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.

Showing 1-10 of 10 results.

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

Views

Author

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

A163102 a(n) = n^2*(n+1)^2/2.

Original entry on oeis.org

0, 2, 18, 72, 200, 450, 882, 1568, 2592, 4050, 6050, 8712, 12168, 16562, 22050, 28800, 36992, 46818, 58482, 72200, 88200, 106722, 128018, 152352, 180000, 211250, 246402, 285768, 329672, 378450, 432450, 492032, 557568, 629442, 708050, 793800, 887112, 988418
Offset: 0

Views

Author

Omar E. Pol, Jul 24 2009

Keywords

Comments

Row sums of triangle A163282.
Also, the number of nonattacking placements of 2 rooks on an (n+1) X (n+1) board. - Thomas Zaslavsky, Jun 26 2010
If P_{k}(n) is the n-th k-gonal number, then a(n) = P_{s}(n+1)*P_{t}(n+1) - P_{s+1}(n+1)*P_{t-1}(n+1) for s=t+1. - Bruno Berselli, Sep 05 2014
Subsequence of A000982, see formula. - David James Sycamore, Jul 31 2018
Number of edges in the (n+1) X (n+1) rook complement graph. - Freddy Barrera, May 02 2019
Number of paths from (0,0) to (n+2,n+2) consisting of exactly three forward horizontal steps and three upward vertical steps. - Greg Dresden and Snezhana Tuneska, Aug 24 2023

References

  • Seth Chaiken, Christopher R. H. Hanusa, and Thomas Zaslavsky, A q-queens problem, in preparation. - Thomas Zaslavsky, Jun 26 2010

Crossrefs

Programs

Formula

a(n) = 2*A000537(n) = A035287(n+1)/2. - Omar E. Pol, Nov 29 2011
G.f.: 2*x*(1+4*x+x^2)/(1-x)^5. - R. J. Mathar, Nov 30 2011
Let t(n) = A000217(n). Then a(n) = (t(n-1)*(t(n)+t(n+1)) + t(n)*(t(n-1)+t(n+1)) + t(n+1)*(t(n-1)+t(n)))/3. - J. M. Bergot, Jun 21 2012
a(n) = A000982(n*(n+1)). - David James Sycamore, Jul 31 2018
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/3 - 6.
Sum_{n>=1} (-1)^(n+1)/a(n) = 6 - 8*log(2). (End)
Another identity: ..., a(4) = 200 = 1*(2+4+6+8) + 3*(4+6+8) + 5*(6+8) + 7*(8), a(5) = 450 = 1*(2+4+6+8+10) + 3*(4+6+8+10) + 5*(6+8+10) + 7*(8+10) + 9*(10) = 30+84+120+126+90, and so on. - J. M. Bergot, Aug 25 2022
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: x*(2 + x)*(2 + 6*x + x^2)*exp(x)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A254371(n)/4 = A060300(n)/8. (End)

A105636 Transform of n^3 by the Riordan array (1/(1-x^2), x).

Original entry on oeis.org

0, 1, 8, 28, 72, 153, 288, 496, 800, 1225, 1800, 2556, 3528, 4753, 6272, 8128, 10368, 13041, 16200, 19900, 24200, 29161, 34848, 41328, 48672, 56953, 66248, 76636, 88200, 101025, 115200, 130816, 147968, 166753, 187272, 209628, 233928, 260281, 288800, 319600
Offset: 0

Views

Author

Paul Barry, Apr 16 2005

Keywords

Comments

Recurrence a(n) = a(n-2) + n^3, starting with a(0)=0, a(1)=1. Also, in physics, a(n)/4 is the trace of the spin operator |S_z|^3 for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of the absolute values of their 3rd powers is 2*28/8 = a(3)/4. - Stanislav Sykora, Nov 07 2013
Also the number of 3-cycles in the (n+1)-triangular honeycomb queen graph. - Eric W. Weisstein, Jul 14 2017
With zero prepended and offset 1, the sequence starts 0,0,1,8,28,... for n=1,2,3,... Call this b(n). Consider the partitions of n into two parts (p,q). Then b(n) is the total volume of the family of cubes with side length |q - p|. - Wesley Ivan Hurt, Apr 14 2018

Crossrefs

Cf. A289705 (4-cycles), A289706 (5-cycles), A289707 (6-cycles).

Programs

  • GAP
    List([0..30], n -> (2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16); # G. C. Greubel, Dec 16 2018
  • Magma
    [(2*n^4+8*n^3+8*n^2-1)/16+(-1)^n/16: n in [0..50]]; // Vincenzo Librandi, Oct 27 2014
    
  • Mathematica
    LinearRecurrence[{4, -5, 0, 5, -4, 1}, {0, 1, 8, 28, 72, 153}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
    CoefficientList[Series[x (1 + 4 x + x^2)/((1 + x) (1 - x)^5), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 26 2012 *)
    Table[((-1)^n + 2 n^2 (n + 2)^2 - 1)/16, {n, 0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
  • PARI
    my(x='x+O('x^99)); concat(0, Vec(x*(1+4*x+x^2)/((1+x)*(1-x)^5))) \\ Altug Alkan, Apr 16 2018
    
  • Sage
    [(2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16 for n in range(30)] # G. C. Greubel, Dec 16 2018
    

Formula

G.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^5).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6).
a(n) = (2*n^4 + 8*n^3 + 8*n^2 - 1 + (-1)^n)/16.
a(n) = Sum_{k=0..floor((n-1)/2)} (n-2*k)^3.
a(n+1) = Sum_{k=0..n} k^3*(1 - (-1)^(n+k-1))/2.
a(n) = ((((x^2 - (x mod 2) - 4)/4)^2 - (((x^2 - (x mod 2) - 4)/4) mod 2))/8) = floor(((floor(x^2/4) - 1)^2)/8) where x = 2*n + 2. Replace x with 2*n - 1 to obtain A050534(n) = 3*A000332(n+1). Note that a(2*n) = A060300(n)/2 and a(2*n + 1) = A002593(n+1). - Raphie Frank, Jan 30 2014
a(n) = floor(1/(exp(2/n^2) - 1)^2)/2. Also a(n) = A007590(n+1)*A074148(n-1)/2. - Richard R. Forberg, Oct 26 2014
Sum_{n>=1} 1/a(n) = -cot(Pi/sqrt(2))*Pi/sqrt(2) - 1/2. - Amiram Eldar, Aug 25 2022

A007204 Crystal ball sequence for D_4 lattice.

Original entry on oeis.org

1, 25, 169, 625, 1681, 3721, 7225, 12769, 21025, 32761, 48841, 70225, 97969, 133225, 177241, 231361, 297025, 375769, 469225, 579121, 707281, 855625, 1026169, 1221025, 1442401, 1692601, 1974025, 2289169, 2640625, 3031081, 3463321, 3940225, 4464769, 5040025
Offset: 0

Views

Author

N. J. A. Sloane and J. H. Conway, Apr 28 1994

Keywords

Comments

Equals binomial transform of [1, 24, 120, 192, 96, 0, 0, 0, ...]. - Gary W. Adamson, Aug 13 2009
Hypotenuse of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + A069074(n-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
Numbers n such that n*x^4 + x^2 + 1 is reducible. - Arkadiusz Wesolowski, Nov 04 2013

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [(2*n^2+2*n+1)^2: n in [0..40]]; // Vincenzo Librandi, Nov 18 2016
    
  • Maple
    A007204:=n->(2*n^2+2*n+1)^2; seq(A007204(n), n=0..30);
  • Mathematica
    Table[(2n^2+2n+1)^2,{n,0,30}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,25,169,625,1681},40] (* Harvey P. Dale, Mar 03 2013 *)
  • PARI
    a(n)=(2*n^2+2*n+1)^2 \\ Charles R Greathouse IV, Feb 08 2017

Formula

G.f.: (1 + 54*x^2 + 20*x + 20*x^3 + x^4)/(1-x)^5.
a(0)=1, a(1)=25, a(2)=169, a(3)=625, a(4)=1681, a(n)=5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Harvey P. Dale, Mar 03 2013
Sum_{n>=0} 1/a(n) = Pi*(sinh(Pi) - Pi)/(2*(cosh(Pi) + 1)) = 1.0487582722070177... - Ilya Gutkovskiy, Nov 18 2016
a(n) = A016754(n) + A060300(n). - Bruce J. Nicholson, Apr 14 2017
a(n) = A001844(n)^2 = (2*n^2+2*n+1)^2. - Bruce J. Nicholson, May 15 2017
a(n) = A000583(n+1) + A099761(n) + 2*A005563(n-1)*A000290(n). - Charlie Marion, Jan 14 2021
E.g.f.: exp(x)*(1 + 24*x + 60*x^2 + 32*x^3 + 4*x^4). - Stefano Spezia, Jun 06 2021

Extensions

More terms from Harvey P. Dale, Mar 03 2013

A171522 Denominator of 1/n^2-1/(n+2)^2.

Original entry on oeis.org

0, 9, 16, 225, 144, 1225, 576, 3969, 1600, 9801, 3600, 20449, 7056, 38025, 12544, 65025, 20736, 104329, 32400, 159201, 48400, 233289, 69696, 330625, 97344, 455625, 132496, 613089, 176400, 808201, 230400, 1046529, 295936, 1334025, 374544, 1677025, 467856
Offset: 0

Views

Author

Paul Curtz, Dec 11 2009

Keywords

Comments

This is the third column in the table of denominators of the hydrogenic spectra (the main diagonal A147560):
0, 0, 0, 0, 0, 0, 0, 0... A000004
1, 4, 9, 16, 25, 36, 49, 64... A000290
1, 36, 16, 100, 9, 196, 64, 324... A061038
1, 144, 225, 12, 441, 576, 81, 900... A061040
1, 400, 144, 784, 64,1296, 400,1936... A061042
1, 900 1225,1600,2025, 100,3025,3600... A061044
1,1764, 576, 324, 225,4356, 48,6084... A061046
1,3136,3969,4900,5929,7056,8281, 196... A061048.

Crossrefs

Cf. A105371. Bisections: A060300, A069075.

Programs

  • Maple
    A171522 := proc(n) if n = 0 then 0 else lcm(n+2,n) ; %^2 ; end if ; end:
    seq(A171522(n),n=0..70) ; # R. J. Mathar, Dec 15 2009
  • Mathematica
    a[n_] := If[EvenQ[n], (n*(n+2))^2/4, (n*(n+2))^2]; Table[a[n], {n, 0, 36}] (* Jean-François Alcover, Jun 13 2017 *)
  • PARI
    concat(0, Vec(x*(x^8+4*x^6+16*x^5+190*x^4+64*x^3+180*x^2+16*x+9) / ((x-1)^5*-(x+1)^5) + O(x^100))) \\ Colin Barker, Nov 05 2014

Formula

a(n) = (A066830(n+1))^2.
a(n) = -((-5+3*(-1)^n)*n^2*(2+n)^2)/8. - Colin Barker, Nov 05 2014
G.f.: x*(x^8+4*x^6+16*x^5+190*x^4+64*x^3+180*x^2+16*x+9) / ((x-1)^5*-(x+1)^5). - Colin Barker, Nov 05 2014

Extensions

Edited and extended by R. J. Mathar, Dec 15 2009

A218485 Positive numbers differing from next greater square by a square.

Original entry on oeis.org

3, 5, 8, 12, 15, 16, 21, 24, 27, 32, 35, 40, 45, 48, 55, 60, 63, 65, 72, 77, 80, 84, 91, 96, 99, 105, 112, 117, 120, 128, 135, 140, 143, 144, 153, 160, 165, 168, 171, 180, 187, 192, 195, 200, 209, 216, 221, 224, 231, 240, 247, 252, 255, 264, 273, 280, 285
Offset: 1

Views

Author

Michel Marcus, Oct 30 2012

Keywords

Comments

Square terms in the sequence are 16, 144, 576, 1600, 3600, 7056, ..., that is, A060300 ((2n(n+1))^2) except 0. And their indices, ind(n), are: 6, 34, 100, 220, 410, 686, ..., that is, ind(n) = 2*A132124(.). - Michel Marcus, suggested by Zak Seidov, Nov 26 2013

Examples

			8 = 3^2 - 1^2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[300],IntegerQ[Sqrt[(1+Floor[Sqrt[#]])^2-#]]&] (* Zak Seidov, Nov 26 2013 *)
  • PARI
    sq1(n) = {for (i=1, n, a = sqrtint(i) + 1; if (issquare(a^2-i), print1(i, ", ")););}

A254371 Sum of cubes of the first n even numbers (A016743).

Original entry on oeis.org

0, 8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800, 352800, 426888, 512072, 609408, 720000, 845000, 985608, 1143072, 1318688, 1513800, 1729800, 1968128, 2230272, 2517768, 2832200, 3175200
Offset: 0

Views

Author

Luciano Ancora, Mar 16 2015

Keywords

Comments

Property: for n >= 2, each (a(n), a(n)+1, a(n)+2) is a triple of consecutive terms that are the sum of two nonzero squares; precisely: a(n) = (n*(n + 1))^2 + (n*(n + 1))^2, a(n)+1 = (n^2+2n)^2 + (n^2-1)^2 and a(n)+2 = (n^2+n+1)^2 + (n^2+n-1)^2 (see Diophante link). - Bernard Schott, Oct 05 2021

Crossrefs

Cf. A000537 (sum of first n cubes); A002593 (sum of first n odd cubes).
Cf. A060300 (2*a(n)).
First bisection of A105636; second bisection of A212892.

Programs

  • GAP
    List([0..35],n->2*(n*(n+1))^2); # Muniru A Asiru, Oct 24 2018
  • Magma
    [2*n^2*(n+1)^2: n in [0..40]]; // Bruno Berselli, Mar 23 2015
    
  • Maple
    A254371:=n->2*n^2*(n + 1)^2: seq(A254371(n), n=0..50); # Wesley Ivan Hurt, Apr 28 2017
  • Mathematica
    Table[2 n^2 (n+1)^2, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 72, 288, 800}, 40]
    Accumulate[Range[0,80,2]^3] (* Harvey P. Dale, Jun 26 2017 *)
  • PARI
    a(n)=sum(i=0, n, 8*i^3); \\ Michael B. Porter, Mar 16 2015
    

Formula

G.f.: 8*x*(1 + 4*x + x^2)/(1 - x)^5.
a(n) = 2*n^2*(n + 1)^2.
a(n) = 2*A035287(n+1) = 2*A002378(n)^2 = 8*A000217(n)^2. - Bruce J. Nicholson, Apr 23 2017
a(n) = 8*A000537(n). - Michel Marcus, Apr 23 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 - 3/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - 2*log(2). (End)
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: 2*x*(2 + x)*(2 + 6*x + x^2)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*A163102(n) = A060300(n)/2. (End)

A300758 a(n) = 2n*(n+1)*(2n+1).

Original entry on oeis.org

0, 12, 60, 168, 360, 660, 1092, 1680, 2448, 3420, 4620, 6072, 7800, 9828, 12180, 14880, 17952, 21420, 25308, 29640, 34440, 39732, 45540, 51888, 58800, 66300, 74412, 83160, 92568, 102660, 113460, 124992, 137280, 150348, 164220, 178920, 194472, 210900, 228228
Offset: 0

Views

Author

Christopher Purcell, Mar 12 2018

Keywords

Comments

The altitude h(n) = a(n)/A001844(n) of the (A005408(n), A046092(n) and A001844(n)) rectangular triangle is an irreducible fraction. - Ralf Steiner, Feb 25 2020
In this case, area A = a(n)/2 = A055112(n). - Bernard Schott, Feb 27 2020

Crossrefs

Formula

a(n) = 12*A000330(n).
G.f.: 12*x*(1+x)/(1-x)^4. - Colin Barker, Mar 12 2018
a(n) = 6*A006331(n) = 4*A059270(n) = 3*A002492(n) = 2*A055112(n). - Omar E. Pol, Apr 04 2018
From Ralf Steiner, Feb 27 2020: (Start)
a(n) = 2*n*A000384(n+1).
a(n) = sqrt(A016754(n)*A060300(n)).
(End)
a(n) = A005408(n) * A046092(n). - Bruce J. Nicholson, Apr 24 2020

Extensions

Edited by N. J. A. Sloane, Aug 01 2019

A272850 a(n) = (n^2 + (n+1)^2)*(n^2 + (n+1)^2 + 2*n*(n+1)).

Original entry on oeis.org

1, 45, 325, 1225, 3321, 7381, 14365, 25425, 41905, 65341, 97461, 140185, 195625, 266085, 354061, 462241, 593505, 750925, 937765, 1157481, 1413721, 1710325, 2051325, 2440945, 2883601, 3383901, 3946645, 4576825, 5279625
Offset: 0

Views

Author

Matthew Badley, May 07 2016

Keywords

Comments

Larger of pair of integers whose Pythagorean means are all integers.
The smaller of the pairs are: (A001844).
The arithmetic means are: (A007204)
The geometric means are: (A005917)
The harmonic means are: (A016754).
Subtracting terms in A016754 from A007204 gives complementary harmonics (A060300).

Crossrefs

Programs

Formula

a(n) = (2*n^2 + 2*n + 1)*(4*n^2 + 4*n + 1).
From Colin Barker, May 24 2016: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>4.
G.f.: (1 + 40*x + 110*x^2 + 40*x^3 + x^4) / (1-x)^5. (End)

A317309 Primes p such that the largest Dyck path of the symmetric representation of sigma(p) has a central valley.

Original entry on oeis.org

3, 5, 11, 13, 23, 37, 41, 43, 59, 61, 79, 83, 89, 107, 109, 113, 137, 139, 149, 151, 173, 179, 181, 211, 223, 227, 229, 257, 263, 269, 271, 307, 311, 313, 317, 353, 359, 367, 373, 409, 419, 421, 431, 433, 467, 479, 487, 491, 541, 547, 557, 599, 601, 607, 613, 617, 619, 673, 677, 683, 691, 701
Offset: 1

Views

Author

Omar E. Pol, Aug 29 2018

Keywords

Comments

Except for the first term 3, primes p such that both Dyck paths of the symmetric representation of sigma(p) have a central valley.
Note that the symmetric representation of sigma of an odd prime consists of two perpendicular bars connected by an irregular zig-zag path (see example).
Odd primes and the terms of this sequence are easily identifiable in the pyramid described in A245092 (see Links section).
For more information about the mentioned Dyck paths see A237593.
Equivalently, primes p such that the largest Dyck path of the symmetric representation of sigma(p) has an even number of peaks.

Examples

			Illustration of initial terms:
-------------------------------------------------
   p  sigma(p)  Diagram of the symmetry of sigma
-------------------------------------------------
                     _   _           _   _
                    | | | |         | | | |
                 _ _|_| | |         | | | |
   3      4     |_ _|  _|_|         | | | |
                 _ _ _|             | | | |
   5      6     |_ _ _|             | | | |
                                 _ _|_| | |
                               _|    _ _|_|
                             _|     |
                            |      _|
                 _ _ _ _ _ _|  _ _|
  11     12     |_ _ _ _ _ _| |
                 _ _ _ _ _ _ _|
  13     14     |_ _ _ _ _ _ _|
.
For the first four terms of the sequence we can see in the above diagram that the largest Dyck path of the symmetric representation of sigma(p) has a central valley.
Compare with A317308.
		

Crossrefs

Primes in A161983.
Except for the first term 3, primes in A317304.
The union of A317308 and this sequence gives A000040.
Primes of the triangle of A060300. - César Aguilera, Nov 12 2020

Programs

  • Python
    from sympy import isprime
    for x in range(1,100):
         for x in range(2*x**2+2*x-(2*x//2),2*x**2+2*x+(2*x//2)+1):
               if isprime(x):
                  print(x, end=', ') # César Aguilera, Nov 12 2020
Showing 1-10 of 10 results.