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 1-4 of 4 results.

A154685 Triangle read by rows: T(n, k) = 2*n*k + n + k + 4.

Original entry on oeis.org

8, 11, 16, 14, 21, 28, 17, 26, 35, 44, 20, 31, 42, 53, 64, 23, 36, 49, 62, 75, 88, 26, 41, 56, 71, 86, 101, 116, 29, 46, 63, 80, 97, 114, 131, 148, 32, 51, 70, 89, 108, 127, 146, 165, 184, 35, 56, 77, 98, 119, 140, 161, 182, 203, 224, 38, 61, 84, 107, 130, 153, 176, 199, 222, 245, 268
Offset: 1

Views

Author

Vincenzo Librandi, Jan 18 2009

Keywords

Comments

The terms form a subset of A153039 because 2*T(n, k) - 7 = (2*n+1)*(2*k+1) are not prime.

Examples

			Triangle begins:
   8;
  11, 16;
  14, 21, 28;
  17, 26, 35, 44;
  20, 31, 42, 53,  64;
  23, 36, 49, 62,  75,  88;
  26, 41, 56, 71,  86, 101, 116;
  29, 46, 63, 80,  97, 114, 131, 148;
  32, 51, 70, 89, 108, 127, 146, 165, 184;
  35, 56, 77, 98, 119, 140, 161, 182, 203, 224;
		

Crossrefs

Cf. A151675 (row sums).
Similar triangle: A155724.
Columns k: A016789 (k=1), A016861 (k=2).
Main diagonal: A137882, A271649.

Programs

Formula

Sum_{k=1..n} T(n, k) = A151675(n). - N. J. A. Sloane, May 31 2009
T(n, k) = A155724(n,k) + 8. - L. Edson Jeffery, Oct 12 2012
From G. C. Greubel, Jan 21 2025: (Start)
T(2*n-1, n) = 4*n^2 + n + 3.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (1/4)*(9*(1-(-1)^n) + 2*(2-3*(-1)^n)*n - 4*(-1)^n*n^2).
G.f.: x*y*(8 - 5*(x+y) + 4*x*y)/((1-x)*(1-y))^2.
E.g.f.: 4 - (4+x)*exp(x) - (4+y)*exp(y) + (4+x+y+2*x*y)*exp(x+y).
(End)

Extensions

Clarified comment. - R. J. Mathar, Jan 24 2009

A271625 a(n) = = 2*(n+1)^2 - 5.

Original entry on oeis.org

3, 13, 27, 45, 67, 93, 123, 157, 195, 237, 283, 333, 387, 445, 507, 573, 643, 717, 795, 877, 963, 1053, 1147, 1245, 1347, 1453, 1563, 1677, 1795, 1917, 2043, 2173, 2307, 2445, 2587, 2733, 2883, 3037, 3195, 3357, 3523, 3693, 3867, 4045, 4227, 4413, 4603, 4797, 4995, 5197, 5403, 5613, 5827
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Apr 11 2016

Keywords

Comments

Numbers n such that 2*n + 10 is a perfect square.

Crossrefs

Numbers h such that 2*h + k is a perfect square: A294774 (k=-9), A255843 (k=-8), A271649 (k=-7), A093328 (k=-6), A097080 (k=-5), A271624 (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), A268581 (k=6), A059993 (k=7), (-1)*A147973 (k=8), A139570 (k=9), this sequence (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

Programs

  • Magma
    [ 2*n^2 + 4*n - 3: n in [1..60]];
    
  • Magma
    [ n: n in [1..6000] | IsSquare(2*n+10)];
    
  • Mathematica
    Table[2 n^2 + 4 n - 3, {n, 53}] (* Michael De Vlieger, Apr 11 2016 *)
    LinearRecurrence[{3,-3,1},{3,13,27},60] (* Harvey P. Dale, Jun 08 2023 *)
    2*Range[2,60]^2 -5 (* G. C. Greubel, Jan 21 2025 *)
  • PARI
    x='x+O('x^99); Vec(x*(3+4*x-3*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016
    
  • Python
    def A271625(n): return 2*pow(n+1,2) - 5
    print([A271625(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

Formula

G.f.: x*(3 + 4*x - 3*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = 13/30 - Pi*cot(sqrt(5/2)*Pi)/(2*sqrt(10)) = 0.5627678459924... . - Vaclav Kotesovec, Apr 11 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 6*x - 3) + 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
a(n) = 2*A000290(n+1) - 5. - G. C. Greubel, Jan 21 2025

Extensions

Name simplified by G. C. Greubel, Jan 21 2025

A271624 a(n) = 2*n^2 - 4*n + 4.

Original entry on oeis.org

2, 4, 10, 20, 34, 52, 74, 100, 130, 164, 202, 244, 290, 340, 394, 452, 514, 580, 650, 724, 802, 884, 970, 1060, 1154, 1252, 1354, 1460, 1570, 1684, 1802, 1924, 2050, 2180, 2314, 2452, 2594, 2740, 2890, 3044, 3202, 3364, 3530, 3700, 3874, 4052, 4234, 4420, 4610, 4804, 5002, 5204, 5410, 5620
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Apr 11 2016

Keywords

Comments

Numbers n such that 2*n - 4 is a perfect square.
For n > 2, the number of square a(n)-gonal numbers is finite. - Muniru A Asiru, Oct 16 2016

Examples

			a(1) = 2*1^2 - 4*1 + 4 = 2.
		

Crossrefs

Cf. A002522, numbers n such that 2*n + k is a perfect square: no sequence (k = -9), A255843 (k = -8), A271649 (k = -7), A093328 (k = -6), A097080 (k = -5), this sequence (k = -4), A051890 (k = -3), A058331 (k = -2), A001844 (k = -1), A001105 (k = 0), A046092 (k = 1), A056222 (k = 2), A142463 (k = 3), A054000 (k = 4), A090288 (k = 5), A268581 (k = 6), A059993 (k = 7), (-1)*A147973 (k = 8), A139570 (k = 9), A271625 (k = 10), A222182 (k = 11), A152811 (k = 12), A181510 (k = 13), A161532 (k = 14), no sequence (k = 15).

Programs

  • Magma
    [ 2*n^2 - 4*n + 4: n in [1..60]];
    
  • Magma
    [ n: n in [1..6000] | IsSquare(2*n-4)];
    
  • Mathematica
    Table[2 n^2 - 4 n + 4, {n, 54}] (* Michael De Vlieger, Apr 11 2016 *)
    LinearRecurrence[{3,-3,1},{2,4,10},60] (* Harvey P. Dale, Jul 18 2023 *)
  • PARI
    x='x+O('x^99); Vec(2*x*(1-x+2*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016
    
  • PARI
    a(n)=2*n^2-4*n+4 \\ Charles R Greathouse IV, Apr 11 2016

Formula

a(n) = 2*A002522(n-1).
G.f.: 2*x*(1 - x + 2*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = (1 + Pi*coth(Pi))/4 = 1.038337023734290587067... . - Vaclav Kotesovec, Apr 11 2016
a(n) = A005893(n-1), n > 1. - R. J. Mathar, Apr 12 2016
a(n) = 2 + 2*(n-1)^2. - Tyler Skywalker, Jul 21 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: 2*(exp(x)*(x^2 - x + 2) - 2).
a(n) = 2*A160457(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A294774 a(n) = 2*n^2 + 2*n + 5.

Original entry on oeis.org

5, 9, 17, 29, 45, 65, 89, 117, 149, 185, 225, 269, 317, 369, 425, 485, 549, 617, 689, 765, 845, 929, 1017, 1109, 1205, 1305, 1409, 1517, 1629, 1745, 1865, 1989, 2117, 2249, 2385, 2525, 2669, 2817, 2969, 3125, 3285, 3449, 3617, 3789, 3965, 4145, 4329, 4517, 4709, 4905
Offset: 0

Views

Author

Bruno Berselli, Nov 08 2017

Keywords

Comments

This is the case k = 9 of 2*n^2 + (1-(-1)^k)*n + (2*k-(-1)^k+1)/4 (similar sequences are listed in Crossrefs section). Note that:
2*( 2*n^2 + (1-(-1)^k)*n + (2*k-(-1)^k+1)/4 ) - k = ( 2*n + (1-(-1)^k)/2 )^2. From this follows an alternative definition for the sequence: Numbers h such that 2*h - 9 is a square. Therefore, if a(n) is a square then its base is a term of A075841.

Crossrefs

1st diagonal of A154631, 3rd diagonal of A055096, 4th diagonal of A070216.
Second column of Mathar's array in A016813 (Comments section).
Subsequence of A001481, A001983, A004766, A020668, A046711 and A057653 (because a(n) = (n+2)^2 + (n-1)^2); A097268 (because it is also a(n) = (n^2+n+3)^2 - (n^2+n+2)^2); A047270; A243182 (for y=1).
Similar sequences (see the first comment): A161532 (k=-14), A181510 (k=-13), A152811 (k=-12), A222182 (k=-11), A271625 (k=-10), A139570 (k=-9), (-1)*A147973 (k=-8), A059993 (k=-7), A268581 (k=-6), A090288 (k=-5), A054000 (k=-4), A142463 or A132209 (k=-3), A056220 (k=-2), A046092 (k=-1), A001105 (k=0), A001844 (k=1), A058331 (k=2), A051890 (k=3), A271624 (k=4), A097080 (k=5), A093328 (k=6), A271649 (k=7), A255843 (k=8), this sequence (k=9).

Programs

  • Maple
    seq(2*n^2 + 2*n + 5, n=0..100); # Robert Israel, Nov 10 2017
  • Mathematica
    Table[2n^2+2n+5,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{5,9,17},50] (* Harvey P. Dale, Sep 18 2023 *)
  • PARI
    Vec((5 - 6*x + 5*x^2) / (1 - x)^3 + O(x^50)) \\ Colin Barker, Nov 13 2017

Formula

O.g.f.: (5 - 6*x + 5*x^2)/(1 - x)^3.
E.g.f.: (5 + 4*x + 2*x^2)*exp(x).
a(n) = a(-1-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 5*A000217(n+1) - 6*A000217(n) + 5*A000217(n-1).
n*a(n) - Sum_{j=0..n-1} a(j) = A002492(n) for n>0.
a(n) = Integral_{x=0..2n+4} |3-x| dx. - Pedro Caceres, Dec 29 2020
Showing 1-4 of 4 results.