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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A190407 Decimal expansion of Sum_{k>=1} (1/2)^A058331(k); based on a diagonal of the natural number array, A000027.

Original entry on oeis.org

1, 2, 6, 9, 5, 5, 0, 3, 2, 4, 6, 5, 0, 4, 8, 5, 7, 8, 4, 1, 6, 2, 5, 0, 5, 4, 3, 6, 3, 5, 7, 2, 5, 6, 7, 8, 8, 0, 6, 9, 6, 2, 1, 6, 8, 1, 9, 0, 1, 4, 6, 8, 0, 0, 2, 3, 1, 5, 0, 6, 1, 7, 8, 4, 9, 2, 5, 0, 9, 9, 2, 2, 7, 6, 2, 2, 7, 3, 0, 7, 5, 3, 8, 2, 1, 6, 5, 1, 3, 8, 3, 2, 3, 0, 9, 6, 1, 4, 3, 1, 3, 9, 1, 4, 3, 1, 4, 5, 8, 3, 2, 5, 4, 2, 1, 3, 0, 3, 3, 2
Offset: 0

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Author

Clark Kimberling, May 10 2011

Keywords

Comments

See A190404.

Examples

			0.12695503246504857842...
		

Crossrefs

Programs

Formula

Equals Sum_{k>=1} (1/2)^V(k), where V=A058331 (1+2*k^2).

A002522 a(n) = n^2 + 1.

Original entry on oeis.org

1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501
Offset: 0

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Author

Keywords

Comments

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).
a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003
Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006
Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005
Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov
1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006
For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007
Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007
{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010
For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010
The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011
If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016
Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016
For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017
Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018
a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019
a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022
From Klaus Purath, Apr 03 2025: (Start)
The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.
It follows from the above that this sequence is a subsequence of A031396. (End)

Examples

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 17*x^4 + 26*x^5 + 37*x^6 + 50*x^7 + 65*x^8 + ...
		

References

  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
  • E. Gura and M. Maschler, Insights into Game Theory: An Alternative Mathematical Experience, Cambridge, 2008; p. 26.
  • Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.

Crossrefs

Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7). - Jason Kimberley, Oct 30 2011
Cf. A002496 (primes).
Cf. A254858.
Subsequence of A031396.

Programs

Formula

O.g.f.: (1-x+2*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
Sequences of the form a(n) = n^2 + K with offset 0 have o.g.f. (K - 2*K*x + K*x^2 + x + x^2)/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a*(n-3). - R. J. Mathar, Apr 28 2008
For n > 0: a(n-1) = A143053(A000290(n)) - 1. - Reinhard Zumkeller, Jul 20 2008
A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n)*a(n-2) = (n-1)^4 + 4. - Reinhard Zumkeller, Feb 12 2009
a(n) = A156798(n)/A087475(n). - Reinhard Zumkeller, Feb 16 2009
From Reinhard Zumkeller, Mar 08 2010: (Start)
a(n) = A170949(A002061(n+1));
A170949(a(n)) = A132411(n+1);
A170950(a(n)) = A002061(n+1). (End)
For n > 1, a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n - 2)^2 + (a(n) + n - 1 + a(n) + n)^2 = (n+1) *(6*n^4 + 18*n^3 + 26*n^2 + 19*n + 6) / 6 = (a(n) + n)^2 + ... + (a(n) + 2*n)^2. - Charlie Marion, Jan 10 2011
From Eric Werley, Jun 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2.
a(n) = a(n-1) + 2*n - 1. (End)
a(n) = (n-1)^2 + 2(n-1) + 2 = 122 read in base n-1 (for n > 3). - Jason Kimberley, Oct 20 2011
a(n)*a(n+1) = a(n*(n+1) + 1) so a(1)*a(2) = a(3). More generally, a(n)*a(n+k) = a(n*(n+k) + 1) + k^2 - 1. - Jon Perry, Aug 01 2012
a(n) = (n!)^2* [x^n] BesselI(0, 2*sqrt(x))*(1+x). - Peter Luschny, Aug 25 2012
a(n) = A070216(n,1) for n > 0. - Reinhard Zumkeller, Nov 11 2012
E.g.f.: exp(x)*(1 + x + x^2). - Geoffrey Critzer, Aug 30 2013
a(n) = A254858(n-2,3) for n > 2. - Reinhard Zumkeller, Feb 09 2015
Sum_{n>=0} (-1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... = A367976 . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674... = A113319. - Vaclav Kotesovec, Apr 10 2016
4*a(n) = A001105(n-1) + A001105(n+1). - Bruno Berselli, Jul 03 2017
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi)*sinh(sqrt(2)*Pi).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi). (End)
Sum_{n>=0} a(n)/n! = 3*e. - Davide Rotondo, Feb 16 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A000124 Central polygonal numbers (the Lazy Caterer's sequence): n(n+1)/2 + 1; or, maximal number of pieces formed when slicing a pancake with n cuts.

Original entry on oeis.org

1, 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, 106, 121, 137, 154, 172, 191, 211, 232, 254, 277, 301, 326, 352, 379, 407, 436, 466, 497, 529, 562, 596, 631, 667, 704, 742, 781, 821, 862, 904, 947, 991, 1036, 1082, 1129, 1177, 1226, 1276, 1327, 1379
Offset: 0

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Author

Keywords

Comments

These are Hogben's central polygonal numbers with the (two-dimensional) symbol
2
.P
1 n
The first line cuts the pancake into 2 pieces. For n > 1, the n-th line crosses every earlier line (avoids parallelism) and also avoids every previous line intersection, thus increasing the number of pieces by n. For 16 lines, for example, the number of pieces is 2 + 2 + 3 + 4 + 5 + ... + 16 = 137. These are the triangular numbers plus 1 (cf. A000217).
m = (n-1)(n-2)/2 + 1 is also the smallest number of edges such that all graphs with n nodes and m edges are connected. - Keith Briggs, May 14 2004
Also maximal number of grandchildren of a binary vector of length n+2. E.g., a binary vector of length 6 can produce at most 11 different vectors when 2 bits are deleted.
This is also the order dimension of the (strong) Bruhat order on the finite Coxeter group B_{n+1}. - Nathan Reading (reading(AT)math.umn.edu), Mar 07 2002
Number of 132- and 321-avoiding permutations of {1,2,...,n+1}. - Emeric Deutsch, Mar 14 2002
For n >= 1 a(n) is the number of terms in the expansion of (x+y)*(x^2+y^2)*(x^3+y^3)*...*(x^n+y^n). - Yuval Dekel (dekelyuval(AT)hotmail.com), Jul 28 2003
Also the number of terms in (1)(x+1)(x^2+x+1)...(x^n+...+x+1); see A000140.
Narayana transform (analog of the binomial transform) of vector [1, 1, 0, 0, 0, ...] = A000124; using the infinite lower Narayana triangle of A001263 (as a matrix), N; then N * [1, 1, 0, 0, 0, ...] = A000124. - Gary W. Adamson, Apr 28 2005
Number of interval subsets of {1, 2, 3, ..., n} (cf. A002662). - Jose Luis Arregui (arregui(AT)unizar.es), Jun 27 2006
Define a number of straight lines in the plane to be in general arrangement when (1) no two lines are parallel, (2) there is no point common to three lines. Then these are the maximal numbers of regions defined by n straight lines in general arrangement in the plane. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Note that a(n) = a(n-1) + A000027(n-1). This has the following geometrical interpretation: Suppose there are already n-1 lines in general arrangement, thus defining the maximal number of regions in the plane obtainable by n-1 lines and now one more line is added in general arrangement. Then it will cut each of the n-1 lines and acquire intersection points which are in general arrangement. (See the comments on A000027 for general arrangement with points.) These points on the new line define the maximal number of regions in 1-space definable by n-1 points, hence this is A000027(n-1), where for A000027 an offset of 0 is assumed, that is, A000027(n-1) = (n+1)-1 = n. Each of these regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n) = a(n-1) + A000027(n-1). Cf. the comments on A000125 for an analogous interpretation. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
When constructing a zonohedron, one zone at a time, out of (up to) 3-d non-intersecting parallelepipeds, the n-th element of this sequence is the number of edges in the n-th zone added with the n-th "layer" of parallelepipeds. (Verified up to 10-zone zonohedron, the enneacontahedron.) E.g., adding the 10th zone to the enneacontahedron requires 46 parallel edges (edges in the 10th zone) by looking directly at a 5-valence vertex and counting visible vertices. - Shel Kaphan, Feb 16 2006
Binomial transform of (1, 1, 1, 0, 0, 0, ...) and inverse binomial transform of A072863: (1, 3, 9, 26, 72, 192, ...). - Gary W. Adamson, Oct 15 2007
If Y is a 2-subset of an n-set X then, for n >= 3, a(n-3) is the number of (n-2)-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
Equals row sums of triangle A144328. - Gary W. Adamson, Sep 18 2008
It appears that a(n) is the number of distinct values among the fractions F(i+1)/F(j+1) as j ranges from 1 to n and, for each fixed j, i ranges from 1 to j, where F(i) denotes the i-th Fibonacci number. - John W. Layman, Dec 02 2008
a(n) is the number of subsets of {1,2,...,n} that contain at most two elements. - Geoffrey Critzer, Mar 10 2009
For n >= 2, a(n) gives the number of sets of subsets A_1, A_2, ..., A_n of n = {1, 2, ..., n} such that Meet_{i = 1..n} A_i is empty and Sum_{j in [n]} (|Meet{i = 1..n, i != j} A_i|) is a maximum. - Srikanth K S, Oct 22 2009
The numbers along the left edge of Floyd's triangle. - Paul Muljadi, Jan 25 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j] = A[i,i]:=1, A[i,i-1] = -1, and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = (-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 24 2010
Also the number of deck entries of Euler's ship. See the Meijer-Nepveu link. - Johannes W. Meijer, Jun 21 2010
(1 + x^2 + x^3 + x^4 + x^5 + ...)*(1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...) = (1 + 2x + 4x^2 + 7x^3 + 11x^4 + ...). - Gary W. Adamson, Jul 27 2010
The number of length n binary words that have no 0-digits between any pair of consecutive 1-digits. - Jeffrey Liese, Dec 23 2010
Let b(0) = b(1) = 1; b(n) = max(b(n-1)+n-1, b(n-2)+n-2) then a(n) = b(n+1). - Yalcin Aktar, Jul 28 2011
Also number of triangular numbers so far, for n > 0: a(n) = a(n-1) + Sum(A010054(a(k)): 0 <= k < n), see also A097602, A131073. - Reinhard Zumkeller, Nov 15 2012
Also number of distinct sums of 1 through n where each of those can be + or -. E.g., {1+2,1-2,-1+2,-1-2} = {3,-1,1,-3} and a(2) = 4. - Toby Gottfried, Nov 17 2011
This sequence is complete because the sum of the first n terms is always greater than or equal to a(n+1)-1. Consequently, any nonnegative number can be written as a sum of distinct terms of this sequence. See A204009, A072638. - Frank M Jackson, Jan 09 2012
The sequence is the number of distinct sums of subsets of the nonnegative integers, and its first differences are the positive integers. See A208531 for similar results for the squares. - John W. Layman, Feb 28 2012
Apparently the number of Dyck paths of semilength n+1 in which the sum of the first and second ascents add to n+1. - David Scambler, Apr 22 2013
Without 1 and 2, a(n) equals the terminus of the n-th partial sum of sequence 1, 1, 2. Explanation: 1st partial sums of 1, 1, 2 are 1, 2, 4; 2nd partial sums are 1, 3, 7; 3rd partial sums are 1, 4, 11; 4th partial sums are 1, 5, 16, etc. - Bob Selcoe, Jul 04 2013
Equivalently, numbers of the form 2*m^2+m+1, where m = 0, -1, 1, -2, 2, -3, 3, ... . - Bruno Berselli, Apr 08 2014
For n >= 2: quasi-triangular numbers; the almost-triangular numbers being A000096(n), n >= 2. Note that 2 is simultaneously almost-triangular and quasi-triangular. - Daniel Forgues, Apr 21 2015
n points in general position determine "n choose 2" lines, so A055503(n) <= a(n(n-1)/2). If n > 3, the lines are not in general position and so A055503(n) < a(n(n-1)/2). - Jonathan Sondow, Dec 01 2015
The digital root is period 9 (1, 2, 4, 7, 2, 7, 4, 2, 1), also the digital roots of centered 10-gonal numbers (A062786), for n > 0, A133292. - Peter M. Chema, Sep 15 2016
Partial sums of A028310. - J. Conrad, Oct 31 2016
For n >= 0, a(n) is the number of weakly unimodal sequences of length n over the alphabet {1, 2}. - Armend Shabani, Mar 10 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) != e(k). [Martinez and Savage, 2.4]
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) < e(j) and e(i) < e(k). [Martinez and Savage, 2.4]
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) != e(k). [Martinez and Savage, 2.4]
(End)
Numbers m such that 8m - 7 is a square. - Bruce J. Nicholson, Jul 24 2017
From Klaus Purath, Jan 29 2020: (Start)
The odd prime factors != 7 occur in an interval of p successive terms either never or exactly twice, while 7 always occurs only once. If a prime factor p appears in a(n) and a(m) within such an interval, then n + m == -1 (mod p). When 7 divides a(n), then 2*n == -1 (mod 7). a(n) is never divisible by the prime numbers given in A003625.
While all prime factors p != 7 can occur to any power, a(n) is never divisible by 7^2. The prime factors are given in A045373. The prime terms of this sequence are given in A055469.
(End)
From Roger Ford, May 10 2021: (Start)
a(n-1) is the greatest sum of arch lengths for the top arches of a semi-meander with n arches. An arch length is the number of arches covered + 1.
/\ The top arch has a length of 3. /\ The top arch has a length of 3.
/ \ Both bottom arches have a //\\ The middle arch has a length of 2.
//\/\\ length of 1. ///\\\ The bottom arch has a length of 1.
Example: for n = 4, a(4-1) = a(3) = 7 /\
//\\
/\ ///\\\ 1 + 3 + 2 + 1 = 7. (End)
a(n+1) is the a(n)-th smallest positive integer that has not yet appeared in the sequence. - Matthew Malone, Aug 26 2021
For n> 0, let the n-dimensional cube {0,1}^n be, provided with the Hamming distance, d. Given an element x in {0,1}^n, a(n) is the number of elements y in {0,1}^n such that d(x, y) <= 2. Example: n = 4. (0,0,0,0), (1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1), (0,0,1,1), (0,1,0,1), (0,1,1,0), (1,0,0,1), (1,0,1,0), (1,1,0,0) are at distance <= 2 from (0,0,0,0), so a(4) = 11. - Yosu Yurramendi, Dec 10 2021
a(n) is the sum of the first three entries of row n of Pascal's triangle. - Daniel T. Martin, Apr 13 2022
a(n-1) is the number of Grassmannian permutations that avoid a pattern, sigma, where sigma is a pattern of size 3 with exactly one descent. For example, sigma is one of the patterns, {132, 213, 231, 312}. - Jessica A. Tomasko, Sep 14 2022
a(n+4) is the number of ways to tile an equilateral triangle of side length 2*n with smaller equilateral triangles of side length n and side length 1. For example, with n=2, there are 22 ways to tile an equilateral triangle of side length 4 with smaller ones of sides 2 and 1, including the one tiling with sixteen triangles of sides 1 and the one tiling with four triangles of sides 2. - Ahmed ElKhatib and Greg Dresden, Aug 19 2024
Define a "hatpin" to be the planar graph consisting of a distinguished point (called the "head") and a semi-infinite line from that point. The maximum number of regions than can be formed by drawing n hatpins is a(n-1). See link for the case n = 4. - N. J. A. Sloane, Jun 25 2025

Examples

			a(3) = 7 because the 132- and 321-avoiding permutations of {1, 2, 3, 4} are 1234, 2134, 3124, 2314, 4123, 3412, 2341.
G.f. = 1 + 2*x + 4*x^2 + 7*x^3 + 11*x^4 + 16*x^5 + 22*x^6 + 29*x^7 + ...
		

References

  • Robert B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 24.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 80.
  • Henry Ernest Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
  • Derrick Niederman, Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed, A Perigee Book, NY, 2009, p. 83.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane, On single-deletion-correcting codes, in Codes and Designs (Columbus, OH, 2000), 273-291, Ohio State Univ. Math. Res. Inst. Publ., 10, de Gruyter, Berlin, 2002.
  • 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. 98.
  • William Allen Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
  • Akiva M. Yaglom and Isaak M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #44 (First published: San Francisco: Holden-Day, Inc., 1964).

Crossrefs

Cf. A000096 (Maximal number of pieces that can be obtained by cutting an annulus with n cuts, for n >= 1).
Slicing a cake: A000125, a bagel: A003600.
Partial sums =(A033547)/2, (A014206)/2.
The first 20 terms are also found in A025732 and A025739.
Cf. also A055469 Quasi-triangular primes, A002620, A000217.
A row of the array in A386478.

Programs

Formula

G.f.: (1 - x + x^2)/(1 - x)^3. - Simon Plouffe in his 1992 dissertation
a(n) = A108561(n+3, 2). - Reinhard Zumkeller, Jun 10 2005
G.f.: (1 - x^6)/((1 - x)^2*(1 - x^2)*(1 - x^3)). a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 04 2006
Euler transform of length 6 sequence [ 2, 1, 1, 0, 0, -1]. - Michael Somos, Sep 04 2006
a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) and a(1) = 1, a(2) = 2, a(3) = 4. - Artur Jasinski, Oct 21 2008
a(n) = A000217(n) + 1.
a(n) = a(n-1) + n. E.g.f.:(1 + x + x^2/2)*exp(x). - Geoffrey Critzer, Mar 10 2009
a(n) = Sum_{k = 0..n + 1} binomial(n+1, 2(k - n)). - Paul Barry, Aug 29 2004
a(n) = binomial(n+2, 1) - 2*binomial(n+1, 1) + binomial(n+2, 2). - Zerinvary Lajos, May 12 2006
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{l_1 = 0..n + 1} Sum_{l_2 = 0..n}...Sum_{l_i = 0..n - i}...Sum_{l_n = 0..1} delta(l_1, l_2, ..., l_i, ..., l_n) where delta(l_1, l_2, ..., l_i, ..., l_n) = 0 if any l_i != l_(i+1) and l_(i+1) != 0 and delta(l_1, l_2, ..., l_i, ..., l_n) = 1 otherwise. (End)
a(n) = A034856(n+1) - A005843(n) = A000217(n) + A005408(n) - A005843(n). - Jaroslav Krizek, Sep 05 2009
a(n) = 2*a(n-1) - a(n-2) + 1. - Eric Werley, Jun 27 2011
E.g.f.: exp(x)*(1+x+(x^2)/2) = Q(0); Q(k) = 1+x/(1-x/(2+x-4/(2+x*(k+1)/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
a(n) = A014132(n, 1) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n) = 1 + floor(n/2) + ceiling(n^2/2) = 1 + A004526(n) + A000982(n). - Wesley Ivan Hurt, Jun 14 2013
a(n) = A228074(n+1, n). - Reinhard Zumkeller, Aug 15 2013
For n > 0: A228446(a(n)) = 3. - Reinhard Zumkeller, Mar 12 2014
a(n) >= A263883(n) and a(n(n-1)/2) >= A055503(n). - Jonathan Sondow, Dec 01 2015
From Ilya Gutkovskiy, Jun 29 2016: (Start)
Dirichlet g.f.: (zeta(s-2) + zeta(s-1) + 2*zeta(s))/2.
Sum_{n >= 0} 1/a(n) = 2*Pi*tanh(sqrt(7)*Pi/2)/sqrt(7) = A226985. (End)
a(n) = (n+1)^2 - A000096(n). - Anton Zakharov, Jun 29 2016
a(n) = A101321(1, n). - R. J. Mathar, Jul 28 2016
a(n) = 2*a(n-1) - binomial(n-1, 2) and a(0) = 1. - Armend Shabani, Mar 10 2017
a(n) = A002620(n+2) + A002620(n-1). - Anton Zakharov, May 11 2017
From Klaus Purath, Jan 29 2020: (Start)
a(n) = (Sum_{i=n-2..n+2} A000217(i))/5.
a(n) = (Sum_{i=n-2..n+2} A002378(i))/10.
a(n) = (Sum_{i=n..n+2} A002061(i)+1)/6.
a(n) = (Sum_{i=n-1..n+2} A000290(i)+2)/8.
a(n) = A060533(n-1) + 10, n > 5.
a(n) = (A002378(n) + 2)/2.
a(n) = A152948(n+2) - 1.
a(n) = A152950(n+1) - 2.
a(n) = (A002061(n) + A002061(n+2))/4.
(End)
Sum_{n>=0} (-1)^n/a(n) = A228918. - Amiram Eldar, Nov 20 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(15)*Pi/2)*sech(sqrt(7)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = 2*Pi*sech(sqrt(7)*Pi/2). (End)
a((n^2-3n+6)/2) + a((n^2-n+4)/2) = a(n^2-2n+6)/2. - Charlie Marion, Feb 14 2023

A001105 a(n) = 2*n^2.

Original entry on oeis.org

0, 2, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418
Offset: 0

Views

Author

Bernd.Walter(AT)frankfurt.netsurf.de

Keywords

Comments

Number of edges of the complete bipartite graph of order 3n, K_{n,2n}. - Roberto E. Martinez II, Jan 07 2002
"If each period in the periodic system ends in a rare gas ..., the number of elements in a period can be found from the ordinal number n of the period by the formula: L = ((2n+3+(-1)^n)^2)/8..." - Nature, Jun 09 1951; Nature 411 (Jun 07 2001), p. 648. This produces the present sequence doubled up.
Let z(1) = i = sqrt(-1), z(k+1) = 1/(z(k)+2i); then a(n) = (-1)*Imag(z(n+1))/Real(z(n+1)). - Benoit Cloitre, Aug 06 2002
Maximum number of electrons in an atomic shell with total quantum number n. Partial sums of A016825. - Jeremy Gardiner, Dec 19 2004
Arithmetic mean of triangular numbers in pairs: (1+3)/2, (6+10)/2, (15+21)/2, ... . - Amarnath Murthy, Aug 05 2005
These numbers form a pattern on the Ulam spiral similar to that of the triangular numbers. - G. Roda, Oct 20 2010
Integral areas of isosceles right triangles with rational legs (legs are 2n and triangles are nondegenerate for n > 0). - Rick L. Shepherd, Sep 29 2009
Even squares divided by 2. - Omar E. Pol, Aug 18 2011
Number of stars when distributed as in the U.S.A. flag: n rows with n+1 stars and, between each pair of these, one row with n stars (i.e., n-1 of these), i.e., n*(n+1)+(n-1)*n = 2*n^2 = A001105(n). - César Eliud Lozada, Sep 17 2012
Apparently the number of Dyck paths with semilength n+3 and an odd number of peaks and the central peak having height n-3. - David Scambler, Apr 29 2013
Sum of the partition parts of 2n into exactly two parts. - Wesley Ivan Hurt, Jun 01 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective odd leg a (A180620); sequence gives values c-a, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
Number of roots in the root systems of type B_n and C_n (for n > 1). - Tom Edgar, Nov 05 2013
Area of a square with diagonal 2n. - Wesley Ivan Hurt, Jun 18 2014
This sequence appears also as the first and second member of the quartet [a(n), a(n), p(n), p(n)] of the square of [n, n, n+1, n+1] in the Clifford algebra Cl_2 for n >= 0. p(n) = A046092(n). See an Oct 15 2014 comment on A147973 where also a reference is given. - Wolfdieter Lang, Oct 16 2014
a(n) are the only integers m where (A000005(m) + A000203(m)) = (number of divisors of m + sum of divisors of m) is an odd number. - Richard R. Forberg, Jan 09 2015
a(n) represents the first term in a sum of consecutive integers running to a(n+1)-1 that equals (2n+1)^3. - Patrick J. McNab, Dec 24 2016
Also the number of 3-cycles in the (n+4)-triangular honeycomb obtuse knight graph. - Eric W. Weisstein, Jul 29 2017
Also the Wiener index of the n-cocktail party graph for n > 1. - Eric W. Weisstein, Sep 07 2017
Numbers represented as the palindrome 242 in number base B including B=2 (binary), 3 (ternary) and 4: 242(2)=18, 242(3)=32, 242(4)=50, ... 242(9)=200, 242(10)=242, ... - Ron Knott, Nov 14 2017
a(n) is the square of the hypotenuse of an isosceles right triangle whose sides are equal to n. - Thomas M. Green, Aug 20 2019
The sequence contains all odd powers of 2 (A004171) but no even power of 2 (A000302). - Torlach Rush, Oct 10 2019
From Bernard Schott, Aug 31 2021 and Sep 16 2021: (Start)
Apart from 0, integers such that the number of even divisors (A183063) is odd.
Proof: every n = 2^q * (2k+1), q, k >= 0, then 2*n^2 = 2^(2q+1) * (2k+1)^2; now, gcd(2, 2k+1) = 1, tau(2^(2q+1)) = 2q+2 and tau((2k+1)^2) = 2u+1 because (2k+1)^2 is square, so, tau(2*n^2) = (2q+2) * (2u+1).
The 2q+2 divisors of 2^(2q+1) are {1, 2, 2^2, 2^3, ..., 2^(2q+1)}, so 2^(2q+1) has 2q+1 even divisors {2^1, 2^2, 2^3, ..., 2^(2q+1)}.
Conclusion: these 2q+1 even divisors create with the 2u+1 odd divisors of (2k+1)^2 exactly (2q+1)*(2u+1) even divisors of 2*n^2, and (2q+1)*(2u+1) is odd. (End)
a(n) with n>0 are the numbers with period length 2 for Bulgarian and Mancala solitaire. - Paul Weisenhorn, Jan 29 2022
Number of points at L1 distance = 2 from any given point in Z^n. - Shel Kaphan, Feb 25 2023
Integer that multiplies (h^2)/(m*L^2) to give the energy of a 1-D quantum mechanical particle in a box whenever it is an integer multiple of (h^2)/(m*L^2), where h = Planck's constant, m = mass of particle, and L = length of box. - A. Timothy Royappa, Mar 14 2025

Examples

			a(3) = 18; since 2(3) = 6 has 3 partitions with exactly two parts: (5,1), (4,2), (3,3).  Adding all the parts, we get: 1 + 2 + 3 + 3 + 4 + 5 = 18. - _Wesley Ivan Hurt_, Jun 01 2013
		

References

  • Peter Atkins, Julio De Paula, and James Keeler, "Atkins' Physical Chemistry," Oxford University Press, 2023, p. 31.
  • Arthur Beiser, Concepts of Modern Physics, 2nd Ed., McGraw-Hill, 1973.
  • Martin Gardner, The Colossal Book of Mathematics, Classic Puzzles, Paradoxes and Problems, Chapter 2 entitled "The Calculus of Finite Differences," W. W. Norton and Company, New York, 2001, pages 12-13.
  • L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 44.
  • Alain M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000, p. 213.

Crossrefs

Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Cf. A058331 and A247375. - Bruno Berselli, Sep 16 2014
Cf. A194715 (4-cycles in the triangular honeycomb obtuse knight graph), A290391 (5-cycles), A290392 (6-cycles). - Eric W. Weisstein, Jul 29 2017
Integers such that: this sequence (the number of even divisors is odd), A028982 (the number of odd divisors is odd), A028983 (the number of odd divisors is even), A183300 (the number of even divisors is even).

Programs

Formula

a(n) = (-1)^(n+1) * A053120(2*n, 2).
G.f.: 2*x*(1+x)/(1-x)^3.
a(n) = A100345(n, n).
Sum_{n>=1} 1/a(n) = Pi^2/12 =A072691. [Jolley eq. 319]. - Gary W. Adamson, Dec 21 2006
a(n) = A049452(n) - A033991(n). - Zerinvary Lajos, Jun 12 2007
a(n) = A016742(n)/2. - Zerinvary Lajos, Jun 20 2008
a(n) = 2 * A000290(n). - Omar E. Pol, May 14 2008
a(n) = 4*n + a(n-1) - 2, n > 0. - Vincenzo Librandi
a(n) = A002378(n-1) + A002378(n). - Joerg M. Schuetze (joerg(AT)cyberheim.de), Mar 08 2010 [Corrected by Klaus Purath, Jun 18 2020]
a(n) = A176271(n,k) + A176271(n,n-k+1), 1 <= k <= n. - Reinhard Zumkeller, Apr 13 2010
a(n) = A007607(A000290(n)). - Reinhard Zumkeller, Feb 12 2011
For n > 0, a(n) = 1/coefficient of x^2 in the Maclaurin expansion of 1/(cos(x)+n-1). - Francesco Daddi, Aug 04 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Artur Jasinski, Nov 24 2011
a(n) = A070216(n,n) for n > 0. - Reinhard Zumkeller, Nov 11 2012
a(n) = A014132(2*n-1,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n) = A000217(n) + A000326(n). - Omar E. Pol, Jan 11 2013
(a(n) - A000217(k))^2 = A000217(2*n-1-k)*A000217(2*n+k) + n^2, for all k. - Charlie Marion, May 04 2013
a(n) = floor(1/(1-cos(1/n))), n > 0. - Clark Kimberling, Oct 08 2014
a(n) = A251599(3*n-1) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n+4)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = A002061(n+1) + A165900(n). - Torlach Rush, Feb 21 2019
E.g.f.: 2*exp(x)*x*(1 + x). - Stefano Spezia, Oct 12 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Jul 03 2020
From Amiram Eldar, Feb 03 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sqrt(2)*sinh(Pi/sqrt(2))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(2)*sin(Pi/sqrt(2))/Pi. (End)

A079978 Characteristic function of multiples of three.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0
Offset: 0

Views

Author

Vladimir Baltic, Feb 17 2003

Keywords

Comments

Period 3: repeat [1, 0, 0].
a(n)=1 if n=3k, a(n)=0 otherwise.
Decimal expansion of 1/999.
Number of permutations satisfying -k <= p(i)-i <= r and p(i)-i not in I, i=1..n, with k=1, r=2, I={0,1}.
a(n) is also the number of partitions of n with every part being three (a(0)=1 because the empty partition has no parts). Hence a(n) is also the number of 2-regular graphs on n vertices with each component having girth 3. - Jason Kimberley, Oct 02 2011
Euler transformation of A185013. - Jason Kimberley, Oct 02 2011
If b(0)=0 and for n > 0, b(n)=a(n), then starting at n=0, b(n) is the number of incongruent equilateral triangles formed from the vertices of a regular n-gon. The number of incongruent isosceles triangles (strictly two equal sides) is A174257(n) and the number of incongruent scalene triangles is A069905(n-3) for n > 2, otherwise 0. The total number of incongruent triangles is A069905(n). - Frank M Jackson, Nov 19 2022

References

  • D. H. Lehmer, Permutations with strongly restricted displacements. Combinatorial theory and its applications, II (Proc. Colloq., Balatonfured, 1969), pp. 755-770. North-Holland, Amsterdam, 1970.

Crossrefs

Essentially the same as A022003.
Partial sums are given by A002264(n+3).
Characteristic function of multiples of g: A000007 (g=0), A000012 (g=1), A059841 (g=2), this sequence (g=3), A121262 (g=4), A079998 (g=5), A079979 (g=6), A082784 (g=7). - Jason Kimberley, Oct 14 2011
Cf. A007908, A011655 (bit flipped).

Programs

Formula

a(n) = a(n-3) for n > 2.
G.f.: 1/(1-x^3) = 1/( (1-x)*(1+x+x^2)).
a(n) = (1 + e^(i*Pi*A002487(n)))/2, i=sqrt(-1). - Paul Barry, Jan 14 2005
Additive with a(p^e) = 1 if p = 3, 0 otherwise.
a(n) = ((n+1) mod 3) mod 2. Also: a(n) = (1/2)*(1 + (-1)^(n + floor((n+1)/3))). - Hieronymus Fischer, May 29 2007
a(n) = 1 - A011655(n). - Reinhard Zumkeller, Nov 30 2009
a(n) = (1 + (-1)^(2*n/3) + (-1)^(-2*n/3))/3. - Jaume Oliver Lafont, May 13 2010
For the general case: the characteristic function of numbers that are multiples of m is a(n) = floor(n/m) - floor((n-1)/m), m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = floor( ((n-1) mod 3)/2 ). - Wesley Ivan Hurt, Jun 29 2013
a(n) = 2^(n mod 3) mod 2. - Olivier Gérard, Jul 04 2013
a(n) = (w^(2*n) + w^n + 1)/3, w = (-1 + i*sqrt(3))/2 (w is a primitive 3rd root of unity). - Bogart B. Strauss, Jul 20 2013
E.g.f.: (exp(x) + 2*exp(-x/2)*cos(sqrt(3)*x/2))/3. - Geoffrey Critzer, Nov 03 2014
a(n) = (sin(Pi*(n+1)/3)^2)*(2/3) + sin(Pi*(n+1)*2/3)/sqrt(3). - Mikael Aaltonen, Jan 03 2015
a(n) = (2*n^2 + 1) mod 3. The characteristic function of numbers that are multiples of 2k+1 is (2*k*n^(2*k) + 1) mod (2k+1). Example: A058331(n) mod 3 for k=1, A211412(n) mod 5 for k=2, ... - Eric Desbiaux, Dec 25 2015
a(n) = floor(2*(n-1)/3) - 2*floor((n-1)/3). - Wesley Ivan Hurt, Jul 25 2016
a(n) == A007908(n+1) (mod 3), n >= 0. See A011655 (bit flipped). - Wolfdieter Lang, Jun 12 2017
a(n) = 1/3 + (2/3)*cos((2/3)*n*Pi). - Ridouane Oudra, Jan 22 2021
a(n) = A000217(n+1) mod 3. - Christopher Adams, Jan 05 2025

Extensions

Name simplified by Ralf Stephan, Nov 22 2010
Name changed by Jason Kimberley, Oct 14 2011

A000125 Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3) + n + 1.

Original entry on oeis.org

1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526, 4992, 5489, 6018, 6580, 7176, 7807, 8474, 9178, 9920, 10701, 11522, 12384, 13288, 14235, 15226
Offset: 0

Views

Author

Keywords

Comments

Note that a(n) = a(n-1) + A000124(n-1). This has the following geometrical interpretation: Define a number of planes in space to be in general arrangement when
(1) no two planes are parallel,
(2) there are no two parallel intersection lines,
(3) there is no point common to four or more planes.
Suppose there are already n-1 planes in general arrangement, thus defining the maximal number of regions in space obtainable by n-1 planes and now one more plane is added in general arrangement. Then it will cut each of the n-1 planes and acquire intersection lines which are in general arrangement. (See the comments on A000124 for general arrangement with lines.) These lines on the new plane define the maximal number of regions in 2-space definable by n-1 straight lines, hence this is A000124(n-1). Each of this regions acts as a dividing wall, thereby creating as many new regions in addition to the a(n-1) regions already there, hence a(n) = a(n-1) + A000124(n-1). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
More generally, we have: A000027(n) = binomial(n,0) + binomial(n,1) (the natural numbers), A000124(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) (the Lazy Caterer's sequence), a(n) = binomial(n,0) + binomial(n,1) + binomial(n,2) + binomial(n,3) (Cake Numbers). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
If Y is a 2-subset of an n-set X then, for n>=3, a(n-3) is the number of 3-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007
a(n) is the number of compositions (ordered partitions) of n+1 into four or fewer parts or equivalently the sum of the first four terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 23 2009
{a(k): 0 <= k < 4} = divisors of 8. - Reinhard Zumkeller, Jun 17 2009
a(n) is also the maximum number of different values obtained by summing n consecutive positive integers with all possible 2^n sign combinations. This maximum is first reached when summing the interval [n, 2n-1]. - Olivier Gérard, Mar 22 2010
a(n) contains only 5 perfect squares > 1: 4, 64, 576, 67600, and 75203584. The incidences of > 0 are given by A047694. - Frank M Jackson, Mar 15 2013
Given n tiles with two values - an A value and a B value - a player may pick either the A value or the B value. The particular tiles are [n, 0], [n-1, 1], ..., [2, n-2] and [1, n-1]. The sequence is the number of different final A:B counts. For example, with n=4, we can have final total [5, 3] = [4, ] + [, 1] + [, 2] + [1, ] = [, 0] + [3, ] + [2, ] + [, 3], so a(4) = 2^4 - 1 = 15. The largest and smallest final A+B counts are given by A077043 and A002620 respectively. - Jon Perry, Oct 24 2014
For n>=3, a(n) is also the number of maximal cliques in the (n+1)-triangular graph (the 4-triangular graph has a(3)=8 maximal cliques). - Andrew Howroyd, Jul 19 2017
a(n) is the number of binary words of length n matching the regular expression 1*0*1*0*. Coincidentally, A000124 counts binary words of the form 0*1*0*. See Alexandersson and Nabawanda for proof. - Per W. Alexandersson, May 15 2021
For n > 0, let the n-dimensional cube, {0,1}^n be provided with the Hamming distance, d. Given an element x in {0,1}^n, a(n) is the number of elements y in {0,1}^n such that d(x, y) <= 3. Example: n = 4. Let x = (0,0,0,0) be in {0,1}^4.
d(x,y) = 0: y in {(0,0,0,0)}.
d(x,y) = 1: y in {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)}.
d(x,y) = 2: y in {(1,1,0,0), (1,0,1,0), (1,0,0,1), (0,1,1,0), (0,1,0,1), (0,0,1,1)}.
d(x,y) = 3: y in {(1,1,1,0), (1,1,0,1), (1,0,1,1), (0,1,1,1)}.
All these y are at a distance <= 3 from (0,0,0,0), so a(4) = 15. (See Peter C. Heinig's formula). - Yosu Yurramendi, Dec 14 2021
For n >= 2, a(n) is the number of distinct least squares regression lines fitted to n points (j,y_j), 1 <= j <= n, where each y_j is 0 or 1. The number of distinct lines with exactly k 1's among y_1, ..., y_n is A077028(n,k). The number of distinct slopes is A123596(n). - Pontus von Brömssen, Mar 16 2024
The only powers of 2 in this sequence are a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 8, and a(7) = 64. - Jianing Song, Jan 02 2025

Examples

			a(4)=15 because there are 15 compositions of 5 into four or fewer parts. a(6)=42 because the sum of the first four terms in the 6th row of Pascal's triangle is 1+6+15+20=42. - _Geoffrey Critzer_, Jan 23 2009
For n=5, (1, 3, 5, 7, 9, 11, 13, 17, 19, 21, 23, 25, 35) and their opposite are the 26 different sums obtained by summing 5,6,7,8,9 with any sign combination. - _Olivier Gérard_, Mar 22 2010
G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 15*x^4 + 26*x^5 + 42*x^6 + 64*x^7 + ... - _Michael Somos_, Jul 07 2022
		

References

  • V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_3.
  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 27.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 80.
  • H. E. Dudeney, Amusements in Mathematics, Nelson, London, 1917, page 177.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • T. H. Stickels, Mindstretching Puzzles. Sterling, NY, 1994 p. 85.
  • W. A. Whitworth, DCC Exercises in Choice and Chance, Stechert, NY, 1945, p. 30.
  • A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #45 (First published: San Francisco: Holden-Day, Inc., 1964)

Crossrefs

Programs

Formula

a(n) = (n+1)*(n^2-n+6)/6 = (n^3 + 5*n + 6) / 6.
G.f.: (1 - 2*x + 2x^2)/(1-x)^4. - [Simon Plouffe in his 1992 dissertation.]
E.g.f.: (1 + x + x^2/2 + x^3/6)*exp(x).
a(n) = binomial(n,3) + binomial(n,2) + binomial(n,1) + binomial(n,0). - Peter C. Heinig (algorithms(AT)gmx.de), Oct 19 2006
Paraphrasing the previous comment: the sequence is the binomial transform of [1,1,1,1,0,0,0,...]. - Gary W. Adamson, Oct 23 2007
From Ilya Gutkovskiy, Jul 18 2016: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = Sum_{k=0..n} A152947(k+1).
Inverse binomial transform of A134396.
Sum_{n>=0} a(n)/n! = 8*exp(1)/3. (End)
a(n) = -A283551(-n). - Michael Somos, Jul 07 2022
a(n) = A046127(n+1)/2 = A033547(n)/2 + 1. - Jianing Song, Jan 02 2025

Extensions

Minor typo in comments corrected by Mauro Fiorentini, Jan 02 2018

A055998 a(n) = n*(n+5)/2.

Original entry on oeis.org

0, 3, 7, 12, 18, 25, 33, 42, 52, 63, 75, 88, 102, 117, 133, 150, 168, 187, 207, 228, 250, 273, 297, 322, 348, 375, 403, 432, 462, 493, 525, 558, 592, 627, 663, 700, 738, 777, 817, 858, 900, 943, 987, 1032, 1078, 1125, 1173, 1222, 1272
Offset: 0

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Author

Barry E. Williams, Jun 14 2000

Keywords

Comments

If X is an n-set and Y a fixed (n-3)-subset of X then a(n-3) is equal to the number of 2-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Bisection of A165157. - Jaroslav Krizek, Sep 05 2009
a(n) is the number of (w,x,y) having all terms in {0,...,n} and w=x+y-1. - Clark Kimberling, Jun 02 2012
Numbers m >= 0 such that 8m+25 is a square. - Bruce J. Nicholson, Jul 26 2017
a(n-1) = 3*(n-1) + (n-1)*(n-2)/2 is the number of connected, loopless, non-oriented, multi-edge vertex-labeled graphs with n edges and 3 vertices. Labeled multigraph analog of A253186. There are 3*(n-1) graphs with the 3 vertices on a chain (3 ways to label the middle graph, n-1 ways to pack edges on one of connections) and binomial(n-1,2) triangular graphs (one way to label the graphs, pack 1 or 2 or ...n-2 on the 1-2 edge, ...). - R. J. Mathar, Aug 10 2017
a(n) is also the number of vertices of the quiver for PGL_{n+1} (see Shen). - Stefano Spezia, Mar 24 2020
Starting from a(2) = 7, this is the 4th column of the array: natural numbers written by antidiagonals downwards. See the illustration by Kival Ngaokrajang and the cross-references. - Andrey Zabolotskiy, Dec 21 2021

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.

Crossrefs

a(n) = A095660(n+1, 2): third column of (1, 3)-Pascal triangle.
Row n=2 of A255961.

Programs

Formula

G.f.: x*(3-2*x)/(1-x)^3.
a(n) = A027379(n), n > 0.
a(n) = A126890(n,2) for n > 1. - Reinhard Zumkeller, Dec 30 2006
a(n) = A000217(n) + A005843(n). - Reinhard Zumkeller, Sep 24 2008
If we define f(n,i,m) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-m-j), then a(n) = -f(n,n-1,3), for n >= 1. - Milan Janjic, Dec 20 2008
a(n) = A167544(n+8). - Philippe Deléham, Nov 25 2009
a(n) = a(n-1) + n + 2 with a(0)=0. - Vincenzo Librandi, Aug 07 2010
a(n) = Sum_{k=1..n} (k+2). - Gary Detlefs, Aug 10 2010
a(n) = A034856(n+1) - 1 = A000217(n+2) - 3. - Jaroslav Krizek, Sep 05 2009
Sum_{n>=1} 1/a(n) = 137/150. - R. J. Mathar, Jul 14 2012
a(n) = 3*n + A000217(n-1) = 3*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=3..n+2} i. - Wesley Ivan Hurt, Jun 28 2013
a(n) = 3*A000217(n) - 2*A000217(n-1). - Bruno Berselli, Dec 17 2014
a(n) = A046691(n) + 1. Also, a(n) = A052905(n-1) + 2 = A055999(n-1) + 3 for n>0. - Andrey Zabolotskiy, May 18 2016
E.g.f.: x*(6+x)*exp(x)/2. - G. C. Greubel, Apr 05 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/5 - 47/150. - Amiram Eldar, Jan 10 2021
From Amiram Eldar, Feb 12 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = -5*cos(sqrt(33)*Pi/2)/(4*Pi).
Product_{n>=1} (1 + 1/a(n)) = 15*cos(sqrt(17)*Pi/2)/(2*Pi). (End)

A001108 a(n)-th triangular number is a square: a(n+1) = 6*a(n) - a(n-1) + 2, with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 8, 49, 288, 1681, 9800, 57121, 332928, 1940449, 11309768, 65918161, 384199200, 2239277041, 13051463048, 76069501249, 443365544448, 2584123765441, 15061377048200, 87784138523761, 511643454094368, 2982076586042449, 17380816062160328, 101302819786919521
Offset: 0

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Author

Keywords

Comments

b(0)=0, c(0)=1, b(i+1)=b(i)+c(i), c(i+1)=b(i+1)+b(i); then a(i) (the number in the sequence) is 2b(i)^2 if i is even, c(i)^2 if i is odd and b(n)=A000129(n) and c(n)=A001333(n). - Darin Stephenson (stephenson(AT)cs.hope.edu) and Alan Koch
For n > 1 gives solutions to A007913(2x) = A007913(x+1). - Benoit Cloitre, Apr 07 2002
If (X,X+1,Z) is a Pythagorean triple, then Z-X-1 and Z+X are in the sequence.
For n >= 2, a(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is A001109. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
This is the r=8 member of the r-family of sequences S_r(n) defined in A092184 where more information can be found.
Also, 1^3 + 2^3 + 3^3 + ... + a(n)^3 = k(n)^4 where k(n) is A001109. - Anton Vrba (antonvrba(AT)yahoo.com), Nov 18 2006
If T_x = y^2 is a triangular number which is also a square, the least number which is both triangular and square and greater than T_x is T_(3*x + 4*y + 1) = (2*x + 3*y + 1)^2 (W. Sierpiński 1961). - Richard Choulet, Apr 28 2009
If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or (a+1) is a perfect square. If (a,b) is a solution of the Diophantine equation 0 + 1 + 2 + ... + x = y^2, then a or a/8 is a perfect square. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c, then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution, too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with a < c < e, then (8*d^2, d*(f-b)) is a solution, too. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation 0 + 1 + 2 + ... + x = y^2 with p < r, then r = 3p + 4q + 1 and s = 2p + 3q + 1. - Mohamed Bouhamida, Sep 02 2009
Also numbers k such that (ceiling(sqrt(k*(k+1)/2)))^2 - k*(k+1)/2 = 0. - Ctibor O. Zizka, Nov 10 2009
From Lekraj Beedassy, Mar 04 2011: (Start)
Let x=a(n) be the index of the associated triangular number T_x=1+2+3+...+x and y=A001109(n) be the base of the associated perfect square S_y=y^2. Now using the identity S_y = T_y + T_{y-1}, the defining T_x = S_y may be rewritten as T_y = T_x - T_{y-1}, or 1+2+3+...+y = y+(y+1)+...+x. This solves the Strand Magazine House Number problem mentioned in A001109 in references from Poo-Sung Park and John C. Butcher. In a variant of the problem, solving the equation 1+3+5+...+(2*x+1) = (2*x+1)+(2*x+3)+...+(2*y-1) implies S_(x+1) = S_y - S_x, i.e., with (x,x+1,y) forming a Pythagorean triple, the solutions are given by pairs of x=A001652(n), y=A001653(n). (End)
If P = 8*n +- 1 is a prime, then P divides a((P-1)/2); e.g., 7 divides a(3) and 41 divides a(20). Also, if P = 8*n +- 3 is prime, then 4*P divides (a((P-1)/2) + a((P+1)/2) + 3). - Kenneth J Ramsey, Mar 05 2012
Starting at a(2), a(n) gives all the dimensions of Euclidean k-space in which the ratio of outer to inner Soddy hyperspheres' radii for k+1 identical kissing hyperspheres is rational. The formula for this ratio is (1+3k+2*sqrt(2k*(k+1)))/(k-1) where k is the dimension. So for a(3) = 49, the ratio is 6 in the 49th dimension. See comment for A010502. - Frank M Jackson, Feb 09 2013
Conjecture: For n>1 a(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015
For n=2*k, k>0, a(n) is divisible by 8 (deficient), so since all proper divisors of deficient numbers are deficient, then a(n) is deficient. For n=2*k+1, k>0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. sigma(a(5)) = 1723 < 3362 = 2*a(5). In either case, a(n) is deficient. - Muniru A Asiru, Apr 14 2016
The squares of NSW numbers (A008843) interleaved with twice squares from A084703, where A008843(n) = A002315(n)^2 and A084703(n) = A001542(n)^2. Conjecture: Also numbers n such that sigma(n) = A000203(n) and sigma(n-th triangular number) = A074285(n) are both odd numbers. - Jaroslav Krizek, Aug 05 2016
For n > 0, numbers for which the number of odd divisors of both n and of n + 1 is odd. - Gionata Neri, Apr 30 2018
a(n) will be solutions to some (A000217(k) + A000217(k+1))/2. - Art Baker, Jul 16 2019
For n >= 2, a(n) is the base for which A058331(A001109(n)) is a length-3 repunit. Example: for n=2, A001109(2)=6 and A058331(6)=73 and 73 in base a(2)=8 is 111. See Grantham and Graves. - Michel Marcus, Sep 11 2020

Examples

			a(1) = ((3 + 2*sqrt(2)) + (3 - 2*sqrt(2)) - 2) / 4 = (3 + 3 - 2) / 4 = 4 / 4 = 1;
a(2) = ((3 + 2*sqrt(2))^2 + (3 - 2*sqrt(2))^2 - 2) / 4 = (9 + 4*sqrt(2) + 8 + 9 - 4*sqrt(2) + 8 - 2) / 4 = (18 + 16 - 2) / 4 = (34 - 2) / 4 = 32 / 4 = 8, etc.
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 204.
  • 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. 10.
  • M. S. Klamkin, "International Mathematical Olympiads 1978-1985," (Supplementary problem N.T.6)
  • W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, pp. 21-22 MR2002669
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 257-258.

Crossrefs

Partial sums of A002315. A000129, A005319.
a(n) = A115598(n), n > 0. - Hermann Stamm-Wilbrandt, Jul 27 2014

Programs

  • Haskell
    a001108 n = a001108_list !! n
    a001108_list = 0 : 1 : map (+ 2)
       (zipWith (-) (map (* 6) (tail a001108_list)) a001108_list)
    -- Reinhard Zumkeller, Jan 10 2012
    
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(1+x)/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018
  • Maple
    A001108:=-(1+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation, without the leading 0
  • Mathematica
    Table[(1/2)(-1 + Sqrt[1 + Expand[8(((3 + 2Sqrt[2])^n - (3 - 2Sqrt[2])^n)/(4Sqrt[2]))^2]]), {n, 0, 100}] (* Artur Jasinski, Dec 10 2006 *)
    Transpose[NestList[{#[[2]],#[[3]],6#[[3]]-#[[2]]+2}&,{0,1,8},20]][[1]] (* Harvey P. Dale, Sep 04 2011 *)
    LinearRecurrence[{7, -7, 1}, {0, 1, 8}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
  • PARI
    a(n)=(real((3+quadgen(32))^n)-1)/2
    
  • PARI
    a(n)=(subst(poltchebi(abs(n)),x,3)-1)/2
    
  • PARI
    a(n)=if(n<0,a(-n),(polsym(1-6*x+x^2,n)[n+1]-2)/4)
    
  • PARI
    x='x+O('x^99); concat(0, Vec(x*(1+x)/((1-x)*(1-6*x+x^2)))) \\ Altug Alkan, May 01 2018
    

Formula

a(0) = 0, a(n+1) = 3*a(n) + 1 + 2*sqrt(2*a(n)*(a(n)+1)). - Jim Nastos, Jun 18 2002
a(n) = floor( (1/4) * (3+2*sqrt(2))^n ). - Benoit Cloitre, Sep 04 2002
a(n) = A001653(k)*A001653(k+n) - A001652(k)*A001652(k+n) - A046090(k)*A046090(k+n). - Charlie Marion, Jul 01 2003
a(n) = A001652(n-1) + A001653(n-1) = A001653(n) - A046090(n) = (A001541(n)-1)/2 = a(-n). - Michael Somos, Mar 03 2004
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Antonio Alberto Olivares, Oct 23 2003
a(n) = Sum_{r=1..n} 2^(r-1)*binomial(2n, 2r). - Lekraj Beedassy, Aug 21 2004
If n > 1, then both A000203(n) and A000203(n+1) are odd numbers: n is either a square or twice a square. - Labos Elemer, Aug 23 2004
a(n) = (T(n, 3)-1)/2 with Chebyshev's polynomials of the first kind evaluated at x=3: T(n, 3) = A001541(n). - Wolfdieter Lang, Oct 18 2004
G.f.: x*(1+x)/((1-x)*(1-6*x+x^2)). Binet form: a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n - 2)/4. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
a(n) = floor(sqrt(2*A001110(n))) = floor(A001109(n)*sqrt(2)) = 2*(A000129(n)^2) - (n mod 2) = A001333(n)^2 - 1 + (n mod 2). - Henry Bottomley, Apr 19 2000, corrected by Eric Rowland, Jun 23 2017
A072221(n) = 3*a(n) + 1. - David Scheers, Dec 25 2006
A028982(a(n)) + 1 = A028982(a(n) + 1). - Juri-Stepan Gerasimov, Mar 28 2011
a(n+1)^2 + a(n)^2 + 1 = 6*a(n+1)*a(n) + 2*a(n+1) + 2*a(n). - Charlie Marion, Sep 28 2011
a(n) = 2*A001653(m)*A053141(n-m-1) + A002315(m)*A046090(n-m-1) + a(m) with m < n; otherwise, a(n) = 2*A001653(m)*A053141(m-n) - A002315(m)*A001652(m-n) + a(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = A048739(2n-2), n > 0. - Richard R. Forberg, Aug 31 2013
From Peter Bala, Jan 28 2014: (Start)
A divisibility sequence: that is, a(n) divides a(n*m) for all n and m. Case P1 = 8, P2 = 12, Q = 1 of the 3-parameter family of linear divisibility sequences found by Williams and Guy.
a(2*n+1) = A002315(n)^2 = Sum_{k = 0..4*n + 1} Pell(n), where Pell(n) = A000129(n).
a(2*n) = (1/2)*A005319(n)^2 = 8*A001109(n)^2.
(2,1) entry of the 2 X 2 matrix T(n,M), where M = [0, -3; 1, 4] and T(n,x) is the Chebyshev polynomial of the first kind. (End)
E.g.f.: exp(x)*(exp(2*x)*cosh(2*sqrt(2)*x) - 1)/2. - Stefano Spezia, Oct 25 2024

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Apr 19 2000
More terms from Lekraj Beedassy, Aug 21 2004

A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

Original entry on oeis.org

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679, 34481, 39775, 45585, 51935, 58849, 66351, 74465, 83215, 92625, 102719, 113521, 125055, 137345, 150415, 164289, 178991
Offset: 1

Views

Author

Keywords

Comments

Final digits of a(n), i.e., a(n) mod 10, are repeated periodically with period of length 5 {1,5,5,5,9}. There is a symmetry in this list since the sum of two numbers equally distant from the ends is equal to 10 = 1 + 9 = 5 + 5 = 2*5. Last two digits of a(n), i.e., a(n) mod 100, are repeated periodically with period of length 50. - Alexander Adamchuk, Aug 11 2006
a(n) = VarScheme(n,2) in the scheme displayed in A128195. - Peter Luschny, Feb 26 2007
If Y is a 3-subset of a 2n-set X then, for n >= 2, a(n-2) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
The numbers are the constant number found in magic squares of order n, where n is an odd number, see the comment in A006003. A Magic Square of side 1 is 1; 3 is 15; 5 is 65 and so on. - David Quentin Dauthier, Nov 07 2008
Two times the area of the triangle with vertices at (0,0), ((n - 1)^2, n^2), and (n^2, (n - 1)^2). - J. M. Bergot, Jun 25 2013
Bisection of A006003. - Omar E. Pol, Sep 01 2018
Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals a(n+1). The first five rows of array M are [1, 7, 17, 31, 49, ...]; [3, 5, 15, 29, 47, ...]; [9, 11, 13, 27, 45, ...]; [19, 21, 23, 25, 43, ...]; [33, 35, 37, 39, 41, ...]. - J. M. Bergot, Jul 16 2013 [This contribution was moved here from A047926 by Petros Hadjicostas, Mar 08 2021.]
For n>=2, these are the primitive sides s of squares of type 2 described in A344332. - Bernard Schott, Jun 04 2021
(a(n) + 1) / 2 = A212133(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 21 2024

References

  • J. H. Conway and R. K. Guy, The Book of Numbers, p. 53.
  • E. Deza and M. M. Deza, Figurate Numbers, World Scientific Publishing, 2012, pp. 123-124.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A063493, A063494, A063495, A063496.
Column k=3 of A047969.

Programs

  • Haskell
    a005917 n = a005917_list !! (n-1)
    a005917_list = map sum $ f 1 [1, 3 ..] where
       f x ws = us : f (x + 2) vs where (us, vs) = splitAt x ws
    -- Reinhard Zumkeller, Nov 13 2014
    
  • Magma
    [n^4 - (n-1)^4: n in [1..50]]; // Vincenzo Librandi, Aug 01 2011
    
  • Mathematica
    Table[n^4-(n-1)^4,{n,40}]  (* Harvey P. Dale, Apr 01 2011 *)
    #[[2]]-#[[1]]&/@Partition[Range[0,40]^4,2,1] (* More efficient than the above Mathematica program because it only has to calculate each 4th power once *) (* Harvey P. Dale, Feb 07 2015 *)
    Differences[Range[0,40]^4] (* Harvey P. Dale, Aug 11 2023 *)
  • PARI
    a(n)=n^4-(n-1)^4 \\ Charles R Greathouse IV, Jul 31 2011
    
  • Python
    A005917_list, m = [], [24, -12, 2, 1]
    for _ in range(10**2):
        A005917_list.append(m[-1])
        for i in range(3):
            m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

Formula

a(n) = (2*n - 1)*(2*n^2 - 2*n + 1).
Sum_{i=1..n} a(i) = n^4 = A000583(n). First differences of A000583.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
More generally, g.f. for n^m - (n - 1)^m is Euler(m, x)/(1 - x)^m, where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292). E.g.f.: x*(exp(y/(1 - x)) - exp(x*y/(1 - x)))/(exp(x*y/(1 - x))-x*exp(y/(1 - x))). - Vladeta Jovovic, May 08 2002
a(n) = sum of the next (2*n - 1) odd numbers; i.e., group the odd numbers so that the n-th group contains (2*n - 1) elements like this: (1), (3, 5, 7), (9, 11, 13, 15, 17), (19, 21, 23, 25, 27, 29, 31), ... E.g., a(3) = 65 because 9 + 11 + 13 + 15 + 17 = 65. - Xavier Acloque, Oct 11 2003
a(n) = 2*n - 1 + 12*Sum_{i = 1..n} (i - 1)^2. - Xavier Acloque, Oct 16 2003
a(n) = (4*binomial(n,2) + 1)*sqrt(8*binomial(n,2) + 1). - Paul Barry, Mar 14 2004
Binomial transform of [1, 14, 36, 24, 0, 0, 0, ...], if the offset is 0. - Gary W. Adamson, Dec 20 2007
Sum_{i=1..n-1}(a(i) + a(i+1)) = 8*Sum_{i=1..n}(i^3 + i) = 16*A002817(n-1) for n > 1. - Bruno Berselli, Mar 04 2011
a(n+1) = a(n) + 2*(6*n^2 + 1) = a(n) + A005914(n). - Vincenzo Librandi, Mar 16 2011
a(n) = -a(-n+1). a(n) = (1/6)*(A181475(n) - A181475(n-2)). - Bruno Berselli, Sep 26 2011
a(n) = A045975(2*n-1,n) = A204558(2*n-1)/(2*n - 1). - Reinhard Zumkeller, Jan 18 2012
a(n+1) = Sum_{k=0..2*n+1} (A176850(n,k) - A176850(n-1,k))*(2*k + 1), n >= 1. - L. Edson Jeffery, Nov 02 2012
a(n) = A005408(n-1) * A001844(n-1) = (2*(n - 1) + 1) * (2*(n - 1)*n + 1) = A000290(n-1)*12 + 2 + a(n-1). - Bruce J. Nicholson, May 17 2017
a(n) = A007588(n) + A007588(n-1) = A000292(2n-1) + A000292(2n-2) + A000292(2n-3) = A002817(2n-1) - A002817(2n-2). - Bruce J. Nicholson, Oct 22 2017
a(n) = A005898(n-1) + 6*A000330(n-1) (cf. Deza, Deza, 2012, p. 123, Section 2.6.2). - Felix Fröhlich, Oct 01 2018
a(n) = A300758(n-1) + A005408(n-1). - Bruce J. Nicholson, Apr 23 2020
G.f.: polylog(-4, x)*(1-x)/x. See the Simon Plouffe formula above (with expanded numerator), and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 10 2021

A000127 Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.

Original entry on oeis.org

1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457, 41449, 46938, 52956, 59536, 66712, 74519, 82993, 92171, 102091, 112792, 124314, 136698
Offset: 1

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Comments

a(n) is the sum of the first five terms in the n-th row of Pascal's triangle. - Geoffrey Critzer, Jan 18 2009
{a(k): 1 <= k <= 5} = divisors of 16. - Reinhard Zumkeller, Jun 17 2009
Equals binomial transform of [1, 1, 1, 1, 1, 0, 0, 0, ...]. - Gary W. Adamson, Mar 02 2010
From Bernard Schott, Apr 05 2021: (Start)
As a(n) = 2^(n-1) for n = 1..5, it is misleading to believe that a(n) = 2^(n-1) for n > 5 (see Patrick Popescu-Pampu link); other curiosities: a(6) = 2^5 - 1 and a(10) = 2^8.
The sequence of the first differences is A000125, the sequence of the second differences is A000124, the sequence of the third differences is A000027 and the sequence of the fourth differences is the all 1's sequence A000012 (see J. H. Conway and R. K. Guy reference, p. 80). (End)
a(n) is the number of binary words of length n matching the regular expression 0*1*0*1*0*. A000124 and A000125 count binary words of the form 0*1*0* and 1*0*1*0*, respectively. - Manfred Scheucher, Jun 22 2023

Examples

			a(7)=99 because the first five terms in the 7th row of Pascal's triangle are 1 + 7 + 21 + 35 + 35 = 99. - _Geoffrey Critzer_, Jan 18 2009
G.f. = x + 2*x^2 + 4*x^3 + 8*x^4 + 16*x^5 + 31*x^6 + 57*x^7 + 99*x^8 + 163*x^9 + ...
		

References

  • R. B. Banks, Slicing Pizzas, Racing Turtles and Further Adventures in Applied Mathematics, Princeton Univ. Press, 1999. See p. 28.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 72, Problem 2.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, Chap. 3.
  • J. H. Conway and R. K. Guy, Le Livre des Nombres, Eyrolles, 1998, p. 80.
  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problem 33 pp. 18; 128 Ellipses Paris 2004.
  • A. Deledicq and D. Missenard, A La Recherche des Régions Perdues, Math. & Malices, No. 22 Summer 1995 issue pp. 22-3 ACL-Editions Paris.
  • M. Gardner, Mathematical Circus, pp. 177; 180-1 Alfred A. Knopf NY 1979.
  • M. Gardner, The Colossal Book of Mathematics, 2001, p. 561.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • M. de Guzman, Aventures Mathématiques, Prob. B pp. 115-120 PPUR Lausanne 1990.
  • Ross Honsberger; Mathematical Gems I, Chap. 9.
  • Ross Honsberger; Mathematical Morsels, Chap. 3.
  • Jeux Mathématiques et Logiques, Vol. 3 pp. 12; 51 Prob. 14 FFJM-SERMAP Paris 1988.
  • J. N. Kapur, Reflections of a Mathematician, Chap.36, pp. 337-343, Arya Book Depot, New Delhi 1996.
  • C. D. Miller, V. E. Heeren, J. Hornsby, M. L. Morrow and J. Van Newenhizen, Mathematical Ideas, Tenth Edition, Pearson, Addison-Wesley, Boston, 2003, Cptr 1, 'The Art of Problem Solving, page 6.
  • I. Niven, Mathematics of Choice, pp. 158; 195 Prob. 40 NML 15 MAA 1965.
  • C. S. Ogilvy, Tomorrow's Math, pp. 144-6 OUP 1972.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 252-255.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers, Prometheus Books, NY, 2007, page 81-87.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 213.
  • 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).

Crossrefs

Programs

  • Haskell
    a000127 = sum . take 5 . a007318_row  -- Reinhard Zumkeller, Nov 24 2012
    
  • Magma
    [(n^4-6*n^3+23*n^2-18*n+24)/24: n in [1..50]]; // Vincenzo Librandi, Feb 16 2015
    
  • Maple
    A000127 := n->(n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24;
    with (combstruct):ZL:=[S, {S=Sequence(U, card=1)}, unlabeled]: seq(count(subs(r=6, ZL), size=m), m=0..41); # Zerinvary Lajos, Mar 08 2008
  • Mathematica
    f[n_] := Sum[Binomial[n, i], {i, 0, 4}]; Table[f@n, {n, 0, 40}] (* Robert G. Wilson v, Jun 29 2007 *)
    Total/@Table[Binomial[n-1,k],{n,50},{k,0,4}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,2,4,8,16},50] (* Harvey P. Dale, Aug 24 2011 *)
    Table[(n^4 - 6 n^3 + 23 n^2 - 18 n + 24) / 24, {n, 100}] (* Vincenzo Librandi, Feb 16 2015 *)
    a[ n_] := Binomial[n, 4] + Binomial[n, 2] + 1; (* Michael Somos, Dec 23 2017 *)
  • PARI
    a(n)=(n^4-6*n^3+23*n^2-18*n+24)/24 \\ Charles R Greathouse IV, Mar 22 2016
    
  • PARI
    {a(n) = binomial(n, 4) + binomial(n, 2) + 1}; /* Michael Somos, Dec 23 2017 */
    
  • Python
    def A000127(n): return n*(n*(n*(n - 6) + 23) - 18)//24 + 1 # Chai Wah Wu, Sep 18 2021

Formula

a(n) = C(n-1, 4) + C(n-1, 3) + ... + C(n-1, 0) = A055795(n) + 1 = C(n, 4) + C(n-1, 2) + n.
a(n) = Sum_{k=0..2} C(n, 2k). - Joel Sanderi (sanderi(AT)itstud.chalmers.se), Sep 08 2004
a(n) = (n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24.
G.f.: (1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^5. (for offset 0) - Simon Plouffe in his 1992 dissertation
E.g.f.: (1 + x + x^2/2 + x^3/6 + x^4/24)*exp(x) (for offset 0). [Typos corrected by Juan M. Marquez, Jan 24 2011]
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 4. - Harvey P. Dale, Aug 24 2011
a(n) = A000124(A000217(n-1)) - n*A000217(n-2) - A034827(n), n > 1. - Melvin Peralta, Feb 15 2016
a(n) = A223718(-n). - Michael Somos, Dec 23 2017
For n > 2, a(n) = n + 1 + sum_{i=2..(n-2)}sum_{j=1..(n-i)}(1+(i-1)(j-1)). - Alec Jones, Nov 17 2019

Extensions

Formula corrected and additional references from torsten.sillke(AT)lhsystems.com
Additional correction from Jonas Paulson (jonasso(AT)sdf.lonestar.org), Oct 30 2003
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