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 1-9 of 9 results.

A214345 Interleaved reading of A073577 and A053755.

Original entry on oeis.org

5, 7, 17, 23, 37, 47, 65, 79, 101, 119, 145, 167, 197, 223, 257, 287, 325, 359, 401, 439, 485, 527, 577, 623, 677, 727, 785, 839, 901, 959, 1025, 1087, 1157, 1223, 1297, 1367, 1445, 1519, 1601, 1679, 1765, 1847, 1937, 2023, 2117, 2207, 2305, 2399, 2501
Offset: 0

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Author

Keywords

Comments

The elements of this sequence satisfy the property that for every n=2k the triple (a(2k-1)^2, a(2k)^2 , a(2k+1)^2) is an arithmetic progression, i.e., 2*a(2k)^2 = a(2k-1)^2 + a(2k+1)^2. In general a triple((x-y)^2,z^2,(x+y)^2) is an arithmetic progression if and only if x^2+y^2=z^2 : in the case of this sequence 7^2, 17^2, and 23^2 is such a triple (i.e. 15-8 =7, 17, 8+15=23, and 8^2+15^2=17^2) .
The first differences of such a sequence is always an interleaved sequence; in this case the interleaved sequence is 2,10,6,14,10,... (A142954).

Examples

			For n = 7, a(7)=2*a(6)-2*a(4)+a(3)=2*65-2*37+23=79
		

Crossrefs

First differences: A142954; 2-element moving average (a(n-1) + a(n))/2: A002378. - Guenther Schrack, Oct 25 2018

Programs

  • GAP
    a:=[7,17];; for n in [3..50] do a[n]:=4*(n+1)+a[n-2]; od; Concatenation([5],a); # Muniru A Asiru, Oct 26 2018
  • Magma
    I:=[5, 7, 17, 23];[n le 4 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..75]];
    
  • Maple
    seq(coeff(series((x^3-3*x^2+3*x-5)/((x-1)^3*(x+1)),x,n+1), x, n), n = 0 .. 50); # Muniru A Asiru, Oct 26 2018
  • Mathematica
    LinearRecurrence[{2,0,-2,1},{5,7,17,23},50] (* Harvey P. Dale, Apr 02 2018 *)
  • Maxima
    A214345(n):=(2*n*(n+4)+3*(-1)^n+7)/2$
    makelist(A214345(n),n,0,30); /* Martin Ettl, Nov 01 2012 */
    

Formula

a(2n+1) = A073577(n+1); a(2n) = A053755(n+1).
a(n+1)-a(n) = A142954(n+1).
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4).
G.f.: (x^3-3*x^2+3*x-5)/((x-1)^3*(x+1)).
a(n) = (2*n*(n+4)+3*(-1)^n+7)/2.
2*a(2n)^2 = a(2n-1)^2 + a(2n+1)^2.
a(n) = 4*(n+1) + a(n-2) for n > 1; a(-n) = a(n-4). - Guenther Schrack, Oct 24 2018
E.g.f.: (5 + 5*x + x^2)*cosh(x) + (2 + 5*x + x^2)*sinh(x). - Stefano Spezia, Feb 22 2024

A293620 Numbers k such that f(k), f(k+1) and f(k+2) are all primes, where f(k) = (2k+1)^2 - 2 (A073577).

Original entry on oeis.org

1, 2, 16, 58, 149, 177, 534, 681, 954, 1045, 1052, 1255, 1367, 1563, 2046, 2074, 2515, 2557, 2564, 2788, 3586, 3593, 3908, 4062, 4552, 5252, 5371, 5385, 6400, 6729, 7443, 7478, 9305, 9375, 9942, 10355, 10411, 10726, 10740, 11286, 11545, 11559, 11832, 11965
Offset: 1

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Author

Amiram Eldar, Oct 13 2017

Keywords

Comments

Sierpiński proved that under Schinzel's hypothesis H this sequence is infinite.
Sierpiński showed that the only quadruple of consecutive primes of the form (2k+1)^2 - 2 are for k = 1 (i.e., 1 and 2 are the only consecutive terms in this sequence).
Numbers k such that the 3 consecutive integers k, k+1 and k+2 belong to A088572. - Michel Marcus, Oct 13 2017

Examples

			The first triples are: k = 1: (7, 23, 47), k = 2: (23, 47, 79), k = 16: (1087, 1223, 1367).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[10^4], AllTrue[{(2#+1)^2-2, (2#+3)^2-2, (2#+5)^2-2},PrimeQ] &]
    SequencePosition[Table[If[PrimeQ[(2k+1)^2-2],1,0],{k,12000}],{1,1,1}][[;;,1]] (* Harvey P. Dale, Feb 09 2025 *)
  • PARI
    f(n) = 4*n^2 + 4*n - 1;
    isok(n) = isprime(f(n)) && isprime(f(n+1)) && isprime(f(n+2)); \\ Michel Marcus, Oct 13 2017

A049598 12 times triangular numbers.

Original entry on oeis.org

0, 12, 36, 72, 120, 180, 252, 336, 432, 540, 660, 792, 936, 1092, 1260, 1440, 1632, 1836, 2052, 2280, 2520, 2772, 3036, 3312, 3600, 3900, 4212, 4536, 4872, 5220, 5580, 5952, 6336, 6732, 7140, 7560, 7992, 8436, 8892, 9360, 9840, 10332, 10836, 11352
Offset: 0

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Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

a(n-1) is the Wiener index of the helm graph H(n) (n>=3). The graph H(n) is obtained from an n-wheel graph (on n+1 nodes) by adjoining a pendant edge at each node of the cycle. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of H(n) is (1/2)*n*t*((n-3)t^3 + 2(n-2)t^2 + (n+3)t + 6). - Emeric Deutsch, Sep 28 2010
Also sequence found by reading the line from 0, in the direction 0, 12, ..., and the same line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized tetradecagonal numbers A195818. Axis perpendicular to A195158 in the same spiral. - Omar E. Pol, Sep 29 2011
Also the Wiener index of the (n+1)-gear graph. - Eric W. Weisstein, Sep 08 2017

Examples

			a(1) = 12*1 + 0 = 12;
a(2) = 12*2 + 12 = 36;
a(3) = 12*3 + 36 = 72.
		

Crossrefs

Programs

  • Mathematica
    12 * Accumulate[Range[0, 50]] (* Harvey P. Dale, Feb 05 2013 *)
    (* Start from Eric W. Weisstein, Sep 08 2017 *)
    Table[6 n (n + 1), {n, 0, 20}]
    12 PolygonalNumber[3, Range[0, 20]]
    12 Binomial[Range[20], 2]
    LinearRecurrence[{3, -3, 1}, {12, 36, 72}, {0, 20}]
    (* End *)
  • PARI
    a(n)=6*n*(n+1) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 6*n*(n+1).
G.f.: 12*x/(1-x)^3.
a(n) = 12*A000217(n). - Omar E. Pol, Dec 11 2008
a(n) = 12*n + a(n-1) (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A003154(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A032528(2*n+1) - 1. - Adriano Caroli, Jul 19 2013
a(n) = A001844(n) + A073577(n). - Bruce J. Nicholson, Aug 06 2017
E.g.f.: 6*x*(x+2)*exp(x). - G. C. Greubel, Aug 23 2017
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/6.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 1/6. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(6/Pi)*cos(sqrt(5/3)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (6/Pi)*cos(Pi/(2*sqrt(3))). (End)

A088572 Numbers n such that (2n+1)^2 - 2 is prime.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 10, 13, 14, 16, 17, 18, 21, 23, 24, 27, 30, 31, 34, 35, 37, 38, 44, 46, 51, 53, 58, 59, 60, 63, 65, 67, 69, 72, 77, 80, 84, 86, 88, 91, 95, 102, 105, 108, 111, 115, 116, 118, 119, 123, 126, 128, 129, 132, 133, 136, 139, 142, 146, 149, 150, 151, 154, 156, 157
Offset: 1

Views

Author

Herman H. Rosenfeld (herm3(AT)pacbell.net), Nov 17 2003

Keywords

Crossrefs

Programs

A030440 Values of Newton-Gregory forward interpolating polynomial (1/3)*(n-1)*(2*n+3)*(2*n-1).

Original entry on oeis.org

1, 0, 7, 30, 77, 156, 275, 442, 665, 952, 1311, 1750, 2277, 2900, 3627, 4466, 5425, 6512, 7735, 9102, 10621, 12300, 14147, 16170, 18377, 20776, 23375, 26182, 29205, 32452, 35931, 39650, 43617, 47840, 52327, 57086, 62125, 67452, 73075, 79002, 85241, 91800, 98687, 105910
Offset: 0

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Author

Ilias.Kotsireas(AT)lip6.fr (Ilias Kotsireas)

Keywords

Comments

Starting at a(2)=7, partial sums of A073577. - J. M. Bergot, Apr 20 2016

References

  • S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234; http://www.scirp.org/journal/am; http://dx.doi.org/10.4236/am.2014.515216

Crossrefs

Cf. A073577, A000384 (trinomial k=2 column), A106734, A127672, A027907.

Programs

  • Magma
    [(1/3)*(n-1)*(2*n-1)*(2*n+3):n in [0..50]]; // Vincenzo Librandi, Apr 20 2018
  • Mathematica
    LinearRecurrence[{4, -6, 4, -1}, {1, 0, 7, 30}, 40] (* Vincenzo Librandi, Apr 20 2018 *)
  • PARI
    a(n) = (n-1)*(2*n-1)*(2*n+3)/3; \\ Altug Alkan, Apr 19 2018
    

Formula

G.f.: (1+13*x^2-2*x^3-4*x)/(1-x)^4. - R. J. Mathar, May 18 2014
a(n) = (1/6) * (A106734(2n) - 1), n > 0. - Mathew Englander, Jun 06 2014
E.g.f.: (3 - 3*x + 12*x^2 + 4*x^3)*exp(x)/3. - Ilya Gutkovskiy, Apr 20 2016
a(n+1) = trinomial(2*n+1, 3) = binomial(2*n+1, 3) + (2*n+1)*(2*n) = n*(2*n+1)*(2*n+5)/3, for n >= 0, with the trinomial irregular triangle A027907. a(n+1) = trinomial(2*n+1, 4*n-1), for n >= 1 (symmetry). a(n+1) = Integral_{x=0..2} (1/sqrt(4 - x^2))*(x^2 - 1)^(2*n+1)*R(4*(n-1), x)/Pi with the R polynomial coefficients given in A127672. [Comtet, p. 77, the integral formula for q=3, n -> 2*n+1, k = 3, rewritten with x = 2*cos(phi)]. The g.f. of {a(n+1)}{n>= 0} is x*(7 + 2*x - x^2)/(1 - x)^4. - _Wolfdieter Lang, Apr 19 2018

A227786 Take squares larger than 1, subtract 3 from even squares and 2 from odd squares; a(n) = a(n-1) + A168276(n+1) (with a(1) = 1).

Original entry on oeis.org

1, 7, 13, 23, 33, 47, 61, 79, 97, 119, 141, 167, 193, 223, 253, 287, 321, 359, 397, 439, 481, 527, 573, 623, 673, 727, 781, 839, 897, 959, 1021, 1087, 1153, 1223, 1293, 1367, 1441, 1519, 1597, 1679, 1761, 1847, 1933, 2023, 2113, 2207, 2301, 2399, 2497, 2599, 2701
Offset: 1

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Author

Antti Karttunen, Jul 31 2013

Keywords

Comments

Conjecture: from n>=2 onward, a(n) gives the positions of 2's in A227761.
a(29) = 897 = 3*13*23 is the first term which is neither prime nor semiprime, that is, has more than two prime divisors.

Crossrefs

Bisections: A082109, A073577. Cf. also A227761.

Formula

a(n) = A000290(n+1) - 2 - (n mod 2).
a(1)=1, and for n>1, a(n) = a(n-1)+A168276(n+1).
a(n) = (1/2) * (2*n^2 + 4*n -3 + (-1)^n) = 2*A116940(n-1) + 1. a(n-1) = 2*ceiling(n^2/2) - 3 = 2*A000985(n) - 3. G.f.: x*(-x^3 - x^2 + 5*x + 1)/((1-x)^3 * (1+x)). - Ralf Stephan, Aug 10 2013

A145009 Array read by antidiagonals: array of odd integers found in |A144912| with axes b = {4, 6, 8, ...} and n = {b^2, b^4, b^6, ...}.

Original entry on oeis.org

7, 13, 13, 19, 23, 19, 25, 33, 33, 25, 31, 43, 47, 43, 31, 37, 53, 61, 61, 53, 37, 43, 63, 75, 79, 75, 63, 43, 49, 73, 89, 97, 97, 89, 73, 49, 55, 83, 103, 115, 119, 115, 103, 83, 55, 61, 93, 117, 133, 141, 141, 133, 117, 93, 61
Offset: 0

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Author

Reikku Kulon, Sep 28 2008

Keywords

Comments

The complete array can be defined as 6(x + y) + 4xy + 7.
Values along the edges are given by 6x + 7 and thus include the larger member of every twin prime pair except 5. The smaller member, 6x + 5, is adjacent in |A144912|.
Taking the origin to be z = 1, the main diagonal is given by 4z^2 + 4z - 1 (A073577).
Sums along antidiagonals are given by z(2z^2 + 12z + 7) / 3.
From Reikku Kulon, Sep 29 2008: (Start)
Any entry in the triangle can be produced from the two terms diagonally above or below and the edges can be found by taking the odd numbers as the "missing" values, starting from 1. If the terms are denoted:
.. a0 .. ...
a1 .. a2 ...
.. a3 .. ...
then:
a0 = (a1 + a2)/2 - 4*(a1 + a2 + 4)/(a2 - a1);
a3 = (a1 + a2)/2 + 4*(a1 + a2 + 4)/(a2 - a1). [Corrected by Jinyuan Wang, Jul 29 2020]
(End)

Examples

			Array A(n,k) begins:
7,  13, 19, 25,  31,  37,  43,  ...
13, 23, 33, 43,  53,  63,  73,  ...
19, 33, 47, 61,  75,  89,  103, ...
25, 43, 61, 79,  97,  115, 133, ...
31, 53, 75, 97,  119, 141, 163, ...
37, 63, 89, 115, 141, 167, 193, ...
...
		

Crossrefs

Formula

A(n, k) = |A144912(2*n+4, (2*n+4)^(2*k+2))| = 6*(n+k) + 4*n*k + 7.

A153167 Numbers n such that n+2 is not a Chen prime.

Original entry on oeis.org

2, 4, 6, 7, 8, 10, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 30, 31, 32, 33, 34, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 88, 89, 90
Offset: 1

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Author

Vincenzo Librandi, Dec 20 2008

Keywords

Comments

Contains all strictly positive even numbers A005843.
For each odd k>1 we can accumulate the numbers == k^2-2 (mod 2k) in a row, the last entry equal to A073577(k):
7; (k=3)
13, 23; (k=5)
19, 33, 47; (k=7)
25, 43, 61, 79; (k=9)
31, 53, 75, 97, 119; (k=11)
7, 63, 89, 115, 141, 167; (k=13)
43, 73, 103, 133, 163, 193,223; (k=17)
49, 83, 17, 151,185, 219, 253, 287; (k=19)
Each element T of this table has the format T= k^2-2-j*2*k, so T+2 is of the form k*(k-2*j), therefore not prime, and consequently all elements T are in the sequence.

Crossrefs

Extensions

Edited, 41, 59 (see A102540) etc. inserted by R. J. Mathar, Oct 16 2009

A338284 a(n) is the smallest nonsquare m such that the second partial quotient in the continued fraction for sqrt(m) equals n.

Original entry on oeis.org

7, 2, 23, 5, 47, 10, 79, 17, 119, 26, 167, 37, 223, 50, 287, 65, 359, 82, 439, 101, 527, 122, 623, 145, 727, 170, 839, 197, 959, 226, 1087, 257, 1223, 290, 1367, 325, 1519, 362, 1679, 401, 1847, 442, 2023, 485, 2207, 530, 2399, 577, 2599, 626, 2807, 677, 3023, 730
Offset: 1

Views

Author

Max Alekseyev, Oct 20 2020

Keywords

Examples

			a(3) = 23, since sqrt(23) = [4; 1, 3, ...] and m=23 is the smallest integer such that sqrt(m) has with second partial quotient equal 3.
		

Crossrefs

Interweaving of A073577 and A002522.
Cf. A013945 (first partial quotient = n).

Programs

  • Mathematica
    CoefficientList[Series[(7 + 2*x + 2*x^2 - x^3 - x^4 + x^5) / ((1-x)^3 * (1+x)^3),{x,0,20}], x] (* Georg Fischer, Aug 18 2021 *)

Formula

For even n, a(n) = A013945(n) = A002522(n/2) = (n/2)^2 + 1.
For odd n, a(n) = A073577((n+1)/2) = n^2 + 4*n + 2.
O.g.f.: (7 + 2*x + 2*x^2 - x^3 - x^4 + x^5) / ((1-x)^3 * (1+x)^3).
Showing 1-9 of 9 results.