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 1-6 of 6 results.

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

1, 0, 4, 7, 1, 9, 7, 5, 5, 1, 1, 9, 6, 5, 9, 7, 7, 4, 6, 1, 5, 4, 2, 1, 4, 4, 6, 1, 0, 9, 3, 1, 6, 7, 6, 2, 8, 0, 6, 5, 7, 2, 3, 1, 3, 3, 1, 2, 5, 0, 3, 5, 2, 7, 3, 6, 5, 8, 3, 1, 4, 8, 6, 4, 1, 0, 2, 6, 0, 5, 4, 6, 8, 7, 6, 2, 0, 6, 9, 6, 6, 6, 2, 0, 9, 3, 4, 4, 9, 4, 1, 7, 8, 0, 7, 0, 5, 6, 8
Offset: 1

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

Examples

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Links

Crossrefs

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.
Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

Programs

Formula

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

A136017 a(n) = 36n^2 - 1.

Original entry on oeis.org

35, 143, 323, 575, 899, 1295, 1763, 2303, 2915, 3599, 4355, 5183, 6083, 7055, 8099, 9215, 10403, 11663, 12995, 14399, 15875, 17423, 19043, 20735, 22499, 24335, 26243, 28223, 30275, 32399, 34595, 36863, 39203, 41615, 44099, 46655, 49283, 51983
Offset: 1

Views

Author

Artur Jasinski, Dec 10 2007

Keywords

Comments

The least common multiple of 6*n+1 and 6*n-1. - Colin Barker, Feb 11 2017

Links

Crossrefs

Cf. A088878, A023208, A136016, A061037, A016910.

Programs

Formula

O.g.f.: x*(-35-38*x+x^2)/(-1+x)^3 = 1-35/(-1+x)-108/(-1+x)^2-72/(-1+x)^3. - R. J. Mathar, Dec 12 2007
a(n) = A061037(12n+10)=(6n-1)*(6n+1). - Paul Curtz, Sep 25 2008
Sum_{k>=1} (-1)^(k+1)/a(k) = (Pi-3)/6. - Jaume Oliver Lafont, Oct 20 2009
E.g.f.: 1 + (36 x^2 + 26 x - 1) exp(x). - Robert Israel, Jun 09 2016
Product_{n >= 1} A016910(n) / a(n) = Pi / 3. - Fred Daniel Kline, Jun 09 2016
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(3)*Pi/12. - Amiram Eldar, Jun 27 2020

A243941 Number of decompositions of 36*n^2 into the sum of two twin prime pairs.

Original entry on oeis.org

1, 2, 2, 5, 5, 3, 6, 3, 6, 8, 5, 6, 7, 6, 10, 10, 9, 8, 15, 10, 13, 8, 23, 5, 16, 21, 10, 20, 13, 30, 12, 14, 26, 16, 35, 16, 21, 22, 23, 38, 17, 28, 20, 36, 37, 16, 30, 27, 35, 33, 35, 29, 25, 34, 43, 51, 32, 44, 28, 39, 51, 40, 49, 31, 76, 31, 30, 52, 36, 103
Offset: 1

Views

Author

Olivier Gérard, Jun 15 2014

Keywords

Comments

Following a remark of M. T. Kong Tong on seqfan, there seems to be always at least one way to partition (6n)^2 into the sum of two prime pairs. This sequence gives the number of different solutions.
If there are only finitely many prime twins, this sequence will contain an infinite number of zeros.

Examples

			A solution is denoted by {p,q} where p,p+2,q,q+2 are all primes and p<=q.
a(10) = 8 because there are 8 ways to partition 3600 in this way.
The solution using the smallest prime numbers is 11+13+1787+1789 = 3600.
All 8 solutions are {11, 1787}, {101, 1697}, {179, 1619}, {191, 1607}, {311, 1487}, {347,1451}, {521, 1277} and {569, 1229}.

References

  • Liang Ding Xiang, Problem 93#, Bulletin of Mathematics (Wuhan), 6 (1992), 41. ISSN 0488-7395.

Links

Crossrefs

Cf. A016910 (36n^2).
Cf. A243940 (decompositions of n^2 into 4 primes).

Programs

  • PARI
    a(n)={my(m=18*n^2, s=0); forprime(p=5, m/2, if(isprime(m-p) && isprime(p-2) && isprime(m-p+2), s++)); s} \\ Andrew Howroyd, Sep 17 2019

Extensions

Liang reference from Alexander R. Povolotsky
Terms a(41) and beyond from Andrew Howroyd, Sep 17 2019

A229488 Conjecturally, possible differences between prime(k)^2 and the previous prime for some k.

Original entry on oeis.org

1, 2, 6, 8, 12, 14, 18, 20, 24, 26, 30, 32, 38, 42, 44, 48, 50, 54, 56, 60, 62, 66, 68, 72, 74, 78, 80, 84, 86, 90, 92, 96, 98, 102, 104, 108, 110, 114, 116, 120, 122, 126, 128, 132, 134, 138, 140, 146, 150, 152, 156, 158, 162, 164, 168, 170, 174, 176, 180
Offset: 1

Views

Author

T. D. Noe, Oct 21 2013

Keywords

Comments

Are there any missing terms? The first 10^7 primes were examined. All these differences occur for some k < 10^5. Note that the first differences of these terms is 1, 2, 4, or 6.
From R. J. Mathar, Oct 29 2013: (Start)
This sequence of possible differences d= prime(k)^2 -q looks similar to A047238; 1 is an exception associated with the single even prime, 1=2^2-3.
[Reason: Otherwise primes are odd, squared primes are also odd, so the differences are even and therefore in the class {0,2,4} mod 6.
Furthermore primes are of the form 3n+1 or 3n+2, squared primes are of the form 9n^2+6n+1 or 9n^2+12n+4, so squared primes are of the form ==1 (mod 3).
The difference prime(k)^2-q is therefore the difference between a number ==1 (mod 3) and a number == {1,2} (mod 3) and therefore a number == {0,2} mod 3. This is never of the form 6n+4 ( == 1 mod 3). So the differences are in the class {0,2} mod 6, demonstrating that this is essentially a subsequence of A047238.]
Furthermore, differences 36, 144, 324,... of the form (6n)^2, A016910, appear in A047238 but not here, because prime(k)^2 -q=(6n)^2 is equivalent to prime(k)^2-(6n)^2 =q =(prime(k)+6n)*(prime(k)-6n), which requires an explicit factorization of the prime q. This is a contradiction if we assure that prime(k)-6n is not equal 1; if we scanned explicitly all primes up to prime(k)=10^7, for example, all (6n)^2 up to 6n<=10^7 are proved not to be in the sequence. (End)

Crossrefs

Cf. A000040 (primes), A001248 (primes squared).
Cf. A004277 (conjecturally, possible gaps between adjacent primes).
Cf. A054270 (prime below prime(n)^2).
Cf. A229489 (possible differences between prime(k)^2 and the next prime).

Programs

  • Mathematica
    t = Table[p2 = Prime[k]^2; p2 - NextPrime[p2, -1], {k, 100000}]; Take[Union[t], 60]

A303302 a(n) = 34*n^2.

Original entry on oeis.org

0, 34, 136, 306, 544, 850, 1224, 1666, 2176, 2754, 3400, 4114, 4896, 5746, 6664, 7650, 8704, 9826, 11016, 12274, 13600, 14994, 16456, 17986, 19584, 21250, 22984, 24786, 26656, 28594, 30600, 32674, 34816, 37026, 39304, 41650, 44064, 46546, 49096, 51714, 54400, 57154, 59976, 62866, 65824, 68850, 71944
Offset: 0

Views

Author

Omar E. Pol, May 13 2018

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 34, ..., in the square spiral whose vertices are the generalized 19-gonal numbers A303813.

Links

Crossrefs

Cf. A005843, A008599, A303813.
Cf. similar sequences of the type k*n^2: A000290 (k=1), A001105 (k=2), A033428 (k=3), A016742 (k=4), A033429 (k=5), A033581 (k=6), A033582 (k=7), A139098 (k=8), A016766 (k=9), A033583 (k=10), A033584 (k=11), A135453 (k=12), A152742 (k=13), A144555 (k=14), A064761 (k=15), A016802 (k=16), A244630 (k=17), A195321 (k=18), A244631 (k=19), A195322 (k=20), A064762 (k=21), A195323 (k=22), A244632 (k=23), A195824 (k=24), A016850 (k=25), A244633 (k=26), A244634 (k=27), A064763 (k=28), A244635 (k=29), A244636 (k=30), A244082 (k=32), this sequence (k=34), A016910 (k=36), A016982 (k=49), A017066 (k=64), A017162 (k=81), A017270 (k=100), A017390 (k=121), A017522 (k=144).

Programs

  • Magma
    [34*n^2: n in [0..50]]; // Vincenzo Librandi Jun 07 2018
  • Mathematica
    Table[34 n^2, {n, 0, 40}]
LinearRecurrence[{3,-3,1},{0,34,136},50] (* Harvey P. Dale, Jul 23 2018 *)
  • PARI
    a(n) = 34*n^2;
  • PARI
    concat(0, Vec(34*x*(1 + x) / (1 - x)^3 + O(x^40))) \\ Colin Barker, Jun 12 2018

Formula

a(n) = 34*A000290(n) = 17*A001105(n) = 2*A244630(n).
G.f.: 34*x*(1 + x)/(1 - x)^3. - Vincenzo Librandi, Jun 07 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 34*x*(1 + x)*exp(x).
a(n) = A005843(n)*A008599(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A235644 Number of decompositions of 12*n into the sum of two (not necessarily distinct) twin prime pairs.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 0, 2, 1, 3, 3, 1, 2, 1, 3, 2, 2, 2, 3, 1, 3, 1, 2, 3, 3, 6, 2, 3, 1, 2, 4, 3, 4, 4, 1, 3, 2, 3, 5, 2, 7, 1, 3, 2, 2, 5, 2, 5, 2, 3, 2, 2, 3, 5, 3, 4, 1, 0, 3, 1, 6, 2, 3, 3, 1, 5, 2, 5, 3, 3, 4, 1, 4
Offset: 1

Views

Author

Lear Young, Jun 16 2014

Keywords

Comments

In the 1980's, Liang conjectured that (6n)^2 = p_1 + p_2 + p_3 + p_4, where (p_1, p_2) and (p_3, p_4) are twin prime pairs. See reference for more details.
It seems there are at least 2 solutions for the decompositions when n > 701.
If the two twin prime pairs are required to be distinct, the sequence is A187759.

Examples

			a(736) = 2 because 12*736 = 197 + 199 + 4217 + 4219 = 857 + 859 + 3557 + 3559, so there are 2 ways of expressing 12*n as the sum of two twin prime pairs.

References

  • Liang Ding Xiang, Problem 93#, Bulletin of Mathematics (Wuhan), 6(1992),41. ISSN 0488-7395.

Crossrefs

Cf. A016910, A243941, A187759, A243914, A243956.

Programs

  • PARI
    v=select(p->isprime(p-2)&&p>5, primes(200))\6; l=List(); for(i=1, #v, if(2*v[i]<100, listput(l, 2*v[i])); for(j=i+1, #v, if((v[i]+v[j])<100, listput(l, v[i]+v[j])))); l1=vecsort(l); k=1; for(i=1, 100, s=sum(j=k, #l1, l1[j]==i); print1(s", "); k+=s) \\ Lear Young, Jun 16 2014
  • PARI
    v=select(p->isprime(p-2)&&p>5,primes(110))\6;for(i=1, 99, print1(sum(j=1,#v,vecsearch(v,i-v[j])>0 && i-v[j]>=v[j])", "))   \\ change i-v[j]>=v[j] to i-v[j]>v[j] is A187759.  Lear Young, Jun 16 2014
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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