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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A016754 Odd squares: a(n) = (2n+1)^2. Also centered octagonal numbers.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025
Offset: 0

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Keywords

Comments

The brown rat (rattus norwegicus) breeds very quickly. It can give birth to other rats 7 times a year, starting at the age of three months. The average number of pups is 8. The present sequence gives the total number of rats, when the intervals are 12/7 of a year and a young rat starts having offspring at 24/7 of a year. - Hans Isdahl, Jan 26 2008
Numbers n such that tau(n) is odd where tau(x) denotes the Ramanujan tau function (A000594). - Benoit Cloitre, May 01 2003
If Y is a fixed 2-subset of a (2n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
Binomial transform of [1, 8, 8, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
All terms of this sequence are of the form 8k+1. For numbers 8k+1 which aren't squares see A138393. Numbers 8k+1 are squares iff k is a triangular number from A000217. And squares have form 4n(n+1)+1. - Artur Jasinski, Mar 27 2008
Sequence arises from reading the line from 1, in the direction 1, 25, ... and the line from 9, in the direction 9, 49, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Equals the triangular numbers convolved with [1, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
First differences: A008590(n) = a(n) - a(n-1) for n>0. - Reinhard Zumkeller, Nov 08 2009
Central terms of the triangle in A176271; cf. A000466, A053755. - Reinhard Zumkeller, Apr 13 2010
Odd numbers with odd abundance. Odd numbers with even abundance are in A088828. Even numbers with odd abundance are in A088827. Even numbers with even abundance are in A088829. - Jaroslav Krizek, May 07 2011
Appear as numerators in the non-simple continued fraction expansion of Pi-3: Pi-3 = K_{k>=1} (1-2*k)^2/6 = 1/(6+9/(6+25/(6+49/(6+...)))), see also the comment in A007509. - Alexander R. Povolotsky, Oct 12 2011
Ulam's spiral (SE spoke). - Robert G. Wilson v, Oct 31 2011
All terms end in 1, 5 or 9. Modulo 100, all terms are among { 1, 9, 21, 25, 29, 41, 49, 61, 69, 81, 89 }. - M. F. Hasler, Mar 19 2012
Right edge of both triangles A214604 and A214661: a(n) = A214604(n+1,n+1) = A214661(n+1,n+1). - Reinhard Zumkeller, Jul 25 2012
Also: Odd numbers which have an odd sum of divisors (= sigma = A000203). - M. F. Hasler, Feb 23 2013
Consider primitive Pythagorean triangles (a^2 + b^2 = c^2, gcd(a, b) = 1) with hypotenuse c (A020882) and respective even leg b (A231100); sequence gives values c-b, sorted with duplicates removed. - K. G. Stier, Nov 04 2013
For n>1 a(n) is twice the area of the irregular quadrilateral created by the points ((n-2)*(n-1),(n-1)*n/2), ((n-1)*n/2,n*(n+1)/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+2)*(n+3)/2,(n+1)*(n+2)/2). - J. M. Bergot, May 27 2014
Number of pairs (x, y) of Z^2, such that max(abs(x), abs(y)) <= n. - Michel Marcus, Nov 28 2014
Except for a(1)=4, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 737", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016
a(n) is the sum of 2n+1 consecutive numbers, the first of which is n+1. - Ivan N. Ianakiev, Dec 21 2016
a(n) is the number of 2 X 2 matrices with all elements in {0..n} with determinant = 2*permanent. - Indranil Ghosh, Dec 25 2016
Engel expansion of Pi*StruveL_0(1)/2 where StruveL_0(1) is A197037. - Benedict W. J. Irwin, Jun 21 2018
Consider all Pythagorean triples (X,Y,Z=Y+1) ordered by increasing Z; the segments on the hypotenuse {p = a(n)/A001844(n), q = A060300(n)/A001844(n) = A001844(n) - p} and their ratio p/q = a(n)/A060300(n) are irreducible fractions in Q\Z. X values are A005408, Y values are A046092, Z values are A001844. - Ralf Steiner, Feb 25 2020
a(n) is the number of large or small squares that are used to tile primitive squares of type 2 (A344332). - Bernard Schott, Jun 03 2021
Also, positive odd integers with an odd number of odd divisors (for similar sequence with 'even', see A348005). - Bernard Schott, Nov 21 2021
a(n) is the least odd number k = x + y, with 0 < x < y, such that there are n distinct pairs (x,y) for which x*y/k is an integer; for example, a(2) = 25 and the two corresponding pairs are (5,20) and (10,15). The similar sequence with 'even' is A016742 (see Comment of Jan 26 2018). - Bernard Schott, Feb 24 2023
From Peter Bala, Jan 03 2024: (Start)
The sequence terms are the exponents of q in the series expansions of the following infinite products:
1) q*Product_{n >= 1} (1 - q^(16*n))*(1 + q^(8*n)) = q + q^9 + q^25 + q^49 + q^81 + q^121 + q^169 + ....
2) q*Product_{n >= 1} (1 + q^(16*n))*(1 - q^(8*n)) = q - q^9 - q^25 + q^49 + q^81 - q^121 - q^169 + + - - ....
3) q*Product_{n >= 1} (1 - q^(8*n))^3 = q - 3*q^9 + 5*q^25 - 7*q^49 + 9*q^81 - 11*q^121 + 13*q^169 - + ....
4) q*Product_{n >= 1} ( (1 + q^(8*n))*(1 - q^(16*n))/(1 + q^(16*n)) )^3 = q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 - 15*q^225 + + - - .... (End)

References

  • L. Lorentzen and H. Waadeland, Continued Fractions with Applications, North-Holland 1992, p. 586.

Crossrefs

Cf. A000447 (partial sums).
Cf. A348005, A379481 [= a(A048673(n)-1)].
Partial sums of A022144.
Positions of odd terms in A341528.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = 1 + Sum_{i=1..n} 8*i = 1 + 8*A000217(n). - Xavier Acloque, Jan 21 2003; Zak Seidov, May 07 2006; Robert G. Wilson v, Dec 29 2010
O.g.f.: (1+6*x+x^2)/(1-x)^3. - R. J. Mathar, Jan 11 2008
a(n) = 4*n*(n + 1) + 1 = 4*n^2 + 4*n + 1. - Artur Jasinski, Mar 27 2008
a(n) = A061038(2+4n). - Paul Curtz, Oct 26 2008
Sum_{n>=0} 1/a(n) = Pi^2/8 = A111003. - Jaume Oliver Lafont, Mar 07 2009
a(n) = A000290(A005408(n)). - Reinhard Zumkeller, Nov 08 2009
a(n) = a(n-1) + 8*n with n>0, a(0)=1. - Vincenzo Librandi, Aug 01 2010
a(n) = A033951(n) + n. - Reinhard Zumkeller, May 17 2009
a(n) = A033996(n) + 1. - Omar E. Pol, Oct 03 2011
a(n) = (A005408(n))^2. - Zak Seidov, Nov 29 2011
From George F. Johnson, Sep 05 2012: (Start)
a(n+1) = a(n) + 4 + 4*sqrt(a(n)).
a(n-1) = a(n) + 4 - 4*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 8.
a(n+1) = 3*a(n) - 3*a(n-1) + a(n-2).
(a(n+1) - a(n-1))/8 = sqrt(a(n)).
a(n+1)*a(n-1) = (a(n)-4)^2.
a(n) = 2*A046092(n) + 1 = 2*A001844(n) - 1 = A046092(n) + A001844(n).
Limit_{n -> oo} a(n)/a(n-1) = 1. (End)
a(n) = binomial(2*n+2,2) + binomial(2*n+1,2). - John Molokach, Jul 12 2013
E.g.f.: (1 + 8*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 23 2016
a(n) = A101321(8,n). - R. J. Mathar, Jul 28 2016
Product_{n>=1} A033996(n)/a(n) = Pi/4. - Daniel Suteu, Dec 25 2016
a(n) = A014105(n) + A000384(n+1). - Bruce J. Nicholson, Nov 11 2017
a(n) = A003215(n) + A002378(n). - Klaus Purath, Jun 09 2020
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=0} a(n)/n! = 13*e.
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/e. (End)
Sum_{n>=0} (-1)^n/a(n) = A006752. - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 28 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi/4 (A003881). (End)
From Leo Tavares, Nov 24 2021: (Start)
a(n) = A014634(n) - A002943(n). See Diamond Triangles illustration.
a(n) = A003154(n+1) - A046092(n). See Diamond Stars illustration. (End)
From Peter Bala, Mar 11 2024: (Start)
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 ))))).
3/2 - 2*log(2) = Sum_{k >= 1} 1/(k*a(k)*a(k-1)) = 1/(9 - 3/(17 - 60/(33 - 315/(57 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*2^2 - ... ))))).
Row 2 of A142992. (End)
From Peter Bala, Mar 26 2024: (Start)
8*a(n) = (2*n + 1)*(a(n+1) - a(n-1)).
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 1/2 - Pi/8 = 1/(9 + (1*3)/(8 + (3*5)/(8 + ... + (4*n^2 - 1)/(8 + ... )))). For the continued fraction use Lorentzen and Waadeland, p. 586, equation 4.7.9 with n = 1. Cf. A057813. (End)

Extensions

Additional description from Terrel Trotter, Jr., Apr 06 2002

A016742 Even squares: a(n) = (2*n)^2.

Original entry on oeis.org

0, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100, 8464
Offset: 0

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4 times the squares.
Number of edges in the complete bipartite graph of order 5n, K_{n,4n} - Roberto E. Martinez II, Jan 07 2002
It is conjectured (I think) that a regular Hadamard matrix of order n exists iff n is an even square (cf. Seberry and Yamada, Th. 10.11). A Hadamard matrix is regular if the sum of the entries in each row is the same. - N. J. A. Sloane, Nov 13 2008
Sequence arises from reading the line from 0, in the direction 0, 16, ... and the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
The entries from a(1) on can be interpreted as pair sums of (2, 2), (8, 8), (18, 18), (32, 32) etc. that arise from a re-arrangement of the subshell orbitals in the periodic table of elements. 8 becomes the maximum number of electrons in the (2s,2p) or (3s,3p) orbitals, 18 the maximum number of electrons in (4s,3d,4p) or (5s,3d,5p) shells, for example. - Julio Antonio Gutiérrez Samanez, Jul 20 2008
The first two terms of the sequence (n=1, 2) give the numbers of chemical elements using only n types of atomic orbitals, i.e., there are a(1)=4 elements (H,He,Li,Be) where electrons reside only on s-orbitals, there are a(2)=16 elements (B,C,N,O,F,Ne,Na,Mg,Al,Si,P,S,Cl,Ar,K,Ca) where electrons reside only on s- and p-orbitals. However, after that, there is 37 (which is one more than a(3)=36) elements (from Sc, Scandium, atomic number 21 to La, Lanthanum, atomic number 57) where electrons reside only on s-, p- and d-orbitals. This is because Lanthanum (with the electron configuration [Xe]5d^1 6s^2) is an exception to the Aufbau principle, which would predict that its electron configuration is [Xe]4f^1 6s^2. - Antti Karttunen, Aug 14 2008.
Number of cycles of length 3 in the king's graph associated with an (n+1) X (n+1) chessboard. - Anton Voropaev (anton.n.voropaev(AT)gmail.com), Feb 01 2009
a(n+1) is the molecular topological index of the n-star graph S_n. - Eric W. Weisstein, Jul 11 2011
a(n) is the sum of two consecutives odd numbers 2*n^2-1 and 2*n^2+1 and the difference of two squares (n^2+1)^2 - (n^2-1)^2. - Pierre CAMI, Jan 02 2012
For n > 3, a(n) is the area of the irregular quadrilateral created by the points ((n-4)*(n-3)/2,(n-3)*(n-2)/2), ((n-2)*(n-1)/2,(n-1)*n/2), ((n+1)*(n+2)/2,n*(n+1)/2), and ((n+3)*(n+4)/2,(n+2)*(n+3)/2). - J. M. Bergot, May 27 2014
Number of terms less than 10^k: 1, 2, 5, 16, 50, 159, 500, 1582, 5000, 15812, 50000, 158114, 500000, ... - Muniru A Asiru, Jan 28 2018
Right-hand side of the binomial coefficient identity Sum_{k = 0..2*n} (-1)^(k+1)* binomial(2*n,k)*binomial(2*n + k,k)*(2*n - k) = a(n). - Peter Bala, Jan 12 2022

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.
  • Seberry, Jennifer and Yamada, Mieko; Hadamard matrices, sequences and block designs, in Dinitz and Stinson, eds., Contemporary design theory, pp. 431-560, Wiley-Intersci. Ser. Discrete Math. Optim., Wiley, New York, 1992.
  • W. D. Wallis, Anne Penfold Street and Jennifer Seberry Wallis, Combinatorics: Room squares, sum-free sets, Hadamard matrices, Lecture Notes in Mathematics, Vol. 292, Springer-Verlag, Berlin-New York, 1972. iv+508 pp.

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.
Cf. sequences listed in A254963.
Other n X n king graph cycle counts: A288918 (4-cycles), A288919 (5-cycles), A288920 (6-cycles).
Cf. A016813.

Programs

Formula

O.g.f.: 4*x*(1+x)/(1-x)^3. - R. J. Mathar, Jul 28 2008
a(n) = A000290(n)*4 = A001105(n)*2. - Omar E. Pol, May 21 2008
a(n) = A155955(n,2) for n > 1. - Reinhard Zumkeller, Jan 31 2009
Sum_{n>=1} 1/a(n) = (1/4)*Pi^2/6 = Pi^2/24. - Ant King, Nov 04 2009
a(n) = a(n-1) + 8*n - 4 (with a(0)=0). - Vincenzo Librandi, Nov 19 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 4, a(2) = 16. - Philippe Deléham, Mar 26 2013
a(n) = A118729(8n+3). - Philippe Deléham, Mar 26 2013
Pi = 2*Product_{n>=1} (1 + 1/(a(n)-1)). - Adriano Caroli, Aug 04 2013
Pi = Sum_{n>=0} 8/(a(2n+1)-1). - Adriano Caroli, Aug 06 2013
E.g.f.: exp(x)*(4x^2 + 4x). - Geoffrey Critzer, Oct 07 2013
a(n) = A000384(n) + A014105(n). - Bruce J. Nicholson, Nov 11 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/48 (A245058). - Amiram Eldar, Oct 10 2020
From Amiram Eldar, Jan 25 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/2)/(Pi/2) (A308716).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/2)/(Pi/2) = 2/Pi (A060294). (End)
a(n) = A016754(n) - A016813(n). - Leo Tavares, Feb 24 2022

Extensions

More terms from Sabir Abdus-Samee (sabdulsamee(AT)prepaidlegal.com), Mar 13 2006

A016802 a(n) = (4*n)^2.

Original entry on oeis.org

0, 16, 64, 144, 256, 400, 576, 784, 1024, 1296, 1600, 1936, 2304, 2704, 3136, 3600, 4096, 4624, 5184, 5776, 6400, 7056, 7744, 8464, 9216, 10000, 10816, 11664, 12544, 13456, 14400, 15376, 16384, 17424, 18496, 19600, 20736, 21904, 23104, 24336, 25600, 26896, 28224
Offset: 0

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A bisection of A016742. Sequence arises from reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Also, sequence found by reading the line from 0, in the direction 0, 16, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Sep 10 2011

Crossrefs

Programs

Formula

a(n) = 16*n^2 = 16*A000290(n). - Omar E. Pol, Dec 11 2008
a(n) = 8*A001105(n) = 4*A016742(n) = 2*A139098(n). - Omar E. Pol, Dec 13 2008
a(n) = a(n-1) + 16*(2*n-1) (with a(0)=0). - Vincenzo Librandi, Nov 20 2010
From Amiram Eldar, Jan 25 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/96.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/192.
Product_{n>=1} (1 + 1/a(n)) = sinh(Pi/4)/(Pi/4).
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/4)/(Pi/4) = 2*sqrt(2)/Pi (A112628). (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 16*x*(1 + x)/(1-x)^3.
E.g.f.: 16*x*(1 + x)*exp(x).
a(n) = n*A008598(n) = A195146(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A173949 a(n) = numerator of (Zeta(2, 1/4) - Zeta(2, n+1/4))/16, where Zeta is the Hurwitz Zeta function.

Original entry on oeis.org

0, 1, 26, 2131, 362164, 105007621, 5156362654, 129102916279, 108696708796264, 13163623138673569, 18033329053484721586, 30330904507928806086691, 30344915637965488890716, 1487479897654682071525709
Offset: 0

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Artur Jasinski, Mar 03 2010

Keywords

Comments

For the Catalan constant see A006752.
The denominators are given in A173948.
a(n+1)/A173948(n+1), for n>= 0, gives the partial sum Sum_{k=0..n} 1/(4*k + 1)^2. For {(4*k + 1)^2}A016814.%20The%20limit%20n%20-%3E%20infinity%20is%20given%20in%20A222183%20as%201.074833072...%20.%20-%20_Wolfdieter%20Lang">{k>=0} see A016814. The limit n -> infinity is given in A222183 as 1.074833072... . - _Wolfdieter Lang, Nov 14 2017

Examples

			The rationals r(n) begin: 0/1, 1/1, 26/25, 2131/2025, 362164/342225, 105007621/98903025, 5156362654/4846248225, 129102916279/121156205625, 108696708796264/101892368930625, 13163623138673569/12328976640605625, ... - _Wolfdieter Lang_, Nov 14 2017
		

Crossrefs

Programs

  • Magma
    [0] cat [Numerator((&+[1/(4*k+1)^2: k in [0..n-1]])): n in [1..20]]; // G. C. Greubel, Aug 23 2018
  • Maple
    r := n -> (Psi(1, 1/4) - Zeta(0, 2, n+1/4))/16:
    seq(numer(simplify(r(n))), n=0..13); # Peter Luschny, Nov 14 2017
  • Mathematica
    Table[Numerator[FunctionExpand[(8*Catalan + Pi^2 - Zeta[2, (4*n + 1)/4])/16]], {n, 0, 20}] (* Vaclav Kotesovec, Nov 14 2017 *)
    Numerator[Table[Sum[1/(4*k + 1)^2, {k, 0, n-1}], {n, 0, 20}]] (* Vaclav Kotesovec, Nov 14 2017 *)
  • PARI
    for(n=0,20, print1(numerator(sum(k=0,n-1, 1/(4*k+1)^2)), ", ")) \\ G. C. Greubel, Aug 23 2018
    

Formula

a(n) = numerator of expression (8*Catalan + Pi^2 - Zeta(2, (4*n + 1)/4))/16.
a(n) = numerator(r(n)) with r(n) = (Zeta(2,1/4) - Zeta(2, n + 1/4))/16, with the Hurwitz Zeta function Z(2, k). With Zeta(2,1/4) = 8 Catalan + Pi^2 this is the preceding formula, and Zeta(2, n + 1/4) = Psi(1, n + 1/4) with the polygamma (trigamma) function Psi(1, k). - Wolfdieter Lang, Nov 14 2017

Extensions

Edited by Wolfdieter Lang, Nov 14 2017
Name changed according to a formula of Lang by Peter Luschny, Nov 14 2017

A016862 a(n) = (5*n + 1)^2.

Original entry on oeis.org

1, 36, 121, 256, 441, 676, 961, 1296, 1681, 2116, 2601, 3136, 3721, 4356, 5041, 5776, 6561, 7396, 8281, 9216, 10201, 11236, 12321, 13456, 14641, 15876, 17161, 18496, 19881, 21316, 22801, 24336, 25921
Offset: 0

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Crossrefs

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), A016814 (m=4), this sequence (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).

Programs

Formula

a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). - Harvey P. Dale, Jul 11 2012
Sum_{n>=0} 1/a(n) = polygamma(1, 1/5)/25 = 1.050695088216... - Amiram Eldar, Oct 02 2020
G.f.: (1 +33*x +16*x^2)/(1-x)^3. - Wesley Ivan Hurt, Oct 02 2020
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A017282(n).
a(2*n+1) = 4*A016886(n).
E.g.f.: (1 + 35*x + 25*x^2)*exp(x). (End)

A016826 a(n) = (4n + 2)^2.

Original entry on oeis.org

4, 36, 100, 196, 324, 484, 676, 900, 1156, 1444, 1764, 2116, 2500, 2916, 3364, 3844, 4356, 4900, 5476, 6084, 6724, 7396, 8100, 8836, 9604, 10404, 11236, 12100, 12996, 13924, 14884, 15876, 16900, 17956
Offset: 0

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Keywords

Comments

A bisection of A016742. Sequence arises from reading the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

Crossrefs

Programs

Formula

a(n) = a(n-1) + 32*n (with a(0)=4). - Vincenzo Librandi, Dec 15 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=4, a(1)=36, a(2)=100. - Harvey P. Dale, Nov 24 2011
G.f.: -((4*(x^2+6*x+1))/(x-1)^3). - Harvey P. Dale, Nov 24 2011
a(n) = A000290(A016825(n)). - Wesley Ivan Hurt, Feb 24 2014
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/32.
Sum_{n>=0} (-1)^n/a(n) = G/4, where G is the Catalan constant (A006752). (End)
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(Pi/4).
Product_{n>=0} (1 - 1/a(n)) = 1/sqrt(2) (A010503). (End)

A016838 a(n) = (4n + 3)^2.

Original entry on oeis.org

9, 49, 121, 225, 361, 529, 729, 961, 1225, 1521, 1849, 2209, 2601, 3025, 3481, 3969, 4489, 5041, 5625, 6241, 6889, 7569, 8281, 9025, 9801, 10609, 11449, 12321, 13225, 14161, 15129, 16129, 17161, 18225
Offset: 0

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Author

Keywords

Comments

If Y is a fixed 2-subset of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007
A bisection of A016754. Sequence arises from reading the line from 9, in the direction 9, 49, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
Using (n,n+1) to generate a Pythagorean triangle with sides of lengths xJ. M. Bergot, Jul 17 2013

Crossrefs

Programs

Formula

Denominators of first differences Zeta[2,(4n-1)/4]-Zeta[2,(4(n+1)-1)/4]. - Artur Jasinski, Mar 03 2010
From George F. Johnson, Oct 03 2012: (Start)
G.f.: (9+22*x+x^2)/(1-x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n)-16)^2; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n+1) = (A004767(n))^2.
(End)
Sum_{n>=0} 1/a(n) = Pi^2/16 - G/2, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=0} (1 - 1/a(n)) = Gamma(3/4)^2/sqrt(Pi) = A068465^2 * A087197. - Amiram Eldar, Feb 01 2021

A017282 a(n) = (10*n + 1)^2.

Original entry on oeis.org

1, 121, 441, 961, 1681, 2601, 3721, 5041, 6561, 8281, 10201, 12321, 14641, 17161, 19881, 22801, 25921, 29241, 32761, 36481, 40401, 44521, 48841, 53361, 58081, 63001, 68121, 73441, 78961, 84681, 90601, 96721, 103041, 109561, 116281, 123201
Offset: 0

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Author

Keywords

Crossrefs

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m=3), A016814 (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), A017078 (m=8), A017174 (m=9), this sequence (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).
Cf. A017281.

Programs

  • Magma
    [(10*n+1)^2: n in [0..35]]; // Vincenzo Librandi, Jul 30 2011
    
  • Mathematica
    (* Programs from Michael De Vlieger, Mar 30 2017 *)
    Table[(10 n+1)^2, {n, 0, 35}]
    FoldList[#1 + 200 #2 - 80 &, 1, Range@ 35]
    CoefficientList[Series[(1+118x+81x^2)/(1-x)^3, {x,0,35}], x] (* End *)
    LinearRecurrence[{3,-3,1},{1,121,441},40] (* Harvey P. Dale, Sep 21 2017 *)
  • PARI
    for(n=0, 35, print1((10*n+1)^2", ")); \\ Bruno Berselli, Jul 30 2011
    
  • SageMath
    [(10*n+1)^2 for n in range(51)] # G. C. Greubel, Dec 24 2022

Formula

G.f.: (1+118*x+81*x^2)/(1-x)^3. - Bruno Berselli, Jul 30 2011
a(n) = a(n-1) + 40*(5*n-2), n > 0; a(0)=1. - Miquel Cerda, Oct 30 2016
a(n) = A017281(n)^2. - Michel Marcus, Oct 30 2016
E.g.f.: (1 +120*x +100*x^2)*exp(x). - G. C. Greubel, Dec 24 2022

Extensions

More terms from Bruno Berselli, Jul 30 2011

A016994 a(n) = (7*n + 1)^2.

Original entry on oeis.org

1, 64, 225, 484, 841, 1296, 1849, 2500, 3249, 4096, 5041, 6084, 7225, 8464, 9801, 11236, 12769, 14400, 16129, 17956, 19881, 21904, 24025, 26244, 28561, 30976, 33489, 36100, 38809, 41616, 44521, 47524
Offset: 0

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Author

Keywords

Crossrefs

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), A016814 (m=4), A016862 (m=5), A016922 (m=6), this sequence (m=7), A017078 (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).

Programs

Formula

G.f.: (1 + 61*x + 36*x^2)/(1-x)^3. - Vincenzo Librandi, Jan 27 2013
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A134934(n).
a(2*n+1) = 4*A017030(n).
E.g.f.: (1 + 63*x + 49*x^2)*exp(x). (End)
Sum_{n>=0} 1/a(n) = Psi'(1/7)/49 = 1.027703498712483534.. - R. J. Mathar, May 07 2024

A017078 a(n) = (8*n + 1)^2.

Original entry on oeis.org

1, 81, 289, 625, 1089, 1681, 2401, 3249, 4225, 5329, 6561, 7921, 9409, 11025, 12769, 14641, 16641, 18769, 21025, 23409, 25921, 28561, 31329, 34225, 37249, 40401, 43681, 47089, 50625, 54289, 58081
Offset: 0

Views

Author

Keywords

Crossrefs

Sequences of the form (m*n+1)^2: A000012 (m=0), A000290 (m=1), A016754 (m=2), A016778 (m-3), A016814 (m=4), A016862 (m=5), A016922 (m=6), A016994 (m=7), this sequence (m=8), A017174 (m=9), A017282 (m=10), A017402 (m=11), A017534 (m=12), A134934 (m=14).

Programs

Formula

G.f.: (1 + 78*x + 49*x^2)/(1-x)^3. - R. J. Mathar, Mar 21 2016
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A016754(8*n).
E.g.f.: (1 + 80*x + 64*x^2)*exp(x). (End)
Sum_{n>=0} 1/a(n) = psi'(1/8)/64 = 1.02168958507793.. - R. J. Mathar, May 07 2024
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