OFFSET
0,3
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
REFERENCES
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
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.
LINKS
Daniel Mondot, Table of n, a(n) for n = 0..10000 (first 1000 terms by T. D. Noe)
S. Barbero, U. Cerruti, and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7., section 4.4.
C. K. Cook and M. R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 341.
Bir Kafle, Florian Luca, and Alain Togbé, Pentagonal and heptagonal repdigits, Annales Mathematicae et Informaticae, pp. 137-145.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
B. Srinivasa Rao, Heptagonal Numbers in the Pell Sequence and Diophantine Equations 2x^2 = y^2(5y - 3)^2 ± 2, Fib. Quarterly, 43 (2005), 194-201.
B. Srinivasa Rao, Heptagonal numbers in the associated Pell sequence and Diophantine equations x^2(5x - 3)^2 = 8y^2 ± 4, Fib. Quarterly, 43 (2005), 302-306.
Leo Tavares, Illustration.
Eric Weisstein's World of Mathematics, Heptagonal Number.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
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) = 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
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
EXAMPLE
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
MAPLE
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 *)
PROG
(Magma) a000566:=func< n | n*(5*n-3) div 2 >; [ a000566(n): n in [0..50] ];
(PARI) a(n) = n * (5*n - 3) / 2
(Maxima) makelist(n*(5*n-3)/2, n, 0, 20); /* Martin Ettl, Dec 11 2012 */
(Haskell)
a000566 n = n * (5 * (n - 1) + 2) `div` 2
a000566_list = scanl (+) 0 a016861_list -- Reinhard Zumkeller, Jun 16 2013
(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
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
Partially edited by Joerg Arndt, Mar 11 2010
STATUS
approved