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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 38.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • Leonard E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 2.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Crossrefs

a(n)= A093562(n+1, 2), (5, 1)-Pascal column.
Cf. sequences listed in A254963.

Programs

  • Haskell
    a000566 n = n * (5 * (n - 1) + 2) `div` 2
    a000566_list = scanl (+) 0 a016861_list  -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    a000566:=func< n | n*(5*n-3) div 2 >; [ a000566(n): n in [0..50] ];
    
  • Maple
    A000566 := proc(n)
            n*(5*n-3)/2 ;
    end proc:
    seq(A000566(n),n=0..30); # R. J. Mathar, Oct 02 2020
  • Mathematica
    Table[n (5n - 3)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 7}, 50] (* Harvey P. Dale, Oct 13 2011 *)
    Join[{0},Accumulate[Range[1,315,5]]] (* Harvey P. Dale, Mar 26 2016 *)
    (* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[7], n], {n, 0, 48}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
    PolygonalNumber[7,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 23 2021 *)
  • Maxima
    makelist(n*(5*n-3)/2, n, 0, 20); /* Martin Ettl, Dec 11 2012 */
    
  • PARI
    a(n) = n * (5*n - 3) / 2
    
  • Python
    # Intended to compute the initial segment of the sequence, not isolated terms.
    def aList():
         x, y = 1, 1
         yield 0
         while True:
             yield x
             x, y = x + y + 5, y + 5
    A000566 = aList()
    print([next(A000566) for i in range(49)]) # Peter Luschny, Aug 04 2019
    
  • Python
    [n*(5*n-3)//2 for n in range(50)] # Gennady Eremin, Mar 24 2022

Formula

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

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A007283 a(n) = 3*2^n.

Original entry on oeis.org

3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368, 1610612736, 3221225472, 6442450944, 12884901888
Offset: 0

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Author

Keywords

Comments

Same as Pisot sequences E(3, 6), L(3, 6), P(3, 6), T(3, 6). See A008776 for definitions of Pisot sequences.
Numbers k such that A006530(A000010(k)) = A000010(A006530(k)) = 2. - Labos Elemer, May 07 2002
Also least number m such that 2^n is the smallest proper divisor of m which is also a suffix of m in binary representation, see A080940. - Reinhard Zumkeller, Feb 25 2003
Length of the period of the sequence Fibonacci(k) (mod 2^(n+1)). - Benoit Cloitre, Mar 12 2003
The sequence of first differences is this sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005
Subsequence of A122132. - Reinhard Zumkeller, Aug 21 2006
Apart from the first term, a subsequence of A124509. - Reinhard Zumkeller, Nov 04 2006
Total number of Latin n-dimensional hypercubes (Latin polyhedra) of order 3. - Kenji Ohkuma (k-ookuma(AT)ipa.go.jp), Jan 10 2007
Number of different ternary hypercubes of dimension n. - Edwin Soedarmadji (edwin(AT)systems.caltech.edu), Dec 10 2005
For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n + 1} -> {1, 2, 3} such that for fixed, different x_1, x_2,...,x_n in {1, 2, ..., n + 1} and fixed y_1, y_2,...,y_n in {1, 2, 3} we have f(x_i) <> y_i, (i = 1,2,...,n). - Milan Janjic, May 10 2007
a(n) written in base 2: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, n times 0 (see A003953). - Jaroslav Krizek, Aug 17 2009
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
Numbers containing the number 3 in their Collatz trajectories. - Reinhard Zumkeller, Feb 20 2012
a(n-1) gives the number of ternary numbers with n digits with no two adjacent digits in common; e.g., for n=3 we have 010, 012, 020, 021, 101, 102, 120, 121, 201, 202, 210 and 212. - Jon Perry, Oct 10 2012
If n > 1, then a(n) is a solution for the equation sigma(x) + phi(x) = 3x-4. This equation also has solutions 84, 3348, 1450092, ... which are not of the form 3*2^n. - Farideh Firoozbakht, Nov 30 2013
a(n) is the upper bound for the "X-ray number" of any convex body in E^(n + 2), conjectured by Bezdek and Zamfirescu, and proved in the plane E^2 (see the paper by Bezdek and Zamfirescu). - L. Edson Jeffery, Jan 11 2014
If T is a topology on a set V of size n and T is not the discrete topology, then T has at most 3 * 2^(n-2) many open sets. See Brown and Stephen references. - Ross La Haye, Jan 19 2014
Comment from Charles Fefferman, courtesy of Doron Zeilberger, Dec 02 2014: (Start)
Fix a dimension n. For a real-valued function f defined on a finite set E in R^n, let Norm(f, E) denote the inf of the C^2 norms of all functions F on R^n that agree with f on E. Then there exist constants k and C depending only on the dimension n such that Norm(f, E) <= C*max{ Norm(f, S) }, where the max is taken over all k-point subsets S in E. Moreover, the best possible k is 3 * 2^(n-1).
The analogous result, with the same k, holds when the C^2 norm is replaced, e.g., by the C^1, alpha norm (0 < alpha <= 1). However, the optimal analogous k, e.g., for the C^3 norm is unknown.
For the above results, see Y. Brudnyi and P. Shvartsman (1994). (End)
Also, coordination sequence for (infinity, infinity, infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015
The average of consecutive powers of 2 beginning with 2^1. - Melvin Peralta and Miriam Ong Ante, May 14 2016
For n > 1, a(n) is the smallest Zumkeller number with n divisors that are also Zumkeller numbers (A083207). - Ivan N. Ianakiev, Dec 09 2016
Also, for n >= 2, the number of length-n strings over the alphabet {0,1,2,3} having only the single letters as nonempty palindromic subwords. (Corollary 21 in Fleischer and Shallit) - Jeffrey Shallit, Dec 02 2019
Also, a(n) is the minimum link-length of any covering trail, circuit, path, and cycle for the set of the 2^(n+2) vertices of an (n+2)-dimensional hypercube. - Marco Ripà, Aug 22 2022
The known fixed points of maps n -> A163511(n) and n -> A243071(n). [See comments in A163511]. - Antti Karttunen, Sep 06 2023
The finite subsequence a(3), a(4), a(5), a(6) = 24, 48, 96, 192 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A000244 (see comment there). - Felix Huber, Feb 15 2024
A level 1 Sierpiński triangle is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles. For n>2, a(n-3) is the radius of the level n Sierpiński triangle graph. - Allan Bickle, Aug 03 2024

References

  • Jason I. Brown, Discrete Structures and Their Interactions, CRC Press, 2013, p. 71.
  • T. Ito, Method, equipment, program and storage media for producing tables, Publication number JP2004-272104A, Japan Patent Office (written in Japanese, a(2)=12, a(3)=24, a(4)=48, a(5)=96, a(6)=192, a(7)=384 (a(7)=284 was corrected)).
  • Kenji Ohkuma, Atsuhiro Yamagishi and Toru Ito, Cryptography Research Group Technical report, IT Security Center, Information-Technology Promotion Agency, JAPAN.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Subsequence of the following sequences: A029744, A029747, A029748, A029750, A362804 (after 3), A364494, A364496, A364289, A364291, A364292, A364295, A364497, A364964, A365422.
Essentially same as A003945 and A042950.
Row sums of (5, 1)-Pascal triangle A093562 and of (1, 5) Pascal triangle A096940.
Cf. Latin squares: A000315, A002860, A003090, A040082, A003191; Latin cubes: A098843, A098846, A098679, A099321.

Programs

Formula

G.f.: 3/(1-2*x).
a(n) = 2*a(n - 1), n > 0; a(0) = 3.
a(n) = Sum_{k = 0..n} (-1)^(k reduced (mod 3))*binomial(n, k). - Benoit Cloitre, Aug 20 2002
a(n) = A118416(n + 1, 2) for n > 1. - Reinhard Zumkeller, Apr 27 2006
a(n) = A000079(n) + A000079(n + 1). - Zerinvary Lajos, May 12 2007
a(n) = A000079(n)*3. - Omar E. Pol, Dec 16 2008
From Paul Curtz, Feb 05 2009: (Start)
a(n) = b(n) + b(n+3) for b = A001045, A078008, A154879.
a(n) = abs(b(n) - b(n+3)) with b(n) = (-1)^n*A084247(n). (End)
a(n) = 2^n + 2^(n + 1). - Jaroslav Krizek, Aug 17 2009
a(n) = A173786(n + 1, n) = A173787(n + 2, n). - Reinhard Zumkeller, Feb 28 2010
A216022(a(n)) = 6 and A216059(a(n)) = 7, for n > 0. - Reinhard Zumkeller, Sep 01 2012
a(n) = (A000225(n) + 1)*3. - Martin Ettl, Nov 11 2012
E.g.f.: 3*exp(2*x). - Ilya Gutkovskiy, May 15 2016
A020651(a(n)) = 2. - Yosu Yurramendi, Jun 01 2016
a(n) = sqrt(A014551(n + 1)*A014551(n + 2) + A014551(n)^2). - Ezhilarasu Velayutham, Sep 01 2019
a(A048672(n)) = A225546(A133466(n)). - Michel Marcus and Peter Munn, Nov 29 2019
Sum_{n>=1} 1/a(n) = 2/3. - Amiram Eldar, Oct 28 2020

A016861 a(n) = 5*n + 1.

Original entry on oeis.org

1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96, 101, 106, 111, 116, 121, 126, 131, 136, 141, 146, 151, 156, 161, 166, 171, 176, 181, 186, 191, 196, 201, 206, 211, 216, 221, 226, 231, 236, 241, 246, 251, 256, 261, 266, 271, 276, 281
Offset: 0

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Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

Numbers ending in 1 or 6.
Apart from initial terms, same as 5n-14.
Complement of A047203; A027445(a(n)) mod 10 = 4. - Reinhard Zumkeller, Oct 23 2006
Campbell reference shows: "A graph on n vertices with at least 4n-9 edges is intrinsically linked. A graph on n vertices with at least 5n-14 edges is intrinsically knotted." - Jonathan Vos Post, Jan 18 2007
Central terms of the triangle in A153125: a(n) = A153125(2*n+1, n+1). - Reinhard Zumkeller, Dec 20 2008
For n > 2, also the number of (not necessarily maximal) cliques in the n-Moebius ladder graph. - Eric W. Weisstein, Nov 29 2017
For n > 3, also the number of (not necessarily maximal) cliques in the n-prism graph. - Eric W. Weisstein, Nov 29 2017
For n >= 1, a(n) is the size of any hexagonal chain graph with n cells. - Christian Barrientos, Sarah Minion, Mar 07 2018
For n >= 1, a(n) is the number of possible outcomes of the summation when using n dice. - Bram Kole, Dec 24 2018
Numbers congruent to 1 (mod 5). - Muniru A Asiru, Jan 01 2019
Numbers k such that the k-th Fibonacci number, A000045(k), and the k-th Lucas number, A000032(k), end with the same decimal digit. - Amiram Eldar, Apr 15 2023

Crossrefs

Cf. A093562 ((5, 1) Pascal, column m=1).
Cf. A000566 (partial sums).

Programs

Formula

G.f.: (1+4*x)/(1-x)^2.
Row sums of triangle A131843. - Gary W. Adamson, Jul 21 2007
a(n) = 2*a(n-1) - a(n-2) with a(0)=1, a(1)=6. - Vincenzo Librandi, Aug 01 2010
a(n) = A017293(n)/2 = A008587(n)+1. - Wesley Ivan Hurt, May 03 2014
E.g.f.: exp(x)*(1 + 5*x). - Stefano Spezia, Mar 23 2021
Sum_{n>=0} (-1)^n/a(n) = sqrt(2+2/sqrt(5))*Pi/10 + log(phi)/sqrt(5) + log(2)/5, where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

Extensions

More terms from Reinhard Zumkeller, Oct 23 2006

A029653 Numbers in (2,1)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285
Offset: 0

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Author

Keywords

Comments

Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x)+1. See the Mathematica section. - Clark Kimberling, Feb 28 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013
The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Feb 25 2018

Examples

			The triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7  8  9 10 ...
0:  1
1:  2  1
2:  2  3  1
3:  2  5  4   1
4:  2  7  9   5   1
5:  2  9 16  14   6   1
6:  2 11 25  30  20   7   1
7:  2 13 36  55  50  27   8   1
8:  2 15 49  91 105  77  35   9  1
9:  2 17 64 140 196 182 112  44 10  1
10: 2 19 81 204 336 378 294 156 54 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1         \/1        \      /1        \
|2 1      ||0 1       ||0 1      |      |2 1      |
|2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    |
|2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  |
|2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1|
|...      ||...       ||...      |      |...      |
- _Peter Bala_, Dec 27 2014
		

References

  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Crossrefs

(d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).

Programs

  • Haskell
    a029653 n k = a029653_tabl !! n !! k
    a029653_row n = a029653_tabl !! n
    a029653_tabl = [1] : iterate
                   (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
    -- Reinhard Zumkeller, Dec 16 2013
    
  • Maple
    A029653 :=  proc(n,k)
    if n = 0 then
      1;
    else
      binomial(n-1, k)+binomial(n, k)
    fi
    end proc: # R. J. Mathar, Jun 30 2013
  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
    v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]  (* A208510 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]  (* A029653 *)
    (* Clark Kimberling, Feb 28 2012 *)
  • Python
    from sympy import Poly
    from sympy.abc import x
    def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
    def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
    def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
    for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017
    
  • Python
    from math import comb, isqrt
    def A029653(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*((r<<1)-a)//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005
Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

Extensions

More terms from James Sellers

A002413 Heptagonal (or 7-gonal) pyramidal numbers: a(n) = n*(n+1)*(5*n-2)/6.

Original entry on oeis.org

0, 1, 8, 26, 60, 115, 196, 308, 456, 645, 880, 1166, 1508, 1911, 2380, 2920, 3536, 4233, 5016, 5890, 6860, 7931, 9108, 10396, 11800, 13325, 14976, 16758, 18676, 20735, 22940, 25296, 27808, 30481, 33320, 36330, 39516, 42883, 46436, 50180, 54120
Offset: 0

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Author

Keywords

Comments

The partial sums of A000566. - R. J. Mathar, Mar 19 2008
A002413(n + 1) is the number of 4-tuples (w, x, y, z) having all terms in {0, ..., n} and w = floor((x + y + z)/2). - Clark Kimberling, May 28 2012
From Ant King, Oct 25 2012: (Start)
For n > 0, the digital roots of this sequence A010888(A002413(n)) form the purely periodic 27-cycle {1, 8, 8, 6, 7, 7, 2, 6, 6, 7, 5, 5, 3, 4, 4, 8, 3, 3, 4, 2, 2, 9, 1, 1, 5, 9, 9}.
For n > 0, the units' digits of this sequence A010879(A002413(n)) form the purely periodic 20-cycle {1, 8, 6, 0, 5, 6, 8, 6, 5, 0, 6, 8, 1, 0, 0, 6, 3, 6, 0, 0}.
(End)

Examples

			For n=7, a(7) = 7*1 + 6*6 + 5*11 + 4*16 + 3*21 + 2*26 + 1*31 = 308. - _Bruno Berselli_, Feb 10 2014
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
  • 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.
  • 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

Cf. A093562 ((5, 1) Pascal, column m = 3).
Cf. similar sequences listed in A237616.

Programs

Formula

a(n) = n*(n + 1)*(5*n - 2)/6.
G.f.: x*(1 + 4*x)/(1 - x)^4. [Suggested by Simon Plouffe in his 1992 dissertation.]
From Ant King, Oct 25 2012: (Start)
a(n) = a(n - 1) + n*(5*n - 3)/2.
a(n) = 3*a(n - 1) - 3*a(n - 2) + a(n - 3) + 5.
a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4)
a(n) = (n + 1)*(2*A000566(n) + n)/6 = (5*n - 2)*A000217(n)/3.
a(n) = A000292(n) + 4*A000292(n - 1)
a(n) = A002412(n) + A000292(n - 1)
a(n) = A000217(n) + 5*A000292(n - 1)
a(n) = binomial(n + 2, 3) + 4*binomial(n + 1, 3) = (5*n - 2) * binomial(n + 1, 2)/3.
Sum_{n >= 1} 1/a(n) = 15*(log(3125) + sqrt(5)*log((3 - sqrt(5))/2) - 2*Pi*sqrt(5*(5 - 2*sqrt(5)))/5 - 8/5)/28 = 1.207293...
(End)
a(n) = Sum_{i=0..n-1} (n-i)*(5*i+1). - Bruno Berselli, Feb 10 2014
a(n) = A080851(5,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: x*(6 + 18*x + 5*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017
a(n) = Sum_{i=0..n-1} (n+2*i)*(n-i). - Leonid Bedratyuk, Jul 09 2024

Extensions

More terms from James Sellers, Dec 23 1999
a(0)=0 prepended by Max Alekseyev, Nov 23 2011

A022095 Fibonacci sequence beginning 1, 5.

Original entry on oeis.org

1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500, 809, 1309, 2118, 3427, 5545, 8972, 14517, 23489, 38006, 61495, 99501, 160996, 260497, 421493, 681990, 1103483, 1785473, 2888956, 4674429, 7563385, 12237814, 19801199, 32039013, 51840212, 83879225, 135719437
Offset: 0

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Author

Keywords

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(5; n-1-k, k), n >= 1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5; n, k), the (5,1) Pascal triangle A093562. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666.
Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28, ...). - Gary W. Adamson, Jul 14 2007
In general, for a Fibonacci sequence beginning with 1,b we have:
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2b) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2b)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Subsequence of primes: 5, 11, 17, 73, 191, 809, 421493, 1103483, ... . - R. J. Mathar, Aug 09 2012
Pisano periods: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 9, 60, ... (differs from A001175). - R. J. Mathar, Aug 10 2012

Crossrefs

Row n=4 of A109754 (shifted).

Programs

  • GAP
    List([0..40],n->4*Fibonacci(n)+Fibonacci(n+1)); # Muniru A Asiru, Mar 04 2018
    
  • Magma
    a0:=1; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013
    
  • Maple
    with(combinat): a:= n-> 4*fibonacci(n)+fibonacci(n+1): seq(a(n), n=0..32); # Zerinvary Lajos, Oct 05 2007
  • Mathematica
    f[n_] := (LucasL[n - 2] + 8*LucasL[n - 1] + 4*LucasL[n] + 2*LucasL[n + 1])/5; Array[f, 38, 0] (* or *)
    LinearRecurrence[{1, 1}, {1, 5}, 38] (* Robert G. Wilson v, Oct 22 2012 *)
  • PARI
    a(n)=fibonacci(n-1)+5*fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011
    
  • SageMath
    A022095=BinaryRecurrenceSequence(1,1,1,5)
    [A022095(n) for n in range(41)] # G. C. Greubel, Jun 02 2025

Formula

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=5.
G.f.: (1+4*x)/(1-x-x^2).
a(n) = 4*Fibonacci(n) + Fibonacci(n+1), n >= 1. - Zerinvary Lajos, Oct 05 2007, corrected by R. J. Mathar, Apr 07 2011
a(n-1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 2*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 4*Fibonacci(n+2) - 3*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = (L(n-2) + 8*L(n-1) + 4*L(n) + 2*L(n+1))/5 for the Lucas numbers L(n). - J. M. Bergot, Oct 22 2012
a(n) = ((2*sqrt(5) - 1)*(((1 + sqrt(5))/2)^(n+1)) + (2*sqrt(5) + 1)*(((1 - sqrt(5))/2)^(n+1)))/(sqrt(5)). - Bogart B. Strauss, Jul 19 2013
a(n) = Lucas(n-1) + Fibonacci(n+3) = Lucas(n+2) - Fibonacci(n-3). - Greg Dresden and Griffin Donaldson, Mar 03 2022

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024
Offset: 1

Views

Author

Boris Putievskiy, Aug 15 2013

Keywords

Comments

The third row is (n^4 - n^2 + 24*n + 24)/12.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

Examples

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;
		

Crossrefs

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).
Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

Programs

  • GAP
    T:= function(n,k)
        if k=0 then return n^2;
        elif k=n then return 2^n;
        else return T(n-1,k-1) + T(n-1,k);
        fi;
      end;
    Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019
  • Maple
    T:= proc(n, k) option remember;
          if k=0 then n^2
        elif k=n then 2^k
        else T(n-1, k-1) + T(n-1, k)
          fi
        end:
    seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
    Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)
  • PARI
    T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019
    
  • Python
    def funcL(n):
       q = n**2
       return q
    def funcR(n):
       q = 2**n
       return q
    for n in range (1,9871):
       t=int((math.sqrt(8*n-7) - 1)/ 2)
       i=n-t*(t+1)/2-1
       j=(t*t+3*t+4)/2-n-1
       sum1=0
       sum2=0
       for m1 in range (1,i+1):
          sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
       for m2 in range (1,j+1):
          sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
       sum=sum1+sum2
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k==0): return n^2
        elif (k==n): return 2^n
        else: return T(n-1, k-1) + T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019
    

Formula

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Extensions

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A002418 4-dimensional figurate numbers: a(n) = (5*n-1)*binomial(n+2,3)/4.

Original entry on oeis.org

0, 1, 9, 35, 95, 210, 406, 714, 1170, 1815, 2695, 3861, 5369, 7280, 9660, 12580, 16116, 20349, 25365, 31255, 38115, 46046, 55154, 65550, 77350, 90675, 105651, 122409, 141085, 161820, 184760, 210056, 237864, 268345, 301665, 337995
Offset: 0

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Author

Keywords

Comments

Partial sums of A002413.
Principal diagonal of the convolution array A213550, for n>0. - Clark Kimberling, Jun 17 2012
Convolution of A000027 with A000566. - Bruno Berselli, Dec 06 2012
Coefficients in the hypergeometric series identity 1 - 9*(x - 1)/(4*x + 1) + 35*(x - 1)*(x - 2)/((4*x + 1)*(4*x + 2)) - 95*(x - 1)*(x - 2)*(x - 3)/((4*x + 1)*(4*x + 2)*(4*x + 3)) + ... = 0, valid for Re(x) > 1. Cf. A000326 and A002412. Column 4 of A103450. - Peter Bala, Mar 14 2019

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
  • 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

Cf. A093562 ((5, 1) Pascal, column m=4).
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

Programs

  • GAP
    List([0..40],n->(5*n-1)*Binomial(n+2,3)/4); # Muniru A Asiru, Mar 18 2019
    
  • Magma
    [(5*n - 1)*Binomial(n + 2, 3)/4: n in [0..40]]; // Vincenzo Librandi, Oct 17 2012
    
  • Magma
    /* A000027 convolved with A000566: */ A000566:=func; [&+[(n-i+1)*A000566(i): i in [0..n]]: n in [0..35]]; // Bruno Berselli, Dec 06 2012
    
  • Mathematica
    Table[(5n-1) Binomial[n+2,3]/4,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,1,9,35,95},40] (* Harvey P. Dale, Oct 16 2012 *)
    CoefficientList[Series[x*(1 + 4*x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Oct 17 2012 *)
  • PARI
    a(n)=(5*n-1)*binomial(n+2,3)/4 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Sage
    [(5*n-1)*binomial(n+2,3)/4 for n in (0..40)] # G. C. Greubel, Jul 03 2019

Formula

G.f.: x*(1+4*x)/(1-x)^5. - Simon Plouffe in his 1992 dissertation.
Starting (1, 9, 35, 95, ...), = A128064 * A000332, (A000332 starting 1, 5, 15, 35, 70, ...), such that a(n) = n*C(n+3,4) - (n-1)*C(n+2,4). E.g., a(5) = 210 = 5*C(8,4) - 4*C(7,4) = 5*70 - 4*35. - Gary W. Adamson, Dec 28 2007
Unit digit, A010879(a(n)), is one of {0,1,9,5,6,4} [Eric Desbiaux] because a(n) mod 5 = 0,1,4,0,0, periodic with period 5. [Proof: A002413(n) mod 5 = 1,3,1,0,0 with period 5 and a(n) are the partial sums of A002413.] - R. J. Mathar, Mar 19 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Oct 16 2012
a(n) = A080852(5,n-1). - R. J. Mathar, Jul 28 2016
a(n) = Sum_{i=0..n} (n-i) * Sum_{j=i..n} j. - J. M. Bergot, May 30 2017
E.g.f.: x*(24 + 84*x + 44*x^2 + 5*x^3)*exp(x)/4!. - G. C. Greubel, Jul 03 2019
Sum_{n>=1} 1/a(n) = (50*sqrt(5)*log(phi) + 125*log(5) - 50*sqrt(1+2/sqrt(5))*Pi - 26)/11, where phi is the golden ratio (A001622). - Amiram Eldar, Feb 11 2022

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x) = (1+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

Examples

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 1];
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.

Programs

  • Haskell
    a093561 n k = a093561_tabl !! n !! k
    a093561_row n = a093561_tabl !! n
    a093561_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
    -- Reinhard Zumkeller, Aug 31 2014
    
  • Python
    from math import comb, isqrt
    def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

Formula

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A027800 a(n) = (n+1)*binomial(n+4, 4).

Original entry on oeis.org

1, 10, 45, 140, 350, 756, 1470, 2640, 4455, 7150, 11011, 16380, 23660, 33320, 45900, 62016, 82365, 107730, 138985, 177100, 223146, 278300, 343850, 421200, 511875, 617526, 739935, 881020, 1042840, 1227600, 1437656, 1675520, 1943865, 2245530, 2583525
Offset: 0

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Author

Thi Ngoc Dinh (via R. K. Guy)

Keywords

Comments

Number of 9-subsequences of [1, n] with just 4 contiguous pairs.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 19 2005
Equals binomial transform of [1, 9, 26, 34, 21, 5, 0, 0, 0, ...]. - Gary W. Adamson, Jul 27 2008
a(n) equals the coefficient of x^4 of the characteristic polynomial of the (n+4) X (n+4) matrix with 2's along the main diagonal and 1's everywhere else (see Mathematica code below). - John M. Campbell, Jul 08 2011
Convolution of triangular numbers (A000217) and heptagonal numbers (A000566). - Bruno Berselli, Jun 27 2013

Examples

			By the fifth comment: A000217(1..6) and A000566(1..6) give the term a(6) = 1*21 + 7*15 + 18*10 + 34*6 + 55*3 + 81*1 = 756. - _Bruno Berselli_, Jun 27 2013
		

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
  • Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-8.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.233, # 9).

Crossrefs

Partial sums of A002418.
Cf. A000332, A093562 ((5, 1) Pascal, column m=5).

Programs

  • GAP
    List([0..40], n-> (n+1)*Binomial(n+4,4)); # G. C. Greubel, Aug 28 2019
  • Magma
    [(n+1)*Binomial(n+4,4): n in [0..40]]; // G. C. Greubel, Aug 28 2019
    
  • Maple
    a:=n->(n+1)^2*(n+2)*(n+3)*(n+4)/24: seq(a(n),n=0..40); # Emeric Deutsch
  • Mathematica
    Table[Coefficient[CharacteristicPolynomial[Array[KroneckerDelta[#1, #2] + 1 &, {n+4, n+4}], x], x^4], {n, 0, 40}] (* John M. Campbell, Jul 08 2011 *)
    Table[(n+1)Binomial[n+4, 4], {n,0,40}] (* or *) CoefficientList[Series[ (1+4x)/(1-x)^6, {x,0,40}], x] (* Michael De Vlieger, Jul 14 2017 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{1,10,45,140,350,756},40] (* Harvey P. Dale, Aug 04 2020 *)
  • PARI
    vector(40, n, n*binomial(n+3,4)) \\ G. C. Greubel, Aug 28 2019
    
  • Sage
    [(n+1)*binomial(n+4,4) for n in (0..40)] # G. C. Greubel, Aug 28 2019
    

Formula

G.f.: (1+4*x)/(1-x)^6.
a(n) = (n+1)*A000332(n+4).
Sum_{n>=0} 1/a(n) = (2/3)*Pi^2 - 49/9. - Jaume Oliver Lafont, Jul 14 2017
E.g.f.: exp(x)*(24 + 216*x + 312*x^2 + 136*x^3 + 21*x^4 + x^5)/24. - Stefano Spezia, May 08 2021
Sum_{n>=0} (-1)^n/a(n) = Pi^2/3 - 80*log(2)/3 + 145/9. - Amiram Eldar, Jan 28 2022
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