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 1-10 of 18 results. Next

A061870 Numbers such that |first digit - second digit + third digit - fourth digit ...| = 1.

Original entry on oeis.org

1, 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 100, 111, 120, 122, 131, 133, 142, 144, 153, 155, 164, 166, 175, 177, 186, 188, 197, 199, 210, 221, 230, 232, 241, 243, 252, 254, 263, 265, 274, 276, 285, 287, 296, 298, 320, 331, 340, 342
Offset: 1

Views

Author

Robert G. Wilson v, May 10 2001

Keywords

Comments

Multiples of 11 plus or minus 1. If 11k+1 is a perfect square (see A219257) then a(n) is the square root of 11k+1. [Gary Detlefs, Feb 22 2010]

Examples

			120 is in the sequence since |1-2+0| = 1.

Links

Crossrefs

Cf. A008593, A060978-A060980, A060982, A061470-A061479, A061870-A061882.
Subsequence of A175885.

Programs

  • Mathematica
    fQ[n_] := Abs[Plus @@ (((-1)^Range[Floor[Log[10, n] + 1]])*IntegerDigits@n)] == 1; Select[ Range@342, fQ@# &]
  • PARI
    altsum(v)=sum(i=1,#v,v[i]*(-1)^i)
is(n)=abs(altsum(digits(n)))==1 \\ Charles R Greathouse IV, May 21 2014
  • Python
    def ok(n): return abs(sum(int(di)*(-1)**i for i, di in enumerate(str(n)))) == 1
print([k for k in range(343) if ok(k)]) # Michael S. Branicky, Jan 26 2023

A132356 a(2*k) = k*(10*k+2), a(2*k+1) = 10*k^2 + 18*k + 8, with k >= 0.

Original entry on oeis.org

0, 8, 12, 36, 44, 84, 96, 152, 168, 240, 260, 348, 372, 476, 504, 624, 656, 792, 828, 980, 1020, 1188, 1232, 1416, 1464, 1664, 1716, 1932, 1988, 2220, 2280, 2528, 2592, 2856, 2924, 3204, 3276, 3572, 3648, 3960, 4040, 4368, 4452, 4796, 4884, 5244, 5336, 5712
Offset: 0

Views

Author

Mohamed Bouhamida, Nov 08 2007

Keywords

Comments

X values of solutions to the equation 10*X^3 + X^2 = Y^2.
Polygonal number connection: 2*H_n + 6S_n, where H_n is the n-th hexagonal number and S_n is the n-th square number. This is the base formula that is expanded upon to achieve the full series. See contributing formula below. - William A. Tedeschi, Sep 12 2010
Equivalently, numbers of the form 2*h*(5*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... . - Bruno Berselli, Feb 02 2017

Links

Crossrefs

Cf. A054000, A056220, A087475, A028347, A046092, A132209.
Cf. numbers m such that k*m+1 is a square: A005563 (k=1), A046092 (k=2), A001082 (k=3), A002378 (k=4), A036666 (k=5), A062717 (k=6), A132354 (k=7), A000217 (k=8), A132355 (k=9), A219257 (k=11), A152749 (k=12), A219389 (k=13), A219390 (k=14), A204221 (k=15), A074378 (k=16), A219394 (k=17), A219395 (k=18), A219396 (k=19), A219190 (k=20), A219391 (k=21), A219392 (k=22), A219393 (k=23), A001318 (k=24), A219259 (k=25), A217441 (k=26), A219258 (k=27), A219191 (k=28).
Cf. A220082 (numbers k such that 10*k-1 is a square).

Programs

  • Mathematica
    CoefficientList[Series[4*x*(2*x^2 + x + 2)/((1 - x)^3*(1 + x)^2), {x, 0, 50}], x] (* G. C. Greubel, Jun 12 2017 *)
LinearRecurrence[{1,2,-2,-1,1},{0,8,12,36,44},50] (* Harvey P. Dale, Dec 15 2023 *)
  • PARI
    my(x='x+O('x^50)); concat([0], Vec(4*x*(2*x^2+x+2)/((1-x)^3*(1+x)^2))) \\ G. C. Greubel, Jun 12 2017
  • PARI
    a(n) = n^2 + n + 6*((n+1)\2)^2 \\ Charles R Greathouse IV, Sep 11 2022

Formula

G.f.: 4*x*(2*x^2+x+2)/((1-x)^3*(1+x)^2). - R. J. Mathar, Apr 07 2008
a(n) = 10*x^2 - 2*x, where x = floor(n/2)*(-1)^n for n >= 1. - William A. Tedeschi, Sep 12 2010
a(n) = ((2*n+1-(-1)^n)*(10*(2*n+1)-2*(-1)^n))/16. - Luce ETIENNE, Sep 13 2014
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4. - Chai Wah Wu, May 24 2016
Sum_{n>=1} 1/a(n) = 5/2 - sqrt(1+2/sqrt(5))*Pi/2. - Amiram Eldar, Mar 15 2022
a(n) = n^2 + n + 6*ceiling(n/2)^2. - Ridouane Oudra, Aug 06 2022

Extensions

More terms from Max Alekseyev, Nov 13 2009

A204221 Integers of the form (N^2 - 1) / 15.

Original entry on oeis.org

0, 1, 8, 13, 17, 24, 45, 56, 64, 77, 112, 129, 141, 160, 209, 232, 248, 273, 336, 365, 385, 416, 493, 528, 552, 589, 680, 721, 749, 792, 897, 944, 976, 1025, 1144, 1197, 1233, 1288, 1421, 1480, 1520, 1581, 1728, 1793, 1837, 1904, 2065, 2136, 2184, 2257, 2432
Offset: 0

Views

Author

Michael Somos, Jan 13 2012

Keywords

Comments

Equivalently, numbers in increasing order of the form m(15m+2) or m(15m+8)+1, where m = 0,-1,1,-2,2,-3,3,... [Bruno Berselli, Nov 27 2012]
The sequence terms occur as exponents in the expansion of the identity Product_{n >= 0} (1 - x^(20*n+1))*(1 - x^(20*n+19))*(1 - x^(20*n+8))*(1 - x^(20*n+12))*(1 - x^(20*n+9))*(1 - x^(20*n+11))*(1 - x^(10*n+10)) = Sum_{n >= 0} x^(n^2+n)*Product_{k >= 2*n+1} 1 - x^k = 1 - x - x^8 + x^13 + x^17 - - + + .... See Andrews et al., p. 591, Exercise 6(c). - Peter Bala, Feb 22 2021.

References

  • George E. Andrews, Richard Askey, and Ranjan Roy, Special Functions, Cambridge University Press, 1999.

Links

Crossrefs

Cf. A204220, A204542 (square roots of 15*a(n)+1), A379210.
Cf. similar sequences listed in A219257.

Programs

  • Magma
    [n: n in [0..2500] | IsSquare(15*n+1)]; // Bruno Berselli, Nov 23 2012
  • Magma
    /* By comment: */ s:=[0, 1] cat &cat[[t*(15*t+2), t*(15*t+8)+1]: t in [-n,n], n in [1..13]]; Sort(s); // Bruno Berselli, Nov 27 2012
  • Maple
    A204221 := proc(q) local n;
for n from 0 to q do
 if type(sqrt(15*n+1), integer) then print(n);
fi; od; end:
A204221(2500); # Peter Bala, Dec 18 2024
  • Mathematica
    Select[Range[0, 2500], IntegerQ[Sqrt[15 # + 1]] &] (* Bruno Berselli, Nov 23 2012 *)
  • PARI
    {a(n) = (15*n^2 + n*[8, 2, 28, 22][n%4 + 1] + 12) \ 16}

Formula

|A204220(n)| is the characteristic function of the numbers in this sequence.
a(-1 - n) = a(n).
G.f. x*(x^2-x+1)*(x^4+8*x^3+12*x^2+8*x+1) / ( (1+x)^2*(1+x^2)^2*(1-x)^3 ). - R. J. Mathar, Jan 28 2012
a(n) = (30*n-10*i^(n(n-1))+3*(-1)^n+7)*(30*n-10*i^(n(n-1))+3*(-1)^n+23)/960, where i=sqrt(-1). - Bruno Berselli, Nov 28 2012
Sum_{n>=1} 1/a(n) = 15/4 - cot(2*Pi/15)*Pi/2 - Pi/(2*sqrt(3)) + sqrt(1+2/sqrt(5))*Pi/2. - Amiram Eldar, Mar 15 2022
From Peter Bala, Dec 17 2024: (Start)
a(n) is quasi-polynomial in n: for n >= 0,
a(4*n+1) = 15*n^2 + 8*n + 1; a(4*n+2) = 15*n^2 + 22*n + 8;
a(4*n+3) = 15*n^2 + 28*n + 13; a(4*n+4) = 15*n^2 + 32*n + 17.
For 1 <= k <= 4, a(4*n+k) = (N_k(n)^2 - 1)/15, where N_1(n) = 15*n + 4, N_2(n) = 15*n + 11, N_3(n) = 15*n + 14 and N_4(n) = 15*n + 16. (End)

A154260 Numbers of the form m*(4*m +- 1)/2.

Original entry on oeis.org

0, 7, 9, 30, 34, 69, 75, 124, 132, 195, 205, 282, 294, 385, 399, 504, 520, 639, 657, 790, 810, 957, 979, 1140, 1164, 1339, 1365, 1554, 1582, 1785, 1815, 2032, 2064, 2295, 2329, 2574, 2610, 2869, 2907, 3180, 3220, 3507, 3549, 3850, 3894, 4209, 4255, 4584
Offset: 1

Views

Author

Vladimir Joseph Stephan Orlovsky, Jan 06 2009

Keywords

Comments

Also integers of the form Sum_{k = 1..j} k/4 = j*(j + 1)/8. - Alonso del Arte, Jan 20 2012
Numbers h such that 32*h + 1 is a square. - Bruno Berselli, Mar 30 2014
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(16*n))*(1 - q^(16*n-7))*(1 - q^(16*n-9)) = 1 - q^7 - q^9 + q^30 + q^34 - q^69 - q^75 + + - - .... - Peter Bala, Dec 24 2024

Links

Crossrefs

Cf. A000217, A001318, A074378, A057569, A057570.
Cf. similar sequences listed in A219257 and A299645.

Programs

  • Magma
    k:=8; f:=func; [0] cat [f(n*m): m in [-1, 1], n in [1..25]]; // Bruno Berselli, Nov 14 2012
  • Mathematica
    Select[Union[Flatten[Table[{n (4n - 1)/2, n (4n + 1)/2}, {n, 0, 199}]]], IntegerQ] (* Alonso del Arte, Jan 20 2012 *)
  • PARI
    print1(0);forstep(n=2,1e2,2,print1(", "n*(4*n-1)/2", "n*(4*n+1)/2)) \\ Charles R Greathouse IV, Jan 20 2012
  • PARI
    print1(s=0);for(n=1,1e3,s+=n/4;if(denominator(s)==1,print1(s", "))) \\ Charles R Greathouse IV, Jan 20 2012
  • Python
    def A154260(n): return (n>>1)*((n<<2)+(-3 if n&1 else -1)) # Chai Wah Wu, Mar 11 2025

Formula

From R. J. Mathar, Jan 07 2009: (Start)
A139274 UNION A139275.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
G.f.: x^2*(7 + 2x + 7x^2)/((1+x)^2*(1-x)^3). (End)
From G. C. Greubel, Sep 08 2016: (Start)
a(n) = (1/4)*(8*n^2 + 6*(-1)^n*n - 8*n - 3*(-1)^n + 3).
E.g.f.: (1/4)*( (3 + 8*x^2)*exp(x) - 3*(1 + 2*x)*exp(-x) ). (End)
From Amiram Eldar, Mar 17 2022: (Start)
Sum_{n>=2} 1/a(n) = 8 - (sqrt(2)+1)*Pi.
Sum_{n>=2} (-1)^n/a(n) = 2*sqrt(2)*log(sqrt(2)+1) - 8*(1-log(2)). (End)
a(n) = (n-1)*(4*n-3)/2 if n is odd and a(n) = n*(4*n-1)/2 if n is even. - Chai Wah Wu, Mar 11 2025

A219191 Numbers of the form k*(7*k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 6, 8, 26, 30, 60, 66, 108, 116, 170, 180, 246, 258, 336, 350, 440, 456, 558, 576, 690, 710, 836, 858, 996, 1020, 1170, 1196, 1358, 1386, 1560, 1590, 1776, 1808, 2006, 2040, 2250, 2286, 2508, 2546, 2780, 2820, 3066, 3108, 3366, 3410, 3680, 3726, 4008
Offset: 1

Views

Author

Bruno Berselli, Nov 14 2012

Keywords

Comments

Equivalently, numbers m such that 28*m+1 is a square.
Also, integer values of h*(h+1)/7.
Let F(r) = Product_{n >= 1} 1 - q^(14*n-r). The sequence terms are the exponents in the expansion of F(0)*F(6)*F(8) = 1 - q^6 - q^8 + q^26 + q^30 - q^60 - q^66 + + - - ... (by the triple product identity).- Peter Bala, Dec 25 2024

Links

Crossrefs

Cf. numbers of the form k*(i*k+1) with k in A001057: i=0, A001057; i=1, A110660; i=2, A000217; i=3, A152749; i=4, A074378; i=5, A219190; i=6, A036498; i=7, this sequence; i=8, A154260.
Cf. A113801 (square roots of 28*a(n)+1, see the comment).
Cf. similar sequences listed in A219257.
Subsequence of A011860.

Programs

  • Magma
    k:=7; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];
  • Magma
    I:=[0,6,8,26,30]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A := proc (q) local n; for n from 0 to q do if type(sqrt(28*n+1), integer) then print(n) fi; od; end: A(4100); # Peter Bala, Dec 25 2024
  • Mathematica
    Rest[Flatten[{# (7 # - 1), # (7 # + 1)} & /@ Range[0, 25]]]
CoefficientList[Series[2 x (3 + x + 3 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,6,8,26,30},50] (* Harvey P. Dale, Sep 14 2022 *)

Formula

G.f.: 2*x^2*(3+x+3*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (14*n*(n-1)+5*(-1)^n*(2*n-1)+5)/8.
a(n) = 2*A057570(n) = (1/7)*A047335(n)*A047274(n+1).
Sum_{n>=2} 1/a(n) = 7 - cot(Pi/7)*Pi. - Amiram Eldar, Mar 17 2022

A219190 Numbers of the form k*(5*k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 4, 6, 18, 22, 42, 48, 76, 84, 120, 130, 174, 186, 238, 252, 312, 328, 396, 414, 490, 510, 594, 616, 708, 732, 832, 858, 966, 994, 1110, 1140, 1264, 1296, 1428, 1462, 1602, 1638, 1786, 1824, 1980, 2020, 2184, 2226, 2398, 2442, 2622, 2668, 2856, 2904, 3100
Offset: 1

Views

Author

Bruno Berselli, Nov 14 2012

Keywords

Comments

Equivalently, numbers m such that 20*m+1 is a square.
Also, integer values of h*(h+1)/5.
More generally, for the numbers of the form n*(k*n+1) with n in A001057, we have:
. generating function (offset 1): x^2*(k-1+2*x+(k-1)*x^2)/((1+x)^2*(1-x)^3);
. n-th term: b(n) = (2*k*n*(n-1)+(k-2)*(-1)^n*(2*n-1)+k-2)/8;
. first differences: (n-1)*((-1)^n*(k-2)+k)/2;
. b(2n+1)-b(2n) = 2*n (independent from k);
. (4*k)*b(n)+1 = (2*k*n+(k-2)*(-1)^n-k)^2/4.

Links

Crossrefs

Subsequence of A011858.
Cf. A090771: square roots of 20*a(n)+1 (see the first comment).
Cf. numbers of the form n*(k*n+1) with n in A001057: k=0, A001057; k=1, A110660; k=2, A000217; k=3, A152749; k=4, A074378; k=5, this sequence; k=6, A036498; k=7, A219191; k=8, A154260.
Cf. similar sequences listed in A219257.

Programs

  • Magma
    k:=5; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];
  • Magma
    I:=[0,4,6,18,22]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Mathematica
    Rest[Flatten[{# (5 # - 1), # (5 # + 1)} & /@ Range[0, 25]]]
CoefficientList[Series[2 x (2 + x + 2 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,4,6,18,22},50] (* Harvey P. Dale, Jan 21 2015 *)

Formula

G.f.: 2*x^2*(2 + x + 2*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (10*n*(n-1) + 3*(-1)^n*(2*n - 1) + 3)/8.
a(n) = 2*A057569(n) = A008851(n+1)*A047208(n)/5.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Jan 21 2015
Sum_{n>=2} 1/a(n) = 5 - sqrt(1+2/sqrt(5))*Pi. - Amiram Eldar, Mar 15 2022
a(n) = A132356(n-1)/2, n >= 1. - Bernard Schott, Mar 15 2022

A219391 Numbers k such that 21*k + 1 is a square.

Original entry on oeis.org

0, 3, 8, 19, 23, 40, 55, 80, 88, 119, 144, 183, 195, 240, 275, 328, 344, 403, 448, 515, 535, 608, 663, 744, 768, 855, 920, 1015, 1043, 1144, 1219, 1328, 1360, 1475, 1560, 1683, 1719, 1848, 1943, 2080, 2120, 2263, 2368, 2519, 2563, 2720, 2835, 3000, 3048, 3219
Offset: 1

Views

Author

Bruno Berselli, Nov 20 2012

Keywords

Comments

Equivalently, numbers in increasing order of the form m*(21*m + 2) or m*(21*m + 16) + 3, where m = 0, -1, 1, -2, 2, -3, 3, ....
Let F(r) = Product_{n >= 1} 1 - q^(28*n-r). The sequence terms are the exponents in the expansion of F(0)*F(3)*F(8)*F(11)*F(14)*F(17)*F(20)*F(25) = 1 - q^3 - q^8 + q^19 + q^23 - q^40 - q^55 + + - - ... (by the quintuple product identity). Cf. A204221. - Peter Bala, Dec 25 2024

Links

Crossrefs

Cf. similar sequences listed in A219257.
Cf. A219721 (square roots of 21*a(n)+1).
Subsequence of A047528.

Programs

  • Magma
    [n: n in [0..3300] | IsSquare(21*n+1)];
  • Magma
    I:=[0,3,8,19,23,40,55,80,88]; [n le 9 select I[n] else Self(n-1)+2*Self(n-4)-2*Self(n-5)-Self(n-8)+Self(n-9): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A219391:=proc(q)
local n;
for n from 1 to q do if type(sqrt(21*n+1), integer) then print(n);
fi; od; end:
A219391(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    Select[Range[0, 3300], IntegerQ[Sqrt[21 # + 1]] &]
CoefficientList[Series[x (3 + 5 x + 11 x^2 + 4 x^3 + 11 x^4 + 5 x^5 + 3 x^6)/((1 + x)^2 (1 - x)^3 (1 + x^2)^2), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,0,0,2,-2,0,0,-1,1},{0,3,8,19,23,40,55,80,88},60] (* Harvey P. Dale, Oct 01 2021 *)
  • Maxima
    makelist((42*n*(n-1)+2*%i^(n*(n+1))*(6*n+(-1)^n-3)+7*(-1)^n*(2*n-1)+11)/32, n, 1, 50);

Formula

G.f.: x^2*(3 + 5*x + 11*x^2 + 4*x^3 + 11*x^4 + 5*x^5 + 3*x^6)/((1 + x)^2*(1 - x)^3*(1 + x^2)^2).
a(n) = a(-n+1) = (42*n*(n-1) + 2*i^(n*(n+1))*(6*n + (-1)^n-3) + 7*(-1)^n*(2*n-1) + 11)/32, where i=sqrt(-1).
Sum_{n>=2} 1/a(n) = 21/4 - cot(2*Pi/21)*Pi/2 + Pi/(2*sqrt(3)) - tan(Pi/14)*Pi/2. - Amiram Eldar, Mar 16 2022

A217441 Numbers k such that 26*k+1 is a square.

Original entry on oeis.org

0, 24, 28, 100, 108, 228, 240, 408, 424, 640, 660, 924, 948, 1260, 1288, 1648, 1680, 2088, 2124, 2580, 2620, 3124, 3168, 3720, 3768, 4368, 4420, 5068, 5124, 5820, 5880, 6624, 6688, 7480, 7548, 8388, 8460, 9348, 9424, 10360, 10440, 11424, 11508, 12540, 12628
Offset: 1

Views

Author

Bruno Berselli, Nov 14 2012

Keywords

Comments

Equivalently, numbers of the form m*(26*m+2), where m = 0,-1,1,-2,2,-3,3,...
Also, integer values of 2*h*(h+1)/13.

Links

Crossrefs

Cf. similar sequences listed in A219257.
Cf. A174768 (the squares A174768^2 belong to the sequence), A175886.

Programs

  • Magma
    [n: n in [0..13000] | IsSquare(26*n+1)];
  • Magma
    I:=[0,24,28,100,108]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A217441:=proc(q)
local n;
for n from 1 to q do if type(sqrt(26*n+1), integer) then print(n);
fi; od; end:
A217441(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    Select[Range[0, 13000], IntegerQ[Sqrt[26 # + 1]] &]
CoefficientList[Series[4 x (6 + x + 6 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,24,28,100,108},50] (* Harvey P. Dale, Nov 03 2019 *)
  • PARI
    a(n)=is(n)=issquare(26*n+1) \\ Charles R Greathouse IV, Oct 16 2015

Formula

G.f.: 4*x^2*(6 + x + 6*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (26*n*(n-1) + 11*(-1)^n*(2*n - 1) - 3)/4 + 3 = (26*n + 11*(-1)^n - 15)*(26*n + 11*(-1)^n - 11)/104.
26*a(2*n-1)+1 = A175886(4*n-3)^2, 26*a(2*n)+1 = A175886(4*n)^2.
Sum_{n>=2} 1/a(n) = 13/2 - cot(Pi/13)*Pi/2. - Amiram Eldar, Mar 17 2022

A219258 Numbers k such that 27*k+1 is a square.

Original entry on oeis.org

0, 25, 29, 104, 112, 237, 249, 424, 440, 665, 685, 960, 984, 1309, 1337, 1712, 1744, 2169, 2205, 2680, 2720, 3245, 3289, 3864, 3912, 4537, 4589, 5264, 5320, 6045, 6105, 6880, 6944, 7769, 7837, 8712, 8784, 9709, 9785, 10760, 10840, 11865, 11949, 13024, 13112
Offset: 1

Views

Author

Bruno Berselli, Nov 19 2012

Keywords

Comments

Equivalently, numbers of the form m*(27*m+2), where m = 0,-1,1,-2,2,-3,3,...
Also, integer values of h*(h+2)/27.
The sequence terms are the exponents in the expansion of Product_{n >= 1} (1 - q^(54*n))*(1 - q^(54*n-25))*(1 - q^(54*n-29)) = 1 - q^25 - q^29 + q^104 + q^112 - q^237 - q^249 + + - - .... - Peter Bala, Dec 24 2024

Links

Crossrefs

Cf. similar sequences listed in A219257.
Cf. A056081 (square roots of 27*a(n)+1).

Programs

  • Magma
    [n: n in [0..14000] | IsSquare(27*n+1)];
  • Magma
    I:=[0,25,29,104,112]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A219258:=proc(q)
local n;
for n from 1 to q do if type(sqrt(27*n+1), integer) then print(n);
fi; od; end:
A219258(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    Select[Range[0, 14000], IntegerQ[Sqrt[27 # + 1]] &]
CoefficientList[Series[x (25 + 4 x + 25 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

Formula

G.f.: x^2*(25 + 4*x + 25*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (54*n*(n-1) + 23*(-1)^n*(2*n - 1) - 1)/8 + 3.
Sum_{n>=2} 1/a(n) = 27/4 - cot(2*Pi/27)*Pi/2. - Amiram Eldar, Mar 17 2022

A219259 Numbers k such that 25*k+1 is a square.

Original entry on oeis.org

0, 23, 27, 96, 104, 219, 231, 392, 408, 615, 635, 888, 912, 1211, 1239, 1584, 1616, 2007, 2043, 2480, 2520, 3003, 3047, 3576, 3624, 4199, 4251, 4872, 4928, 5595, 5655, 6368, 6432, 7191, 7259, 8064, 8136, 8987, 9063, 9960, 10040, 10983, 11067, 12056, 12144
Offset: 1

Views

Author

Bruno Berselli, Nov 19 2012

Keywords

Comments

Equivalently, numbers of the form m*(25*m+2), where m = 0,-1,1,-2,2,-3,3,...
Also, integer values of h*(h+2)/25.
Exponents in the expansion of Product_{n >= 1} (1 - q^(50*n))*(1 - q^(50*n-23))*(1 - q^(50*n-27)) = 1 - q^23 - q^27 + q^96 + q^104 - q^219 - q^231 + + - - .... - Peter Bala, Dec 18 2024

Links

Crossrefs

Cf. similar sequences listed in A219257.
Cf. A047209, A197652.

Programs

  • Magma
    [n: n in [0..13000] | IsSquare(25*n+1)];
  • Magma
    I:=[0,23,27,96,104]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A219259:=proc(q)
local n;
for n from 1 to q do if type(sqrt(25*n+1), integer) then print(n);
fi; od; end:
A219259(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    Select[Range[0, 13000], IntegerQ[Sqrt[25 # + 1]] &]
CoefficientList[Series[x (23 + 4 x + 23 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

Formula

G.f.: x^2*(23 + 4*x + 23*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (50*n*(n-1) + 21*(-1)^n*(2*n - 1) + 5)/8 + 2.
25*a(n)+1 = A047209(A197652(n+1))^2.
Sum_{n>=2} 1/a(n) = 25/4 - cot(2*Pi/25)*Pi/2. - Amiram Eldar, Mar 17 2022
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