INVERT transform of A078050 (signed - cf. comments); and row sums of triangle A144106. - Gary W. Adamson, Sep 11 2008
Odd numbers(n) = 2*n+1 = square pyramidal number(3*n+1) / triangular number(3*n+1). - Pierre CAMI, Sep 27 2008
A000035(a(n))=1, A059841(a(n))=0. - Reinhard Zumkeller, Sep 29 2008
Multiplicative closure of A065091. - Reinhard Zumkeller, Oct 14 2008
a(n) is also the maximum number of triangles that n+2 points in the same plane can determine. 3 points determine max 1 triangle; 4 points can give 3 triangles; 5 points can give 5; 6 points can give 7 etc. - Carmine Suriano, Jun 08 2009
Binomial transform of A130706, inverse binomial transform of A001787(without the initial 0). - Philippe Deléham, Sep 17 2009
Also the 3-rough numbers: positive integers that have no prime factors less than 3. - Michael B. Porter, Oct 08 2009
Or n without 2 as prime factor. - Juri-Stepan Gerasimov, Nov 19 2009
Given an L(2,1) labeling l of a graph G, let k be the maximum label assigned by l. The minimum k possible over all L(2,1) labelings of G is denoted by lambda(G). For n > 0, this sequence gives lambda(K_{n+1}) where K_{n+1} is the complete graph on n+1 vertices. - K.V.Iyer, Dec 19 2009
A176271 = odd numbers seen as a triangle read by rows: a(n) = A176271(A002024(n+1), A002260(n+1)). - Reinhard Zumkeller, Apr 13 2010
For n >= 1, a(n-1) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n-1)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010
Union of A179084 and A179085. - Reinhard Zumkeller, Jun 28 2010
For n>0, continued fraction [1,1,n] = (n+1)/a(n); e.g., [1,1,7] = 8/15. - Gary W. Adamson, Jul 15 2010
Numbers that are the sum of two sequential integers. - Dominick Cancilla, Aug 09 2010
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h and n in A000027), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 4). Also a(n)^2 - 1 == 0 (mod 8). - Bruno Berselli, Nov 17 2010
A004767 = a(a(n)). - Reinhard Zumkeller, Jun 27 2011
A001227(a(n)) = A000005(a(n)); A048272(a(n)) < 0. - Reinhard Zumkeller, Jan 21 2012
a(n) is the minimum number of tosses of a fair coin needed so that the probability of more than n heads is at least 1/2. In fact, Sum_{k=n+1..2n+1} Pr(k heads|2n+1 tosses) = 1/2. - Dennis P. Walsh, Apr 04 2012
A007814(a(n)) = 0; A037227(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2012
1/N (i.e., 1/1, 1/2, 1/3, ...) = Sum_{j=1,3,5,...,infinity} k^j, where k is the infinite set of constants 1/exp.ArcSinh(N/2) = convergents to barover(N). The convergent to barover(1) or [1,1,1,...] = 1/phi = 0.6180339..., whereas c.f. barover(2) converges to 0.414213..., and so on. Thus, with k = 1/phi we obtain 1 = k^1 + k^3 + k^5 + ..., and with k = 0.414213... = (sqrt(2) - 1) we get 1/2 = k^1 + k^3 + k^5 + .... Likewise, with the convergent to barover(3) = 0.302775... = k, we get 1/3 = k^1 + k^3 + k^5 + ..., etc. - Gary W. Adamson, Jul 01 2012
Conjecture on primes with one coach (A216371) relating to the odd integers: iff an integer is in A216371 (primes with one coach either of the form 4q-1 or 4q+1, (q > 0)); the top row of its coach is composed of a permutation of the first q odd integers. Example: prime 19 (q = 5), has 5 terms in each row of its coach: 19: [1, 9, 5, 7, 3] ... [1, 1, 1, 2, 4]. This is interpreted: (19 - 1) = (2^1 * 9), (19 - 9) = (2^1 * 5), (19 - 5) = (2^1 - 7), (19 - 7) = (2^2 * 3), (19 - 3) = (2^4 * 1). - Gary W. Adamson, Sep 09 2012
A005408 is the numerator 2n-1 of the term (1/m^2 - 1/n^2) = (2n-1)/(mn)^2, n = m+1, m > 0 in the Rydberg formula, while A035287 is the denominator (mn)^2. So the quotient a(A005408)/a(A035287) simulates the Hydrogen spectral series of all hydrogen-like elements. - Freimut Marschner, Aug 10 2013
This sequence has unique factorization. The primitive elements are the odd primes (A065091). (Each term of the sequence can be expressed as a product of terms of the sequence. Primitive elements have only the trivial factorization. If the products of terms of the sequence are always in the sequence, and there is a unique factorization of each element into primitive elements, we say that the sequence has unique factorization. So, e.g., the composite numbers do not have unique factorization, because for example 36 = 4*9 = 6*6 has two distinct factorizations.) - Franklin T. Adams-Watters, Sep 28 2013
These are also numbers k such that (k^k+1)/(k+1) is an integer. - Derek Orr, May 22 2014
a(n-1) gives the number of distinct sums in the direct sum {1,2,3,..,n} + {1,2,3,..,n}. For example, {1} + {1} has only one possible sum so a(0) = 1. {1,2} + {1,2} has three distinct possible sums {2,3,4} so a(1) = 3. {1,2,3} + {1,2,3} has 5 distinct possible sums {2,3,4,5,6} so a(2) = 5. - Derek Orr, Nov 22 2014
The number of partitions of 4*n into at most 2 parts. - Colin Barker, Mar 31 2015
a(n) is representable as a sum of two but no fewer consecutive nonnegative integers, e.g., 1 = 0 + 1, 3 = 1 + 2, 5 = 2 + 3, etc. (see A138591). - Martin Renner, Mar 14 2016
Unique solution a( ) of the complementary equation a(n) = a(n-1)^2 - a(n-2)*b(n-1), where a(0) = 1, a(1) = 3, and a( ) and b( ) are increasing complementary sequences. - Clark Kimberling, Nov 21 2017
Also the number of maximal and maximum cliques in the n-centipede graph. - Eric W. Weisstein, Dec 01 2017
Lexicographically earliest sequence of distinct positive integers such that the average of any number of consecutive terms is always an integer. (For opposite property see A042963.) - Ivan Neretin, Dec 21 2017
Maximum number of non-intersecting line segments between vertices of a convex (n+2)-gon. - Christoph B. Kassir, Oct 21 2022
a(n) is the number of parking functions of size n+1 avoiding the patterns 123, 132, and 231. - Lara Pudwell, Apr 10 2023

Examples

			G.f. = q + 3*q^3 + 5*q^5 + 7*q^7 + 9*q^9 + 11*q^11 + 13*q^13 + 15*q^15 + ...

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 28.
  • T. Dantzig, The Language of Science, 4th Edition (1954) page 276.
  • H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology, p. 264.
  • D. Hök, Parvisa mönster i permutationer [Swedish], (2007).
  • E. Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 203-205.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000027, A005843, A065091.
See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A128200, A000290, A078050, A144106, A109613, A167875.
Cf. A001651 (n=1 or 2 mod 3), A047209 (n=1 or 4 mod 5).
Cf. A003558, A216371, A179480 (relating to the Coach theorem).
Cf. A000754 (boustrophedon transform).

Programs

Formula

a(n) = 2*n + 1. a(-1 - n) = -a(n). a(n+1) = a(n) + 2.
G.f.: (1 + x) / (1 - x)^2.
E.g.f.: (1 + 2*x) * exp(x).
G.f. with interpolated zeros: (x^3+x)/((1-x)^2 * (1+x)^2); e.g.f. with interpolated zeros: x*(exp(x)+exp(-x))/2. - Geoffrey Critzer, Aug 25 2012
a(n) = L(n,-2)*(-1)^n, where L is defined as in A108299. - Reinhard Zumkeller, Jun 01 2005
Euler transform of length 2 sequence [3, -1]. - Michael Somos, Mar 30 2007
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = v * (1 + 2*u) * (1 - 2*u + 16*v) - (u - 4*v)^2 * (1 + 2*u + 2*u^2). - Michael Somos, Mar 30 2007
a(n) = b(2*n + 1) where b(n) = n if n is odd is multiplicative. [This seems to say that A000027 is multiplicative? - R. J. Mathar, Sep 23 2011]
From Hieronymus Fischer, May 25 2007: (Start)
a(n) = (n+1)^2 - n^2.
G.f. g(x) = Sum_{k>=0} x^floor(sqrt(k)) = Sum_{k>=0} x^A000196(k). (End)
a(0) = 1, a(1) = 3, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
a(n) = A000330(A016777(n))/A000217(A016777(n)). - Pierre CAMI, Sep 27 2008
a(n) = A034856(n+1) - A000217(n) = A005843(n) + A000124(n) - A000217(n) = A005843(n) + 1. - Jaroslav Krizek, Sep 05 2009
a(n) = (n - 1) + n (sum of two sequential integers). - Dominick Cancilla, Aug 09 2010
a(n) = 4*A000217(n)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. - Bruno Berselli, Nov 17 2010
n*a(2n+1)^2+1 = (n+1)*a(2n)^2; e.g., 3*15^2+1 = 4*13^2. - Charlie Marion, Dec 31 2010
arctanh(x) = Sum_{n>=0} x^(2n+1)/a(n). - R. J. Mathar, Sep 23 2011
a(n) = det(f(i-j+1))A113311(n);%20for%20n%20%3C%200%20we%20have%20f(n)=0.%20-%20_Mircea%20Merca">{1<=i,j<=n}, where f(n) = A113311(n); for n < 0 we have f(n)=0. - _Mircea Merca, Jun 23 2012
G.f.: Q(0), where Q(k) = 1 + 2*(k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
a(n) = floor(sqrt(2*A000384(n+1))). - Ivan N. Ianakiev, Jun 17 2013
a(n) = 3*A000330(n)/A000217(n), n > 0. - Ivan N. Ianakiev, Jul 12 2013
a(n) = Product_{k=1..2*n} 2*sin(Pi*k/(2*n+1)) = Product_{k=1..n} (2*sin(Pi*k/(2*n+1)))^2, n >= 0 (undefined product = 1). See an Oct 09 2013 formula contribution in A000027 with a reference. - Wolfdieter Lang, Oct 10 2013
Noting that as n -> infinity, sqrt(n^2 + n) -> n + 1/2, let f(n) = n + 1/2 - sqrt(n^2 + n). Then for n > 0, a(n) = round(1/f(n))/4. - Richard R. Forberg, Feb 16 2014
a(n) = Sum_{k=0..n+1} binomial(2*n+1,2*k)*4^(k)*bernoulli(2*k). - Vladimir Kruchinin, Feb 24 2015
a(n) = Sum_{k=0..n} binomial(6*n+3, 6*k)*Bernoulli(6*k). - Michel Marcus, Jan 11 2016
a(n) = A000225(n+1) - A005803(n+1). - Miquel Cerda, Nov 25 2016
O.g.f.: Sum_{n >= 1} phi(2*n-1)*x^(n-1)/(1 - x^(2*n-1)), where phi(n) is the Euler totient function A000010. - Peter Bala, Mar 22 2019
Sum_{n>=0} 1/a(n)^2 = Pi^2/8 = A111003. - Bernard Schott, Dec 10 2020
Sum_{n >= 1} (-1)^n/(a(n)*a(n+1)) = Pi/4 - 1/2 = 1/(3 + (1*3)/(4 + (3*5)/(4 + ... + (4*n^2 - 1)/(4 + ... )))). Cf. A016754. - Peter Bala, Mar 28 2024
a(n) = A055112(n)/oblong(n) = A193218(n+1)/Hex number(n). Compare to the Sep 27 2008 comment by Pierre CAMI. - Klaus Purath, Apr 23 2024
a(k*m) = k*a(m) - (k-1). - Ya-Ping Lu, Jun 25 2024
a(n) = A000217(a(n))/n for n > 0. - Stefano Spezia, Feb 15 2025

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010
Peripheral comments deleted by N. J. A. Sloane, May 09 2022

A001477 The nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Although this is a list, and lists normally have offset 1, it seems better to make an exception in this case. - N. J. A. Sloane, Mar 13 2010
The subsequence 0,1,2,3,4 gives the known values of n such that 2^(2^n)+1 is a prime (see A019434, the Fermat primes). - N. J. A. Sloane, Jun 16 2010
Also: The identity map, defined on the set of nonnegative integers. The restriction to the positive integers yields the sequence A000027. - M. F. Hasler, Nov 20 2013
The number of partitions of 2n into exactly 2 parts. - Colin Barker, Mar 22 2015
The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 8960 or 168.- Philippe A.J.G. Chevalier, Dec 29 2015
Partial sums give A000217. - Omar E. Pol, Jul 26 2018
First differences are A000012 (the "all 1's" sequence). - M. F. Hasler, May 30 2020
See A061579 for the transposed infinite square matrix, or triangle with rows reversed. - M. F. Hasler, Nov 09 2021
This is the unique sequence (a(n)) that satisfies the inequality a(n+1) > a(a(n)) for all n in N. This simple and surprising result comes from the 6th problem proposed by Bulgaria during the second day of the 19th IMO (1977) in Belgrade (see link and reference). - Bernard Schott, Jan 25 2023

Examples

			Triangular view:
   0
   1   2
   3   4   5
   6   7   8   9
  10  11  12  13  14
  15  16  17  18  19  20
  21  22  23  24  25  26  27
  28  29  30  31  32  33  34  35
  36  37  38  39  40  41  42  43  44
  45  46  47  48  49  50  51  52  53  54

References

  • Maurice Protat, Des Olympiades à l'Agrégation, suite vérifiant f(n+1) > f(f(n)), Problème 7, pp. 31-32, Ellipses, Paris 1997.

Links

Crossrefs

Cf. A000027 (n>=1).
Cf. A000012 (first differences).
Partial sums of A057427. - Jeremy Gardiner, Sep 08 2002
Cf. A038608 (alternating signs), A001787 (binomial transform).
Cf. A055112.
Cf. Boustrophedon transforms: A231179, A000737.
Cf. A245422.
Number of orbits of Aut(Z^7) as function of the infinity norm A000579, A154286, A102860, A002412, A045943, A115067, A008586, A008585, A005843, A000217.
When written as an array, the rows/columns are A000217, A000124, A152948, A152950, A145018, A167499, A166136, A167487... and A000096, A034856, A055998, A046691, A052905, A055999... (with appropriate offsets); cf. analogous lists for A000027 in A185787.
Cf. A000290.
Cf. A061579 (transposed matrix / reversed triangle).

Programs

Formula

a(n) = n.
a(0) = 0, a(n) = a(n-1) + 1.
G.f.: x/(1-x)^2.
Multiplicative with a(p^e) = p^e. - David W. Wilson, Aug 01 2001
When seen as array: T(k, n) = n + (k+n)*(k+n+1)/2. Main diagonal is 2*n*(n+1) (A046092), antidiagonal sums are n*(n+1)*(n+2)/2 (A027480). - Ralf Stephan, Oct 17 2004
Dirichlet generating function: zeta(s-1). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: x*e^x. - Franklin T. Adams-Watters, Sep 11 2005
a(0)=0, a(1)=1, a(n) = 2*a(n-1) - a(n-2). - Jaume Oliver Lafont, May 07 2008
Alternating partial sums give A001057 = A000217 - 2*(A008794). - Eric Desbiaux, Oct 28 2008
a(n) = 2*A080425(n) + 3*A008611(n-3), n>1. - Eric Desbiaux, Nov 15 2009
a(n) = A007966(n)*A007967(n). - Reinhard Zumkeller, Jun 18 2011
a(n) = Sum_{k>=0} A030308(n,k)*2^k. - Philippe Deléham, Oct 20 2011
a(n) = 2*A028242(n-1) + (-1)^n*A000034(n-1). - R. J. Mathar, Jul 20 2012
a(n+1) = det(C(i+1,j), 1 <= i, j <= n), where C(n,k) are binomial coefficients. - Mircea Merca, Apr 06 2013
a(n-1) = floor(n/e^(1/n)) for n > 0. - Richard R. Forberg, Jun 22 2013
a(n) = A000027(n) for all n>0.
a(n) = floor(cot(1/(n+1))). - Clark Kimberling, Oct 08 2014
a(0)=0, a(n>0) = 2*z(-1)^[( |z|/z + 3 )/2] + ( |z|/z - 1 )/2 for z = A130472(n>0); a 1 to 1 correspondence between integers and naturals. - Adriano Caroli, Mar 29 2015
G.f. as triangle: x*(1 + (x^2 - 5*x + 2)*y + x*(2*x - 1)*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Jul 22 2025

A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Examples

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 813.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • L. 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.
  • M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
  • M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
  • Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
  • 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. 126.

Links

Crossrefs

Cf. A000217, A000292, A000537, A005408, A006003, A006331, A033994, A033999, A046092, A050409, A050446, A050447, A060493, A100157, A132124, A132112, A156921, A157702, A258708, A351105, A351830.
Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

Programs

  • GAP
    List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
  • Haskell
    a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
-- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
  • Magma
    [n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014
  • Magma
    [0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
  • Maple
    A000330 := n -> n*(n+1)*(2*n+1)/6;
a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009
  • Mathematica
    Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)
  • Maxima
    A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
  • PARI
    {a(n) = n * (n+1) * (2*n+1) / 6};
  • PARI
    upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024
  • Python
    a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
  • Sage
    [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019

Formula

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A033996 8 times triangular numbers: a(n) = 4*n*(n+1).

Original entry on oeis.org

0, 8, 24, 48, 80, 120, 168, 224, 288, 360, 440, 528, 624, 728, 840, 960, 1088, 1224, 1368, 1520, 1680, 1848, 2024, 2208, 2400, 2600, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6888, 7224, 7568, 7920, 8280
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Write 0, 1, 2, ... in a clockwise spiral; sequence gives numbers on one of 4 diagonals.
Also, least m > n such that T(m)*T(n) is a square and more precisely that of A055112(n). {T(n) = A000217(n)}. - Lekraj Beedassy, May 14 2004
Also sequence found by reading the line from 0, in the direction 0, 8, ... and the same line from 0, in the direction 0, 24, ..., in the square spiral whose vertices are the generalized decagonal numbers A074377. Axis perpendicular to A195146 in the same spiral. - Omar E. Pol, Sep 18 2011
Number of diagonals with length sqrt(5) in an (n+1) X (n+1) square grid. Every 1 X 2 rectangle has two such diagonals. - Wesley Ivan Hurt, Mar 25 2015
Imagine a board made of squares (like a chessboard), one of whose squares is completely surrounded by square-shaped layers made of adjacent squares. a(n) is the total number of squares in the first to n-th layer. a(1) = 8 because there are 8 neighbors to the unit square; adding them gives a 3 X 3 square. a(2) = 24 = 8 + 16 because we need 16 more squares in the next layer to get a 5 X 5 square: a(n) = (2*n+1)^2 - 1 counting the (2n+1) X (2n+1) square minus the central square. - R. J. Cano, Sep 26 2015
The three platonic solids (the simplex, hypercube, and cross-polytope) with unit side length in n dimensions all have rational volume if and only if n appears in this sequence, after 0. - Brian T Kuhns, Feb 26 2016
The number of active (ON, black) cells in the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 645", based on the 5-celled von Neumann neighborhood. - Robert Price, May 19 2016
The square root of a(n), n>0, has continued fraction [2n; {1,4n}] with whole number part 2n and periodic part {1,4n}. - Ron Knott, May 11 2017
Numbers k such that k+1 is a square and k is a multiple of 4. - Bruno Berselli, Sep 28 2017
a(n) is the number of vertices of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - Emeric Deutsch, May 13 2018
a(n) is the number of vertices in conjoined n X n octagons which are arranged into a square array, a.k.a. truncated square tiling. - Donghwi Park, Dec 20 2020
a(n-2) is the number of ways to place 3 adjacent marks in a diagonal, horizontal, or vertical row on an n X n tic-tac-toe grid. - Matej Veselovac, May 28 2021

Examples

			Spiral with 0, 8, 24, 48, ... along lower right diagonal:
.
  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
   |   |   |   | \ |   |   |
  32  13   2---1   8  23  46
   |   |           | \ |   |
  31  12--11--10---9  24  47
   |                   | \ |
  30--29--28--27--26--25  48
                            \
[Reformatted by _Jon E. Schoenfield_, Dec 25 2016]

References

  • Stuart M. Ellerstein, J. Recreational Math. 29 (3) 188, 1998.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
  • Stephen Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Links

Crossrefs

Cf. A000217, A016754, A002378, A024966, A027468, A028895, A028896, A045943, A046092, A049598, A088538, A124080, A008590 (first differences), A130809 (partial sums).
Sequences from spirals: A001107, A002939, A002943, A007742, A033951, A033952, A033953, A033954, A033988, A033989, A033990, A033991, A033996. - Omar E. Pol, Dec 12 2008
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.

Programs

  • Magma
    [ 4*n*(n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014
  • Maple
    seq(8*binomial(n+1, 2), n=0..46); # Zerinvary Lajos, Nov 24 2006
[seq((2*n+1)^2-1, n=0..46)];
  • Mathematica
    Table[(2n - 1)^2 - 1, {n, 50}] (* Alonso del Arte, Mar 31 2013 *)
  • PARI
    nsqm1(n) = { forstep(x=1,n,2, y = x*x-1; print1(y, ", ") ) }

Formula

a(n) = 4*n^2 + 4*n = (2*n+1)^2 - 1.
G.f.: 8*x/(1-x)^3.
a(n) = A016754(n) - 1 = 2*A046092(n) = 4*A002378(n). - Lekraj Beedassy, May 25 2004
a(n) = A049598(n) - A046092(n); a(n) = A124080(n) - A002378(n). - Zerinvary Lajos, Mar 06 2007
a(n) = 8*A000217(n). - Omar E. Pol, Dec 12 2008
a(n) = A005843(n) * A163300(n). - Juri-Stepan Gerasimov, Jul 26 2009
a(n) = a(n-1) + 8*n (with a(0)=0). - Vincenzo Librandi, Nov 17 2010
For n > 0, a(n) = A058031(n+1) - A062938(n-1). - Charlie Marion, Apr 11 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Mar 25 2015
a(n) = A000578(n+1) - A152618(n). - Bui Quang Tuan, Apr 01 2015
a(n) - a(n-1) = A008590(n), n > 0. - Altug Alkan, Sep 26 2015
From Ilya Gutkovskiy, May 19 2016: (Start)
E.g.f.: 4*x*(2 + x)*exp(x).
Sum_{n>=1} 1/a(n) = 1/4. (End)
Product_{n>=1} a(n)/A016754(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A056220(n) + A056220(n+1). - Bruce J. Nicholson, May 29 2017
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^2. - Seiichi Manyama, Dec 23 2018
a(n)*a(n+k) + 4*k^2 = m^2 where m = (a(n) + a(n+k))/2 - 2*k^2; for k=1, m = 4*n^2 + 8*n + 2 = A060626(n). - Ezhilarasu Velayutham, May 22 2019
Sum_{n>=1} (-1)^n/a(n) = 1/4 - log(2)/2. - Vaclav Kotesovec, Dec 21 2020
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(4/Pi)*cos(Pi/sqrt(2)).
Product_{n>=1} (1 + 1/a(n)) = 4/Pi (A088538). (End)

A083374 a(n) = n^2 * (n^2 - 1)/2.

Original entry on oeis.org

0, 6, 36, 120, 300, 630, 1176, 2016, 3240, 4950, 7260, 10296, 14196, 19110, 25200, 32640, 41616, 52326, 64980, 79800, 97020, 116886, 139656, 165600, 195000, 228150, 265356, 306936, 353220, 404550, 461280, 523776, 592416, 667590, 749700, 839160, 936396
Offset: 1

Views

Author

Alan Sutcliffe (alansut(AT)ntlworld.com), Jun 05 2003

Keywords

Comments

Triangular numbers t_n as n runs through the squares.
Partial sums of A055112: If one generated Pythagorean primitive triangles from n, n+1, then the collective areas of n of them would be equal to the numbers in this sequence. The sum of the first three triangles is 6+30+84 = 120 which is the third nonzero term of the sequence. - J. M. Bergot, Jul 14 2011
Second leg of Pythagorean triangles with smallest side a cube: A000578(n)^2 + a(n)^2 = A037270(n)^2. - Martin Renner, Nov 12 2011
a(n) is the number of segments on an n X n grid or geoboard. - Martin Renner, Apr 17 2014
Consider the partitions of 2n into two parts (p,q). Then a(n) is the total volume of the family of rectangular prisms with dimensions p, q and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 55.

Links

Crossrefs

Cf. A000217, A000578, A002415, A006002, A006011, A008911, A037270, A047928, A055112.

Programs

Formula

a(n) = (n + 1) * A006002(n).
a(n) = A047928(n)/2 = A002415(n+1)*6 = A006011(n+1)*2 = A008911(n+1)*3. - Zerinvary Lajos, May 09 2007
a(n) = binomial(n^2,2), n>=1. - Zerinvary Lajos, Jan 07 2008
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>5. - R. J. Mathar, Apr 10 2009
G.f.: -6*x^2*(1+x)/(x-1)^5. - R. J. Mathar, Apr 10 2009
Sum_{n>1} 1/a(n) = (21 - 2*Pi^2)/6. - Enrique Pérez Herrero, Apr 01 2013
a(n) = Sum_{k=0..n-1} k*A000217(2*k+1). - Bruno Berselli, Sep 04 2013
a(n) = 2*A000217(n-1)*A000217(n). - Gionata Neri, May 04 2015
a(n) = Sum_{i=1..n^2-1} i. - Wesley Ivan Hurt, Nov 24 2015
E.g.f.: exp(x)*x^2*(6 + 6*x + x^2)/2. - Stefano Spezia, Jun 06 2021
Sum_{n>=2} (-1)^n/a(n) = Pi^2/6 - 3/2. - Amiram Eldar, Nov 02 2021

Extensions

Corrected and extended by T. D. Noe, Oct 25 2006

A059270 a(n) is both the sum of n+1 consecutive integers and the sum of the n immediately higher consecutive integers.

Original entry on oeis.org

0, 3, 15, 42, 90, 165, 273, 420, 612, 855, 1155, 1518, 1950, 2457, 3045, 3720, 4488, 5355, 6327, 7410, 8610, 9933, 11385, 12972, 14700, 16575, 18603, 20790, 23142, 25665, 28365, 31248, 34320, 37587, 41055, 44730, 48618, 52725, 57057, 61620
Offset: 0

Views

Author

Henry Bottomley, Jan 24 2001

Keywords

Comments

Group the non-multiples of n as follows, e.g., for n = 4: (1,2,3), (5,6,7), (9,10,11), (13,14,15), ... Then a(n) is the sum of the members of the n-th group. Or, the sum of (n-1)successive numbers preceding n^2. - Amarnath Murthy, Jan 19 2004
Convolution of odds (A005408) and multiples of three (A008585). G.f. is the product of the g.f. of A005408 by the g.f. of A008585. - Graeme McRae, Jun 06 2006
Sums of rows of the triangle in A126890. - Reinhard Zumkeller, Dec 30 2006
Corresponds to the Wiener indices of C_{2n+1} i.e., the cycle on 2n+1 vertices (n > 0). - K.V.Iyer, Mar 16 2009
Also the product of the three numbers from A005843(n) up to A163300(n), divided by 8. - Juri-Stepan Gerasimov, Jul 26 2009
Partial sums of A033428. - Charlie Marion, Dec 08 2013
For n > 0, sum of multiples of n and (n+1) from 1 to n*(n+1). - Zak Seidov, Aug 07 2016
A generalization of Ianakiev's formula, a(n) = A005408(n)*A000217(n), follows. A005408(n+k)*A000217(n) is the sum of n+1 consecutive integers and, after skipping k integers, the sum of the n immediately higher consecutive integers. For example, for n = 3 and k = 2, 9*6 = 54 = 12+13+14+15 = 17+18+19. - Charlie Marion, Jan 25 2022

Examples

			a(5) = 25 + 26 + 27 + 28 + 29 + 30 = 31 + 32 + 33 + 34 + 35 = 165.

Links

Crossrefs

Cf. A059255 for analog for sum of squares.
Cf. A222716 for the analogous sum of triangular numbers.
Cf. A234319 for nonexistence of analogs for sums of n-th powers, n > 2. - Jonathan Sondow, Apr 23 2014
Cf. A098737 (first subdiagonal).
Bisection of A109900.

Programs

  • Magma
    I:=[0, 3, 15, 42]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 23 2012
  • Maple
    A059270 := proc(n) n*(n+1)*(2*n+1)/2 ; end proc: # R. J. Mathar, Jul 10 2011
  • Mathematica
    # (#+1)(2#+1)/2 &/@ Range[0,39] (* Ant King, Jan 03 2011 *)
CoefficientList[Series[3 x (1 + x)/(x - 1)^4, {x, 0, 39}], x]
LinearRecurrence[{4,-6,4,-1},{0,3,15,42},50] (* Vincenzo Librandi, Jun 23 2012 *)
  • PARI
    a(n) = n*(n+1)*(2*n+1)/2 \\ Charles R Greathouse IV, Mar 08 2013
  • Sage
    [bernoulli_polynomial(n+1,3) for n in range(0, 41)] # Zerinvary Lajos, May 17 2009

Formula

a(n) = n*(n+1)*(2*n+1)/2.
a(n) = A000330(n)*3 = A006331(n)*3/2 = A055112(n)/2 = A000217(A002378(n)) - A000217(A005563(n-1)) = A000217(A005563(n)) - A000217(A002378(n)).
a(n) = A110449(n+1, n-1) for n > 1.
a(n) = Sum_{k=A000290(n) .. A002378(n)} k = Sum_{k=n^2..n^2+n} k.
a(n) = Sum_{k=n^2+n+1 .. n^2+2*n} k = Sum_{k=A002061(n+1) .. A005563(n)} k.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6 = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Ant King, Jan 03 2011
G.f.: 3*x*(1+x)/(1-x)^4. - Ant King, Jan 03 2011
a(n) = A000578(n+1) - A000326(n+1). - Ivan N. Ianakiev, Nov 29 2012
a(n) = A005408(n)*A000217(n) = a(n-1) + 3*A000290(n). -Ivan N. Ianakiev, Mar 08 2013
a(n) = n^3 + n^2 + A000217(n). - Charlie Marion, Dec 04 2013
From Ilya Gutkovskiy, Aug 08 2016: (Start)
E.g.f.: x*(6 + 9*x + 2*x^2)*exp(x)/2.
Sum_{n>=1} 1/a(n) = 2*(3 - 4*log(2)) = 0.4548225555204375246621... (End)
a(n) = Sum_{k=0..2*n} A001318(k). - Jacob Szlachetka, Dec 20 2021
a(n) = Sum_{k=0..n} A000326(k) + A005449(k). - Jacob Szlachetka, Dec 21 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(Pi-3). - Amiram Eldar, Sep 17 2022

A055522 Largest area of a Pythagorean triangle with n as length of one of the three sides (in fact as a leg).

Original entry on oeis.org

6, 6, 30, 24, 84, 60, 180, 120, 330, 210, 546, 336, 840, 504, 1224, 720, 1710, 990, 2310, 1320, 3036, 1716, 3900, 2184, 4914, 2730, 6090, 3360, 7440, 4080, 8976, 4896, 10710, 5814, 12654, 6840, 14820, 7980, 17220, 9240, 19866, 10626, 22770, 12144, 25944
Offset: 3

Views

Author

Henry Bottomley, May 22 2000

Keywords

Crossrefs

Cf. A009112, A046079, A046080, A046081, A054435, A054436, A055523, A055524, A055525, A055526, A055527.

Programs

  • Maple
    seq(piecewise(n mod 2 = 0,n*(n^2-4)/8,n*(n^2-1)/4),n=3..60); # C. Ronaldo
  • Mathematica
    Table[n*(3*(n^2 - 2) - (n^2 + 2)*(-1)^n)/16, {n, 3, 50}] (* Wesley Ivan Hurt, Apr 27 2017 *)

Formula

a(n) = n*A055523(n)/2.
a(2k) = k*(k+1)*(k-1), a(2k+1) = k*(k+1)*(2k+1).
O.g.f.: 6*x^3*(x+1+x^2)/((1-x)^4*(1+x)^4). a(2k+1)=A055112(k). a(2k)=A007531(k+1). [R. J. Mathar, Aug 06 2008]
a(n) = n*(3*(n^2-2)-(n^2+2)*(-1)^n)/16. - Luce ETIENNE, Jul 17 2015

A106846 a(n) = Sum_{k + l*m <= n} (k + l*m), with 0 <= k,l,m <= n.

Original entry on oeis.org

0, 4, 22, 64, 144, 269, 461, 720, 1072, 1522, 2092, 2774, 3626, 4614, 5776, 7126, 8694, 10445, 12461, 14684, 17204, 19997, 23077, 26412, 30156, 34206, 38600, 43352, 48532, 54042, 60072, 66458, 73338, 80664, 88450, 96710, 105638, 114999
Offset: 0

Views

Author

Ralf Stephan, May 06 2005

Keywords

Links

Crossrefs

Cf. A106633, A106634, A106847.
Cf. A055112, A006218, A143127, A319085, A059270, A143127.

Programs

  • Maple
    A106846 := proc(n)
    local a,k,l,m ;
    a := 0 ;
    for k from 0 to n do
        for l from 0 to n do
            if l = 0 then
                m := n ;
            else
                m := floor((n-k)/l) ;
            end if;
            if m >=0 then
                m := min(m,n) ;
                a := a+(m+1)*k+l*m*(m+1)/2 ;
            end if;
        end do:
    end do:
    a ;
end proc: # R. J. Mathar, Oct 17 2012
  • Mathematica
    A106846[n_] := Module[{a, k, l, m }, a = 0; For[k = 0, k <= n, k++, For[l = 0, l <= n, l++, If[l == 0, m = n, m = Floor[(n - k)/l]]; If[m >= 0, m = Min[m, n]; a = a + (m + 1)*k + l*m*(m + 1)/2 ]]]; a];
Table[A106846[n], {n, 0, 40}] (* Jean-François Alcover, Apr 04 2024, after R. J. Mathar *)

Formula

From Ridouane Oudra, Jun 24 2024: (Start)
a(n) = (1/2) * (n*(n+1)*(2*n+1) + Sum_{k=1..n} (n^2 + n + k - k^2) * tau(k)).
a(n) = (1/2) * (A055112(n) + (n^2 + n) * A006218(n) + A143127(n) - A319085(n)).
a(n) = A059270(n) + A143127(n) + A106847(n). (End)

A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

Original entry on oeis.org

0, 6, 30, 35, 84, 180, 204, 210, 297, 330, 546, 840, 1170, 1189, 1224, 1710, 2310, 2940, 2970, 3036, 3230, 3900, 4914, 6090, 6930, 7134, 7140, 7245, 7440, 8976, 10710, 12654, 14175, 14820, 16296, 16380, 17220, 19866, 22770, 25172, 25944, 29103
Offset: 1

Views

Author

Zak Seidov, May 30 2010

Keywords

Comments

From Robert G. Wilson v, Jul 24 2010: (Start)
Terms in the i-th row are products contributed with a factor A000217(i):
(1) 0, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, ...
(2) 30, 297, 2940, 29103, 288090, 2851797, 28229880, ...
(3) 84, 1170, 16296, 226974, 3161340, ...
(4) 180, 3230, 57960, 1040050, 18662940, ...
(5) 330, 7245, 159060, 3492075, 76666590, ...
(6) 546, 14175, 368004, 9553929, ...
(7) 840, 25172, 754320, 22604428, ...
(8) 210, 1224, 7134, 41580, 242346, 1412496, 8232630, 47983284, ...
(9) 1710, 64935, 2465820, 93636225, ...
(10) 2310, 96965, 4070220, ...
(11) 3036, 139590, 6418104, ...
(12) 3900, 194922, 9742200, ...
(13) 4914, 265265, 14319396, ...
(14) 6090, 353115, 20474580, ...
(15) 7440, 461160, 28584480, ...
(End)
Numbers m with property that m^2 is a product of two distinct triangular numbers T(i) and T(j) such that i and j are in the same row of the square array A(n, k) defined in A322699. - Onur Ozkan, Mar 17 2023

Links

Crossrefs

From Robert G. Wilson v, Jul 24 2010: (Start)
A001109 (with the exception of 1), A011945, A075848 and A055112 are all proper subsets.
Many terms are in common with A147779.
Cf. A001108, A132596, A007654, A001108, A132593. (End)
Cf. A152005 (two distinct tetrahedral numbers).

Programs

  • Maple
    isA175497 := proc(n)
    local i,Ti,Tj;
    if n = 0 then
        return true;
    end if;
    for i from 1 do
        Ti := i*(i+1)/2 ;
        if Ti > n^2 then
            return false;
        else
            Tj := n^2/Ti ;
            if Tj <> Ti and type(Tj,'integer') then
                if isA000217(Tj) then  # code in A000217
                    return true;
                end if;
            end if;
        end if;
    end do:
end proc:
for n from 0 do
    if isA175497(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, May 26 2016
  • Mathematica
    triangularQ[n_] := IntegerQ[Sqrt[8n + 1]];
okQ[n_] := Module[{i, Ti, Tj}, If[n == 0, Return[True]]; For[i = 1, True, i++, Ti = i(i+1)/2; If[Ti > n^2, Return[False], Tj = n^2/Ti; If[Tj != Ti && IntegerQ[Tj], If[ triangularQ[Tj], Return[True]]]]]];
Reap[For[k = 0, k < 30000, k++, If[okQ[k], Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Jun 13 2023, after R. J. Mathar *)
  • Python
    from itertools import count, islice, takewhile
from sympy import divisors
from sympy.ntheory.primetest import is_square
def A175497_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda k:not k or any(map(lambda d: is_square((d<<3)+1) and is_square((k**2//d<<3)+1), takewhile(lambda d:d**2A175497_list = list(islice(A175497_gen(),20)) # Chai Wah Wu, Mar 13 2023
  • Python
    def A175497_list(n):
    def A322699_A(k, n):
        p, q, r, m = 0, k, 4*k*(k+1), 0
        while m < n:
            p, q, r = q, r, (4*k+3)*(r-q) + p
            m += 1
        return p
    def a(k, n, j):
        if n == 0: return 0
        p = A322699_A(k, n)*(A322699_A(k, n)+1)*(2*k+1) - a(k, n-1, 1)
        q = (4*k+2)*p - A322699_A(k, n)*(A322699_A(k, n)+1)//2
        m = 1
        while m < j: p, q = q, (4*k+2)*q - p; m += 1
        return p
    A = set([a(k, 1, 1) for k in range(n+1)])
    k, l, m = 1, 1, 2
    while True:
        x = a(k, l, m)
        if x < max(A):
            A |= {x}
            A  = set(sorted(A)[:n+1])
            m += 1
        else:
            if m == 1 and l == 1:
                if k > n:
                    return sorted(A)
                k += 1
            elif m > 1:
                l += 1; m = 1
            elif l > 1:
                k += 1; l, m = 1, 1
# Onur Ozkan, Mar 15 2023

Formula

a(n)^2 = A169836(n). - R. J. Mathar, Mar 12 2023

A316224 a(n) = n*(2*n + 1)*(4*n + 1).

Original entry on oeis.org

0, 15, 90, 273, 612, 1155, 1950, 3045, 4488, 6327, 8610, 11385, 14700, 18603, 23142, 28365, 34320, 41055, 48618, 57057, 66420, 76755, 88110, 100533, 114072, 128775, 144690, 161865, 180348, 200187, 221430, 244125, 268320, 294063, 321402, 350385, 381060, 413475, 447678, 483717
Offset: 0

Views

Author

Bruno Berselli, Jun 27 2018

Keywords

Comments

Sums of the consecutive integers from A000384(n) to A000384(n+1)-1. This is the case s=6 of the formula n*(n*(s-2) + 1)*(n*(s-2) + 2)/2 related to s-gonal numbers.
The inverse binomial transform is 0, 15, 60, 48, 0, ... (0 continued).

Examples

			Row sums of the triangle:
|  0 |  ................................................................. 0
|  1 |  2  3  4  5  .................................................... 15
|  6 |  7  8  9 10 11 12 13 14  ........................................ 90
| 15 | 16 17 18 19 20 21 22 23 24 25 26 27  ........................... 273
| 28 | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44  ............... 612
| 45 | 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65  .. 1155
...
where:
. first column is A000384,
. second column is A130883 (without 1),
. third column is A033816,
. diagonal is A014106,
. 0, 2, 8, 18, 32, 50, ... are in A001105.

Links

Crossrefs

First bisection of A059270 and subsequence of A034828, A047866, A109900, A290168.
Sums of the consecutive integers from P(s,n) to P(s,n+1)-1, where P(s,k) is the k-th s-gonal number: A027480 (s=3), A055112 (s=4), A228888 (s=5).

Programs

  • GAP
    List([0..40], n -> n*(2*n+1)*(4*n+1));
  • Julia
    [n*(2*n+1)*(4*n+1) for n in 0:40] |> println
  • Magma
    [n*(2*n+1)*(4*n+1): n in [0..40]];
  • Maple
    seq(n*(2*n+1)*(4*n+1),n=0..40); # Muniru A Asiru, Jun 27 2018
  • Mathematica
    Table[n (2 n + 1) (4 n + 1), {n, 0, 40}]
  • Maxima
    makelist(n*(2*n+1)*(4*n+1), n, 0, 40);
  • PARI
    vector(40, n, n--; n*(2*n+1)*(4*n+1))
  • Python
    [n*(2*n+1)*(4*n+1) for n in range(40)]
  • Sage
    [n*(2*n+1)*(4*n+1) for n in (0..40)]

Formula

O.g.f.: 3*x*(5 + 10*x + x^2)/(1 - x)^4.
E.g.f.: x*(15 + 30*x + 8*x^2)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 3*A258582(n).
a(n) = -3*A100157(-n).
Sum_{n>0} 1/a(n) = 2*(3 - log(4)) - Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) + 2*sqrt(2)*log(1+sqrt(2)) + (sqrt(2)-1/2)*Pi - 6. - Amiram Eldar, Sep 17 2022
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