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 1-2 of 2 results.

A000466 a(n) = 4*n^2 - 1.

Original entry on oeis.org

-1, 3, 15, 35, 63, 99, 143, 195, 255, 323, 399, 483, 575, 675, 783, 899, 1023, 1155, 1295, 1443, 1599, 1763, 1935, 2115, 2303, 2499, 2703, 2915, 3135, 3363, 3599, 3843, 4095, 4355, 4623, 4899, 5183, 5475, 5775, 6083, 6399, 6723, 7055, 7395
Offset: 0

Views

Author

Chan Siu Kee (skchan5(AT)hkein.ie.cuhk.hk)

Keywords

Comments

Sum_{n>=1} (-1)^n*a(n)/n! = 1 - 1/e = A068996. - Gerald McGarvey, Nov 06 2007
Sequence arises from reading the line from -1, in the direction -1, 15, ... and the same line from 3, in the direction 3, 35, ..., in the square spiral whose nonnegative vertices are the squares A000290. - Omar E. Pol, May 24 2008
a(n) is the product of the consecutive odd integers 2n-1 and 2n+1 (cf. A005408). - Doug Bell, Mar 08 2009
For n>0: a(n) = A176271(2*n,n); cf. A016754, A053755. - Reinhard Zumkeller, Apr 13 2010
a(n+1) gives the curvature c(n) of the n-th circle touching the two equal semicircles of the symmetric arbelos (1/2, 1/2) and the (n-1)-st circle, with input c(0) = 3 = A059100(1) (referring to the second circle of the Pappus chain), for n >= 0. - Wolfdieter Lang and Kival Ngaokrajang, Jul 03 2015
After 3, a(n) is pseudoprime to base 2n. For example: (2*2)^(a(2)-1) == 1 (mod a(2)), in fact 4^14 = 15*17895697+1. - Bruno Berselli, Sep 24 2015
Numbers m such that m+1 and (m+1)/4 are squares. - Bruno Berselli, Mar 03 2016
After -1, the least common multiple of 2*m+1 and 2*m-1. - Colin Barker, Feb 11 2017
This sequence contains all products of the twin prime pairs (see A037074). - Charles Kusniec, Oct 03 2019

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
  • L. B. W. Jolley, Summation of Series, Dover, 2nd ed., 1961.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.
  • A. Languasco and A. Zaccagnini, Manuale di Crittografia, Ulrico Hoepli Editore (2015), p. 259.

Links

Crossrefs

Cf. A000290, A001539, A016286, A016742.
Factor of A160466. Superset of A037074.
Cf. A059100 (curvatures for a Pappus chain).

Programs

Formula

O.g.f.: ( 1-6*x-3*x^2 ) / (x-1)^3 . - R. J. Mathar, Mar 24 2011
E.g.f.: (-1 + 4*x + 4*x^2)*exp(x). - Ilya Gutkovskiy, May 26 2016
Sum_{n>=1} 1/a(n) = 1/2 [Jolley eq. 233]. - Benoit Cloitre, Apr 05 2002
Sum_{n>=1} 2/a(n) = 1 = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143, ..., with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15, ... - Gary W. Adamson, Jun 16 2003
1/3 + Sum_{n>=2} 4/a(n) = 1 = 1/3 + 4/15 + 4/35 + 4/63, ..., with partial sums: 1/3, 3/5, 5/7, 7/9, 9/11, ..., (2n+1)/(2n+3). - Gary W. Adamson, Jun 18 2003
Sum_{n>=0} 2/a(2*n+1) = Pi/4 = 2/3 + 2/35 + 2/99, ... = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11) + ... = Sum_{n>=0} (-1)^n/(2*n+1). - Gary W. Adamson, Jun 22 2003
Product(n>=1, (a(n)+1)/a(n)) = Pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
a(n)+2 = A053755(n). - Zak Seidov, Jan 16 2007
a(n)^2 + A008586(n)^2 = A053755(n)^2 (Pythagorean triple). - Zak Seidov, Jan 16 2007
a(n) = a(n-1) + 8*n - 4 for n > 0, a(0)=-1. - Vincenzo Librandi, Dec 17 2010
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - 1/2 = (A019669-1)/2. [Jolley eq (366)]. - R. J. Mathar, Mar 24 2011
For n>0, a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^3*(cos(x))^(2*n-2)). - Francesco Daddi, Aug 02 2011
Nonlinear recurrence for c(n) = a(n+1) (see the arbelos comment above) from Descartes' three circle theorem (see the links under A259555): c(n) = 4 + c(n-1) + 4*sqrt(c(n-1) + 1), with input c(0) = 3 = A059100(1), for n >= 0. The appropriate solution of this recurrence is c(n-1) + 1 = 4*n^2. - Wolfdieter Lang, Jul 03 2015
a(n) = 3*Pochhammer(5/2,n-1)/Pochhammer(1/2,n-1). Hence, the e.g.f. for a(n+1), i.e., dropping the first term, is 3* 1F1(5/2;1/2;x), with 1F1 being the confluent hypergeometric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
Product_{n>=1} (1 - 1/a(n)) = sin(Pi/sqrt(2))/sqrt(2). - Amiram Eldar, Feb 04 2021

A001539 a(n) = (4*n+1)*(4*n+3).

Original entry on oeis.org

3, 35, 99, 195, 323, 483, 675, 899, 1155, 1443, 1763, 2115, 2499, 2915, 3363, 3843, 4355, 4899, 5475, 6083, 6723, 7395, 8099, 8835, 9603, 10403, 11235, 12099, 12995, 13923, 14883, 15875, 16899, 17955, 19043, 20163, 21315, 22499, 23715, 24963, 26243, 27555, 28899
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Sequence arises from reading the line from 3, in the direction 3, 35, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
log(sqrt(2)+1)/sqrt(2) = 0.62322524... = 2/3 - 2/35 + 2/99 - 2/195 + 2/323, ... = (1 - 1/3) + (1/7 - 1/5) + (1/9 - 1/11) + (1/15 - 1/13) + (1/17 - 1/19) + (1/23 - 1/21) + ... - Gary W. Adamson, Mar 01 2009
Numbers k such that k+1 is a square and k+5 is divisible by 8. - Bruno Berselli, Sep 27 2017
The concatenation of 8*A000217(n) and 99 is a term of the sequence. Example: for A000217(5) = 15, 8*15 = 120 and 12099 = a(27). In general, a(5*n+2) = 800*A000217(n) + 99. - Bruno Berselli, Sep 29 2017

Links

Crossrefs

Bisection of A000466.
Cf. A000217, A000290, A001533, A001538, A004767, A016286, A016813, A016826, A157142, A133766, A154633.

Programs

  • Mathematica
    CoefficientList[Series[(3 + 26 x + 3 x^2)/(1 - x)^3, {x, 0, 41}], x] (* or *) Table[(4 n + 1) (4 n + 3), {n, 0, 41}] (* Michael De Vlieger, Sep 29 2017 *)
  • Maxima
    makelist((4*n+1)*(4*n+3), n, 0, 30); /* Martin Ettl, Nov 12 2012 */
  • PARI
    a(n)=(4*n+1)*(4*n+3) \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = A016826(n) - 1 = (A001533(n)+5)/4 = (A001538(n)+16)/9.
Sum_{k>=0} 1/a(k) = Pi/8. - Benoit Cloitre, Aug 20 2002
G.f.: (3 + 26*x + 3*x^2)/(1 - x)^3. - Jaume Oliver Lafont, Mar 07 2009
a(n) = 32*n + a(n-1) for n > 0, a(0)=3. - Vincenzo Librandi, Nov 12 2010
a(n) = a(m) + 16*(n-m)*(n+m+1). The previous formula is obtained for m = n-1. - Bruno Berselli, Sep 29 2017
From Amiram Eldar, Feb 19 2023: (Start)
a(n) = A016813(n)*A004767(n).
Product_{n>=0} (1 - 1/a(n)) = sqrt(2)*cos(Pi/(2*sqrt(2))).
Product_{n>=0} (1 + 1/a(n)) = sqrt(2). (End)
From Elmo R. Oliveira, Oct 23 2024: (Start)
E.g.f.: exp(x)*(3 + 16*x*(2 + x)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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