A142999 -id:A142999 - cp's OEIS frontend

A013661 Decimal expansion of Pi^2/6 = zeta(2) = Sum_{m>=1} 1/m^2.

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

Offset: 1

N. J. A. Sloane

"In 1736 he [Leonard Euler, 1707-1783] discovered the limit to the infinite series, Sum 1/n^2. He did it by doing some rather ingenious mathematics using trigonometric functions that proved the series summed to exactly Pi^2/6. How can this be? ... This demonstrates one of the most startling characteristics of mathematics - the interconnectedness of, seemingly, unrelated ideas." - Clawson [See Hardy and Wright, Theorems 332 and 333. - N. J. A. Sloane, Jan 20 2017]
Also dilogarithm(1). - Rick L. Shepherd, Jul 21 2004
Also Integral_{x>=0} x/(exp(x)-1) dx. [Abramowitz-Stegun, 23.2.7., for s=2, p. 807]
For the partial sums see the fractional sequence A007406/A007407.
Pi^2/6 is also the length of the circumference of a circle whose diameter equals the ratio of volume of an ellipsoid to the circumscribed cuboid. Pi^2/6 is also the length of the circumference of a circle whose diameter equals the ratio of surface area of a sphere to the circumscribed cube. - Omar E. Pol, Oct 07 2011
1 < n^2/(eulerphi(n)*sigma(n)) < zeta(2) for n > 1. - Arkadiusz Wesolowski, Sep 04 2012
Volume of a sphere inscribed in a cube of volume Pi. More generally, Pi^x/6 is the volume of an ellipsoid inscribed in a cuboid of volume Pi^(x-1). - Omar E. Pol, Feb 17 2016
Surface area of a sphere inscribed in a cube of surface area Pi. More generally, Pi^x/6 is the surface area of a sphere inscribed in a cube of surface area Pi^(x-1). - Omar E. Pol, Feb 19 2016
zeta(2)+1 is a weighted average of the integers, n > 2, using zeta(n)-1 as the weights for each n. We have: Sum_{n >= 2} (zeta(n)-1) = 1 and Sum_{n >= 2} n*(zeta(n)-1) = zeta(2)+1. - Richard R. Forberg, Jul 14 2016
zeta(2) is the expected value of sigma(n)/n. - Charlie Neder, Oct 22 2018
Graham shows that a rational number x can be expressed as a finite sum of reciprocals of distinct squares if and only if x is in [0, Pi^2/6-1) U [1, Pi^2/6). See section 4 for other results and Theorem 5 for the underlying principle. - Charles R Greathouse IV, Aug 04 2020
From Peter Bala, Aug 24 2025: (Start)
By definition, zeta(2) = lim_{n -> oo} s(n), where s(n) = Sum_{k = 1..n} 1/k^2. The convergence is slow. For example, s(50) = 1.6(25...) is only correct to 1 decimal digit.
Let S(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*s(n+k). It appears that S(n) converges to zeta(2) much more rapidly. For example, S(50) = 1.64493406684822643647241516664602(137...) gives zeta(2) correct to 32 decimal digits. Cf. A073004. (End)

			1.6449340668482264364724151666460251892189499012067984377355582293700074704032...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.
F. Aubonnet, D. Guinin and B. Joppin, Précis de Mathématiques, Analyse 2, Classes Préparatoires, Premier Cycle Universitaire, Bréal, 1990, Exercice 908, pages 82 and 91-92.
Calvin C. Clawson, Mathematical Mysteries, The Beauty and Magic of Numbers, Perseus Books, 1996, p. 97.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 262.
W. Dunham, Euler: The Master of Us All, The Mathematical Association of America, Washington, D.C., 1999, p. xxii.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 1.4.1 and 5.16, pp. 20, 365.
Hardy and Wright, 'An Introduction to the Theory of Numbers'. See Theorems 332 and 333.
A. A. Markoff, Mémoire sur la transformation de séries peu convergentes en séries très convergentes, Mém. de l'Acad. Imp. Sci. de St. Pétersbourg, XXXVII, 1890.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 10, 161-162.
G. F. Simmons, Calculus Gems, Section B.15, B.24, pp. 270-271, 323-325, McGraw Hill, 1992.
Arnold Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, Deutscher Verlag der Wissenschaften, Berlin, 1963, p. 99, Satz 1.
A. Weil, Number theory: an approach through history; from Hammurapi to Legendre, Birkhäuser, Boston, 1984; see p. 261.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, page 23.

Harry J. Smith, Table of n, a(n) for n = 1..20000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
D. H. Bailey, J. M. Borwein, and D. M. Bradley, Experimental determination of Apéry-like identities for zeta(4n+2), arXiv:math/0505270 [math.NT], 2005-2006.
Peter Bala, New series for old functions.
Peter Bala, Formulas for A013661.
David Benko and John Molokach, The Basel Problem as a Rearrangement of Series, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 171-176.
R. Calinger, Leonhard Euler: The First St. Petersburg Years (1727-1741), Historia Mathematica, Vol. 23, 1996, pp. 121-166.
Robin Chapman, Evaluating Zeta(2).
R. W. Clickery, Probability of two numbers being coprime.
Alessio Del Vigna, On a solution to the Basel problem based on the fundamental theorem of calculus, arXiv:2104.01710 [math.HO], 2021.
Leonhard Euler, On the sums of series of reciprocals, arXiv:math/0506415 [math.HO], 2005-2008.
Leonhard Euler, De summis serierum reciprocarum, E41.
R. L. Graham, On finite sums of unit fractions, Proceedings of the London Mathematical Society, s3-14 (1964), pp. 193-207. doi:10.1112/plms/s3-14.2.193
Michael D. Hirschhorn, A simple proof that zeta(2) = Pi^2/6, The Mathematical Intelligencer 33:3 (2011), pp 81-82.
Melissa Larson, Verifying and discovering BBP-type formulas, 2008.
Alain Lasjaunias and Jean-Paul Tran, A note on the equality Pi^2/6 = Sum_{n>=1} 1/n^2, arXiv:2312.02245 [math.HO], 2023.
Math. Reference Project, The Zeta Function, Zeta(2).
Math. Reference Project, The Zeta Function, Odds That Two Numbers Are Coprime".
Romeo Mestrovic, Euclid's theorem on the infinitude of primes: a historical survey of its proofs (300 BC--2012) and another new proof, arXiv preprint arXiv:1202.3670 [math.HO], 2012.
Jon Perry, Prime Product Paradox
Simon Plouffe, Plouffe's Inverter, Zeta(2) or Pi**2/6 to 100000 digits.
Simon Plouffe, Zeta(2) or Pi**2/6 to 10000 places.
Simon Plouffe, Zeta(2) to Zeta(4096) to 2048 digits each (gzipped file)
A. L. Robledo, value of the Riemann zeta function at s=2, PlanetMath.org.
E. Sandifer, How Euler Did It, Estimating the Basel Problem.
E. Sandifer, How Euler Did It, Basel Problem with Integrals.
C. Tooth, Pi squared over six.
Eric Weisstein's World of Mathematics, Dilogarithm.
Eric Weisstein's World of Mathematics, Riemann Zeta Function zeta(2).
Wikipedia, Basel Problem.
Wikipedia, Bailey-Borwein-Plouffe formula.
Herbert S. Wilf, Accelerated series for universal constants, by the WZ method, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).
Index entries for transcendental numbers.
Index entries for zeta function.

Cf. A001008 (H(n): numerators), A002805 (denominators), A013679 (continued fraction), A002117 (zeta(3)), A013631 (cont.frac. for zeta(3)), A013680 (cont.frac. for zeta(4)), 1/A059956, A108625, A142995, A142999.

Cf. A000290.

pi:=Pi(RealField(110)); Reverse(Intseq(Floor(10^105*pi^2/6))); // Vincenzo Librandi, Oct 13 2015

evalf(Pi^2/6,120); # Muniru A Asiru, Oct 25 2018
# Calculates an approximation with n exact decimal places (small deviation
# in the last digits are possible). Goes back to ideas of A. A. Markoff 1890.
zeta2 := proc(n) local q, s, w, v, k; q := 0; s := 0; w := 1; v := 4;
for k from 2 by 2 to 7*n/2 do
    w := w*v/k;
    q := q + v;
    v := v + 8;
    s := s + 1/(w*q);
od; 12*s; evalf[n](%) end:
zeta2(1000); # Peter Luschny, Jun 10 2020

RealDigits[N[Pi^2/6, 100]][[1]]
RealDigits[Zeta[2],10,120][[1]] (* Harvey P. Dale, Jan 08 2021 *)

fpprec : 100$ ev(bfloat(zeta(2)))$ bfloat(%); /* Martin Ettl, Oct 21 2012 */

PARI
```
default(realprecision, 200); Pi^2/6
```
PARI
```
default(realprecision, 200); dilog(1)
```
PARI
```
default(realprecision, 200); zeta(2)
```

A013661(n)={localprec(n+2); Pi^2/.6\10^n%10} \\ Corrected and improved by M. F. Hasler, Apr 20 2021

default(realprecision, 20080); x=Pi^2/6; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b013661.txt", n, " ", d)); \\ Harry J. Smith, Apr 29 2009

sumnumrat(1/x^2, 1) \\ Charles R Greathouse IV, Jan 20 2022

# Use some guard digits when computing.
# BBP formula (3 / 16) P(2, 64, 6, (16, -24, -8, -6,  1, 0)).
from decimal import Decimal as dec, getcontext
def BBPzeta2(n: int) -> dec:
    getcontext().prec = n
    s = dec(0); f = dec(1); g = dec(64)
    for k in range(int(n * 0.5536546824812272) + 1):
        sixk = dec(6 * k)
        s += f * ( dec(16) / (sixk + 1) ** 2 - dec(24) / (sixk + 2) ** 2
                 - dec(8)  / (sixk + 3) ** 2 - dec(6)  / (sixk + 4) ** 2
                 + dec(1)  / (sixk + 5) ** 2 )
        f /= g
    return (s * dec(3)) / dec(16)
print(BBPzeta2(2000))  # Peter Luschny, Nov 01 2023

Limit_{n->oo} (1/n)*(Sum_{k=1..n} frac((n/k)^(1/2))) = zeta(2) and in general we have lim_{n->oo} (1/n)*(Sum_{k=1..n} frac((n/k)^(1/m))) = zeta(m), m >= 2. - Yalcin Aktar, Jul 14 2005
Equals Integral_{x=0..1} (log(x)/(x-1)) dx or Integral_{x>=1} (log(x/(x-1))/x) dx. - Jean-François Alcover, May 30 2013
For s >= 2 (including Complex), zeta(s) = Product_{n >= 1} prime(n)^s/(prime(n)^s - 1). - Fred Daniel Kline, Apr 10 2014
Also equals 1 + Sum_{n>=0} (-1)^n*StieltjesGamma(n)/n!. - Jean-François Alcover, May 07 2014
zeta(2) = Sum_{n>=1} ((floor(sqrt(n)) - floor(sqrt(n-1)))/n). - Mikael Aaltonen, Jan 10 2015
zeta(2) = Sum_{n>=1} (((sqrt(5)-1)/2/sqrt(5))^n/n^2) + Sum_{n>=1} (((sqrt(5)+1)/2/sqrt(5))^n/ n^2) + log((sqrt(5)-1)/2/sqrt(5))log((sqrt(5)+1)/2/sqrt(5)). - Seiichi Kirikami, Oct 14 2015
The above formula can also be written zeta(2) = dilog(x) + dilog(y) + log(x)*log(y) where x = (1-1/sqrt(5))/2 and y=(1+1/sqrt(5))/2. - Peter Luschny, Oct 16 2015
zeta(2) = Integral_{x>=0} 1/(1 + e^x^(1/2)) dx, because (1 - 1/2^(s-1))*Gamma[1 + s]*Zeta[s] = Integral_{x>=0} 1/(1 + e^x^(1/s)) dx. After Jean-François Alcover in A002162. - Mats Granvik, Sep 12 2016
zeta(2) = Product_{n >= 1} (144*n^4)/(144*n^4 - 40*n^2 + 1). - Fred Daniel Kline, Oct 29 2016
zeta(2) = lim_{n->oo} (1/n) * Sum_{k=1..n} A017665(k)/A017666(k). - Dimitri Papadopoulos, May 10 2019 [See the Walfisz reference, and a comment in A284648, citing also the Sándor et al. Handbook. - Wolfdieter Lang, Aug 22 2019]
Equals Sum_{k>=1} H(k)/(k*(k+1)), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number. - Amiram Eldar, Aug 16 2020
Equals (8/3)*(1/2)!^4 = (8/3)*Gamma(3/2)^4. - Gary W. Adamson, Aug 17 2021
Equals ((m+1)/m) * Integral_{x=0..1} log(Sum {k=0..m} x^k )/x dx, m > 0 (Aubonnet reference). - _Bernard Schott, Feb 11 2022
Equals 1 + Sum_{n>=1} n*(zeta(n+2)-1). - Richard R. Forberg, Jun 04 2023; improved by Natalia L. Skirrow, Jul 25 2025
Equals Psi'(1) where Psi'(x) is the trigamma function (by Abramowitz Stegun 6.4.2). - Andrea Pinos, Oct 22 2024
Equals Integral_{x=0..1} Integral_{y=0..1} 1/(1 - x*y) dy dx. - Kritsada Moomuang, May 22 2025
Equals 1 + Sum_{n>=1} 1/(n^2*(n+1)). - Natalia L. Skirrow, Jul 25 2025

Edited by N. J. A. Sloane, Nov 22 2023

A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1

Offset: 0

Paul D. Hanna, Jun 12 2005

Compare to the corresponding array A108553 of crystal ball sequences for D_n lattice.
From Peter Bala, Jul 18 2008: (Start)
Row reverse of A099608.
This array has a remarkable relationship with the constant zeta(2). The row, column and diagonal entries of the array occur in series acceleration formulas for zeta(2).
For the entries in row n we have zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k)). For example, n = 4 gives zeta(2) = 2*(1 - 1/4 + 1/9 - 1/16) + 1/(1*21) + 1/(4*21*131) + 1/(9*131*471) + ... . See A142995 for further details.
For the entries in column k we have zeta(2) = (1 + 1/4 + 1/9 + ... + 1/k^2) + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k)). For example, k = 4 gives zeta(2) = (1 + 1/4 + 1/9 + 1/16) + 2*(1/(1*9) - 1/(4*9*61) + 1/(9*61*309) - ... ). See A142999 for further details.
Also, as consequence of Apery's proof of the irrationality of zeta(2), we have a series acceleration formula along the main diagonal of the table: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n,n)*T(n-1,n-1)) = 5*(1/3 - 1/(2^2*3*19) + 1/(3^2*19*147) - ...).
There also appear to be series acceleration results along other diagonals. For example, for the main subdiagonal, calculation supports the result zeta(2) = 2 - Sum_{n >= 1} (-1)^(n+1)*(n^2+(2*n+1)^2)/(n^2*(n+1)^2*T(n,n-1)*T(n+1,n)) = 2 - 10/(2^2*7) + 29/(6^2*7*55) - 58/(12^2*55*471) + ..., while for the main superdiagonal we appear to have zeta(2) = 1 + Sum_{n >= 1} (-1)^(n+1)*((n+1)^2 + (2*n+1)^2)/(n^2*(n+1)^2*T(n-1,n)*T(n,n+1)) = 1 + 13/(2^2*5) - 34/(6^2*5*37) + 65/(12^2*37*309) - ... .
Similar series acceleration results hold for Apery's constant zeta(3) involving the crystal ball sequences for the product lattices A_n x A_n; see A143007 for further details. Similar results also hold between the constant log(2) and the crystal ball sequences of the hypercubic lattices A_1 x...x A_1 and between log(2) and the crystal ball sequences for lattices of type C_n ; see A008288 and A142992 respectively for further details. (End)
This array is the Hilbert transform of triangle A008459 (see A145905 for the definition of the Hilbert transform). - Peter Bala, Oct 28 2008

			Square array begins:
  1,   1,    1,     1,      1,       1,       1, ... A000012;
  1,   3,    5,     7,      9,      11,      13, ... A005408;
  1,   7,   19,    37,     61,      91,     127, ... A003215;
  1,  13,   55,   147,    309,     561,     923, ... A005902;
  1,  21,  131,   471,   1251,    2751,    5321, ... A008384;
  1,  31,  271,  1281,   4251,   11253,   25493, ... A008386;
  1,  43,  505,  3067,  12559,   39733,  104959, ... A008388;
  1,  57,  869,  6637,  33111,  124223,  380731, ... A008390;
  1,  73, 1405, 13237,  79459,  350683, 1240399, ... A008392;
  1,  91, 2161, 24691, 176251,  907753, 3685123, ... A008394;
  1, 111, 3191, 43561, 365751, 2181257, ...      ... A008396;
  ...
As a triangle:
  [0]  1
  [1]  1,  1
  [2]  1,  3,   1
  [3]  1,  7,   5,    1
  [4]  1, 13,  19,    7,    1
  [5]  1, 21,  55,   37,    9,    1
  [6]  1, 31, 131,  147,   61,   11,   1
  [7]  1, 43, 271,  471,  309,   91,  13,   1
  [8]  1, 57, 505, 1281, 1251,  561, 127,  15,  1
  [9]  1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
       ...
Inverse binomial transform of rows yield rows of triangle A063007:
  1;
  1,  2;
  1,  6,   6;
  1, 12,  30,  20;
  1, 20,  90, 140,  70;
  1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
  1;
  1,  1;
  1,  4,   1;
  1,  9,   9,   1;
  1, 16,  36,  16,  1;
  1, 25, 100, 100, 25, 1;
  ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
Joseph T. Iosue, T. C. Mooney, Adam Ehrenberg, and Alexey V. Gorshkov, Projective toric designs, difference sets, and quantum state designs, arXiv:2311.13479 [quant-ph], 2023. See page 6.
Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.
A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report. Math. Intelligencer 1 (1978/79), no 4, 195-203.
Eric Weisstein's World of Mathematics, Apéry number.

Rows include: A003215 (row 2), A005902 (row 3), A008384 (row 4), A008386 (row 5), A008388 (row 6), A008390 (row 7), A008392 (row 8), A008394 (row 9), A008396 (row 10).
Cf. A063007, A099601 (n-th term of A_{2n} lattice), A108553.
Cf. A008459 (h-vectors type B associahedra), A145904, A145905.
Cf. A005258 (main diagonal), A108626 (antidiagonal sums).
Cf. A108628, A208675, A363867, A363868, A363869, A363870, A363871.

T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array
A108625:= func< n,k | T(n-k,k) >; // antidiagonals
[A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023

T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1):
seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018

T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *)

T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))

def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array
def A108625(n,k): return T(n-k, k) # antidiagonals
flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
The square array = A063007 * transpose(A007318).
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A142995 a(0) = 0, a(1) = 1, a(n+1) = (2n^2 + 2n + 3)a(n) - n^4a(n-1), n >= 1.

Original entry on oeis.org

0, 1, 7, 89, 1836, 56164, 2390832, 135213840, 9809203968, 888117094656, 98167241088000, 13010123816064000, 2036436482119680000, 371699564417796096000, 78251077775510986752000

Offset: 0

Peter Bala, Jul 18 2008

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + m^2 + m + 1)*a(n) - n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^2 for the constant zeta(2). For other cases see A001819 (m=0), A142996 (m=2), A142997 (m=3) and A142998 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^2*p_m(n)*Sum_{k = 1..n} 1/(k^2*p_m(k-1)*p_m(k)), where p_m(x) := Sum_{k = 0..m} C(m,k)^2*C(x+k,m) = Sum_{k = 0..m} C(m,k)*C(m+k,k)*C(x,k) is the Ehrhart polynomial of the polytope formed from the convex hull of a root system of type A_m (equivalently, the polynomial that generates the crystal ball sequence for the A_m lattice [Bacher et al.]).
The first few are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 3*x^2 + 3*x + 1 and p_3(x) = (10*x^3 + 15*x^2 + 11*x + 3)/3. The o.g.f. for the p_m(x) is ((1-t^2)^x/(1-t)^(2x+1))*Legendre_P(x,(1+t^2)/(1-t^2)) = 1 + (2*x+1)*t + (3*x^2+3*x+1)*t^2 + ... [Gogin & Hirvensalo, Theorem 1 with N = -1].
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^2*f(x+1) + x^2*f(x-1) = (2*x^2 + 2*x + m^2 + m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^2*p_m(n) with initial conditions b(0) = 1, b(1) = m^2 + m + 1. Hence the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = Sum_{k>=1} 1/(k^2*p_m(k-1)*p_m(k)) = 1/((m^2 + m + 1) - 1^4/((m^2 + m + 5) - 2^4/((m^2 + m + 13) - ... - n^4/((2*n^2 + 2*n + m^2 + m + 1) - ...)))) = 2*Sum_{k>=1} (-1)^(k+1)/(m+k)^2. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).
For related results see A142999. For corresponding results for the constants e, log(2) and zeta(3) see A000522, A142979 and A143003 respectively.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 0..252
R. Bacher, P. de la Harpe and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.
N. Gogin and M. Hirvensalo, On the generating function of discrete Chebyshev polynomials, Turku Centre for Computer Science Technical Report No. 819, (2007), 1-8.

Cf. A000522, A001819, A003215 (A_2 lattice), A005902 (A_3 lattice), A008384 (A_4 lattice), A008386 (A_5 lattice), A108625, A142979, A142996, A142997, A142998, A143003.

p := n -> 2*n+1: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

a[n_] := -1/6*n!^2*(2*n*(Pi^2-12) + Pi^2 - 6*(2*n+1)*PolyGamma[1, n+1]) // Simplify; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Mar 06 2013 *)

a(n) = n!^2*p(n)*Sum_{k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = 2*n+1. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 3. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - (n-1)^4/(2*n^2 - 2*n + 3))))), for n >= 2. Lim_{n -> infinity} a(n)/b(n) = 1/(3 - 1^4/(7 - 2^4/(15 - 3^4/(27 - ... - n^4/((2*n^2 + 2*n + 3) - ...))))) = Sum_{k>=1} 1/(k^2*(4*k^2 - 1)) = 2 - zeta(2).

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+1 )*a(n) + n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} (-1)^(k+1)/k^2 for the constant 1/2*zeta(2). For other cases see A142999 (m=0), A143001 (m=2) and A143002 (m=3).

This is the case m = 2 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+1 )*a(n) + n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k >= 1} (-1)^(k+1)/k^2 for the constant (1/2)*zeta(2). For other cases see A142999 (m=0), A143000(m=1) and A143002 (m=3).

This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+1)*a(n) + n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} (-1)^(k+1)/k^2 for the constant 1/2*zeta(2). For other cases see A142999 (m=0), A143000 (m=1) and A143001 (m=2).

cp's OEIS Frontend

A013661 Decimal expansion of Pi^2/6 = zeta(2) = Sum_{m>=1} 1/m^2.

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A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.

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A142995 a(0) = 0, a(1) = 1, a(n+1) = (2*n^2 + 2*n + 3)*a(n) - n^4*a(n-1), n >= 1.

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A143000 a(0) = 0, a(1) = 1, a(n+1) = 3*(2*n+1)*a(n) + n^4*a(n-1).

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A143001 a(0) = 0, a(1) = 1, a(n+1) = 5*(2*n+1)*a(n) + n^4*a(n-1).

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A143002 a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1).

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A142995 a(0) = 0, a(1) = 1, a(n+1) = (2n^2 + 2n + 3)a(n) - n^4a(n-1), n >= 1.

A143000 a(0) = 0, a(1) = 1, a(n+1) = 3(2n+1)a(n) + n^4a(n-1).

A143001 a(0) = 0, a(1) = 1, a(n+1) = 5(2n+1)a(n) + n^4a(n-1).

A143002 a(0) = 0, a(1) = 1, a(n+1) = 7(2n+1)a(n) + n^4a(n-1).