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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A014771 Squares of odd hexagonal numbers.

Original entry on oeis.org

1, 225, 2025, 8281, 23409, 53361, 105625, 189225, 314721, 494209, 741321, 1071225, 1500625, 2047761, 2732409, 3575881, 4601025, 5832225, 7295401, 9018009, 11029041, 13359025, 16040025, 19105641, 22591009, 26532801, 30969225, 35940025, 41486481, 47651409
Offset: 1

Views

Author

Mohammad K. Azarian

Keywords

Links

Crossrefs

Cf. A003215, (hex numbers), A014634 (odd hex numbers), A006752.

Programs

  • Magma
    [(2*n-1)^2*(4*n-3)^2 : n in [1..50]]; // Wesley Ivan Hurt, Jul 31 2016
  • Maple
    A014771:=n->(2*n-1)^2*(4*n-3)^2: seq(A014771(n), n=1..50); # Wesley Ivan Hurt, Jul 31 2016
  • Mathematica
    (Select[Table[n(2n-1), {n,60}], OddQ])^2 (* or *) LinearRecurrence[ {5,-10,10,-5,1}, {1,225,2025,8281,23409}, 30] (* Harvey P. Dale, Jun 23 2011 *)

Formula

G.f.: x*(1+220*x+910*x^2+396*x^3+9*x^4)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; checked and corrected by R. J. Mathar, Sep 16 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5. - Harvey P. Dale, Jun 23 2011
a(n) = (2*n-1)^2*(4*n-3)^2. - Wesley Ivan Hurt, Jul 31 2016
Sum_{n>=1} 1/a(n) = 2*G + 3*Pi^2/8 - Pi - 2*log(2), where G is Catalan's constant (A006752). - Amiram Eldar, Feb 27 2022

Extensions

More terms from Erich Friedman

A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1).

Original entry on oeis.org

1, 3, 12, 20, 15, 21, 56, 72, 45, 55, 132, 156, 91, 105, 240, 272, 153, 171, 380, 420, 231, 253, 552, 600, 325, 351, 756, 812, 435, 465, 992, 1056, 561, 595, 1260, 1332, 703, 741, 1560, 1640, 861, 903, 1892, 1980, 1035, 1081, 2256, 2352, 1225, 1275, 2652
Offset: 0

Views

Author

Paul Curtz, Jul 14 2015

Keywords

Comments

Consider, for n >= 0, a sequence s(n). A useful transform is wi(n) = s(0), s(2), s(3), ..., i.e., s(n) without s(1).
For s(n) = 1/(n+1), wi(n)= 1, 1/3, 1/4, 1/5, ..., whose inverse binomial transform is f(n) = 1, -2/3, 7/12, -11/20, 8/15, -11/21, 29/56, -37/72, 23/45, -28/55, 67/132, -79/156, 46/91, -53/105, 121/240, -137/272, ...
The denominator of f(n) is a(n), for n >= 0.
If the numerator of f(n) is b(n), then it can be seen that b(n+1) = -(-1)^n* A226089(n).
Alternating a(n) - b(n) with a(n) + b(n) yields 0, 1, 5, 9, ... = A160050(n+1).
a(4n+1) is linked to the Rydberg-Ritz spectra of hydrogen.
h(n) = 0, 0, 1, 1, 4, 4, 3, 3, 8, 8, 5, 5, ... = duplicated A022998(n).
A022998(n) is linked to the Balmer series (see A246943(n)).
With an initial 0 and offset=0, a(-n) = a(n). Then (a(n+10) - a(n-10))/10 = 1, 6, 10, 7, 9, 22, 26, ... = g(n). a(n) mod 9 is of period 20.

Links

Crossrefs

Cf. A002378(n+1), A014695(n+1)=A130658(n+2), A014634, A033567(n+1), A104188(n+1), 4*A007742(n+1), A160050 (in A226089), A022998, A109613, A000217(n+1), A246943.

Programs

  • Maple
    A257942:=n->(n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2): seq(A257942(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2015
  • Mathematica
    CoefficientList[Series[-(x^6 + 9 x^4 - 8 x^3 + 9 x^2 + 1)/((x - 1)^3 (x^2 + 1)^3), {x, 0, 50}], x] (* Michael De Vlieger, Jul 14 2015 *)
(* Using inverse binomial transform *) s[0]=1; s[n_] := 1/(n+2); f[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*s[k], {k, 0, n}]; Table[f[n]//Denominator, {n, 0, 50}] (* Jean-François Alcover, Jul 14 2015 *)
LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 3, 12, 20, 15, 21, 56, 72, 45}, 55] (* Vincenzo Librandi, Jul 15 2015 *)
  • PARI
    Vec(-(x^6+9*x^4-8*x^3+9*x^2+1)/((x-1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jul 14 2015
  • PARI
    a(n)=(n+1)*(n+2)/if(n%4<2,2,1) \\ Charles R Greathouse IV, Jul 14 2015

Formula

a(4n) = (2*n+1)*(4*n+1).
a(4n+1) = (2*n+1)*(4*n+3).
a(4n+2) = (4*n+3)*(4*n+4).
a(4n+3) = (4*n+4)*(4*n+5).
a(n) = A064038(n+2) * (period 4: repeat 1, 1, 4, 4).
From Colin Barker, Jul 14 2015: (Start)
a(n) = (-1/8+i/8)*(((-3-i*3)+i*(-i)^n+i^n)*(2+3*n+n^2)) where i=sqrt(-1).
G.f.: -(x^6+9*x^4-8*x^3+9*x^2+1) / ((x-1)^3*(x^2+1)^3). (End)
a(n) = h(n+2) * A109613(n+1).
a(n) = (n+1)*(n+2) * period 4:repeat (1, 1, 2, 2) /2.
From Wesley Ivan Hurt, Jul 18 2015: (Start)
a(n) = (n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2).
a(n) = 3*a(n-1)-6*a(n-2)+10*a(n-3)-12*a(n-4)+12*a(n-5)-10*a(n-6)+6*a(n-7)-3*a(n-8)+a(n-9), n>9. (End)
Sum_{n>=0} 1/a(n) = Pi/4 + 1. - Amiram Eldar, Aug 14 2022

A380472 a(n) = gcd_{primes P >= prime(n+1)} Product_{i=1..n} (P^2-i^2).

Original entry on oeis.org

1, 24, 360, 40320, 1814400, 479001600, 43589145600, 20922789888000, 3201186852864000, 2432902008176640000, 562000363888803840000, 620448401733239439360000, 201645730563302817792000000, 304888344611713860501504000000, 132626429906095529318154240000000, 263130836933693530167218012160000000
Offset: 0

Views

Author

Lily Bétaz, Martin Chevalier, and Basile Fusil, Jun 23 2025

Keywords

Comments

a(n) is the GCD of all numbers of the form Product_{i=1..n} (P^2-i^2) where P is a prime larger than or equal to the (n+1)-th prime.

Examples

			a(1) = 24 because 24 = GCD{P^2-1^2} GCD is taken on all numbers of the form P^2-1^2 with P a prime and P>3. This implies that for all primes P>3, P^2-1 is divisible by 24.
a(2) = 360 because 360 = GCD{(P^2-1^2)(P^2-2^2)} GCD is taken on all numbers of the form (P^2-1^2)(P^2-2^2) with P a prime and P>5. This implies that for all primes P>5, (P^2-1^2)(P^2-2^2) is divisible by 360.
a(3) = 40320 because 40320 = GCD{(P^2-1^2)(P^2-2^2)(P^2-3^2)}.
b(5) = 231 = a(7)/a(6).
c(2) = 112 = a(3)/a(2).

Crossrefs

Cf. A014634 (odd ratio), A014635 (even ratio, multiplied by 4), A084920.

Programs

  • Maple
    seq((2*n + 2)!*(3/4 - (-1)^n/4), n = 0..20)
  • Mathematica
    Table[(2*n + 2)!*(3/4 - (-1)^n/4), {n, 0, 20}]

Formula

E.g.f.: Sum_{n >= 0} a(n)/(2*n)!*z^(2*n) = (1 + 12*z^2 + 12*z^4 + 20*z^6 + 3*z^8)/(1 - z^4)^3.
a(n) = (2*n+2)!*(3/4-(-1)^n/4).
b(n) = (2*n+1)*(4*n+1) = a(2n)/a(2n-1) for n>=1 gives the odd ratios of a(n) (A014634).
c(n) = 4*2*n*(4*n-1) = a(2n-1)/a(2n-2) for n>=1 gives the even ratios of a(n) (4 times A014635).
Sum_{n>=0} 1/a(n) = 3*cosh(1)/2 - cos(1)/2 - 1. - Amiram Eldar, Jul 03 2025

A343578 a(n) = 32*n^2 - 40*n + 10.

Original entry on oeis.org

2, 58, 178, 362, 610, 922, 1298, 1738, 2242, 2810, 3442, 4138, 4898, 5722, 6610, 7562, 8578, 9658, 10802, 12010, 13282, 14618, 16018, 17482, 19010, 20602, 22258, 23978, 25762, 27610, 29522, 31498, 33538, 35642, 37810, 40042, 42338, 44698, 47122, 49610, 52162, 54778, 57458
Offset: 1

Views

Author

Gavin Lupo, Apr 20 2021

Keywords

Comments

a(n) is the sum of cross multiplying integers in groups of 4, a(n) = (4n-4)*(4n-1) + (4n-3)*(4n-2). For example, the group 4,5,6,7 yields the sum 4*7 + 5*6 = 58 = a(2).
Sequence found by reading the line from 2, in the direction 2, 58, ..., in the square spiral whose vertices are the generalized 18-gonal numbers A274979. - Omar E. Pol, Apr 20 2021

Links

Crossrefs

Cf. A274979 (generalized 18-gonal numbers).

Programs

  • Mathematica
    Table[32*n^2 - 40*n + 10, {n, 50}] (* Wesley Ivan Hurt, May 02 2021 *)

Formula

G.f.: 2*x*(1 + 26*x + 5*x^2)/(1 - x)^3. - Stefano Spezia, Apr 22 2021
a(n) = 4*A014634(n-1) - 2 = 8*A033954(n-1) + 2. - Hugo Pfoertner, Apr 24 2021
a(n) = determinant(matrix[4*n-1, -4*n+2, 4*n-3, 4*n-4]). - Peter Luschny, Apr 24 2021
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Wesley Ivan Hurt, May 02 2021
Previous Showing 21-24 of 24 results.

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