A302330 -id:A302330 - cp's OEIS frontend

A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

1, 1, 1, 1, 13, 1, 1, 181, 33, 1, 1, 2521, 1121, 61, 1, 1, 35113, 38081, 3781, 97, 1, 1, 489061, 1293633, 234361, 9505, 141, 1, 1, 6811741, 43945441, 14526601, 931393, 20021, 193, 1, 1, 94875313, 1492851361, 900414901, 91267009, 2842841, 37441, 253, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=-1 results shown here; y=1 results are A188647.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/n) * T_{2*k+1}(n), with the Chebyshev polynomials of the first kind (type T). - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,  13,    181,      2521,        35113, ...
   3 | 1,  33,   1121,     38081,      1293633, ...
   4 | 1,  61,   3781,    234361,     14526601, ...
   5 | 1,  97,   9505,    931393,     91267009, ...
   6 | 1, 141,  20021,   2842841,    403663401, ...
   7 | 1, 193,  37441,   7263361,   1409054593, ...
   8 | 1, 253,  64261,  16322041,   4145734153, ...
   9 | 1, 321, 103361,  33281921,  10716675201, ...
  10 | 1, 397, 158005,  62885593,  25028308009, ...
  11 | 1, 481, 231841, 111746881,  53861764801, ...
  12 | 1, 573, 328901, 188788601, 108364328073, ...
  13 | 1, 673, 453601, 305726401, 206059140673, ...
  14 | 1, 781, 610741, 477598681, 373481557801, ...
  15 | 1, 897, 805505, 723342593, 649560843009, ...
  ...

Row 1-8 give A000012, A001570, A077420, A302329, A302330, A302331, A302332, A253880.
Column 1 is A082109(n-1).
Cf. A188644, A188647 (f(x, y) as above with y=1).
Diagonal gives A322904.

A[n_, k_] := 1/n ChebyshevT[2k+1, n];
Table[A[n-k, k], {n, 1, 9}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jan 02 2019, after Seiichi Manyama *)

A(n,k) = 2 * A188644(n,k) - A(n,k-1).

A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j+1)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

Original entry on oeis.org

1, 61, 3781, 234361, 14526601, 900414901, 55811197261, 3459393815281, 214426605350161, 13290990137894701, 823826961944121301, 51063980650397625961, 3165142973362708688281, 196187800367837541047461, 12160478479832564836254301, 753753477949251182306719201

Offset: 0

Bruno Berselli, Apr 05 2018

Centered hexagonal numbers (A003215) with index in A145607. Example: 35 is a member of A145607, therefore A003215(35) = 3781 is a term of this sequence.
Also, centered 10-gonal numbers (A062786) with index in A182432. Example: 28 is a member of A182432 and A062786(28) = 3781.
a(n) is a solution to the Pell equation (4*a(n))^2 - 15*b(n)^2 = 1. The corresponding b(n) are A258684(n). - Klaus Purath, Jul 19 2025

Colin Barker, Table of n, a(n) for n = 0..500
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (62,-1).

Cf. A003215, A145607; A062786, A182432.
Fourth row of the array A188646.
First bisection of A041449, A042859.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k: A000012 (k=1), A001570 (k=2), A077420 (k=3), this sequence (k=4), A302330 (k=5), A302331 (k=6), A302332 (k=7), A253880 (k=8).

Mathematica
```
LinearRecurrence[{62, -1}, {1, 61}, 20]
```

x='x+O('x^99); Vec((1-x)/(1-62*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 62*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(4))/4.
a(n) = ((4 + sqrt(15))^(2*n + 1) + 1/(4 + sqrt(15))^(2*n + 1))/8.
a(n) = (1/4)*T(2*n+1, 4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022
E.g.f.: exp(31*x)*(4*cosh(8*sqrt(15)*x) + sqrt(15)*sinh(8*sqrt(15)*x))/4. - Stefano Spezia, Jul 25 2025

A046173 Indices of square numbers that are also pentagonal.

Original entry on oeis.org

1, 99, 9701, 950599, 93149001, 9127651499, 894416697901, 87643708742799, 8588189040096401, 841554882220704499, 82463790268588944501, 8080609891439495856599, 791817305570802005002201, 77590015336047156994359099, 7603029685627050583442189501

Offset: 1

Eric W. Weisstein

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^4 = 49 + 20 * sqrt(6). - Ant King, Nov 07 2011
a(n)^2 is of the form (2*m-1)*(3*m-2), and the corresponding values of m are 1, 41, 3961, 388081, 38027921, 3726348121, 365144087881, ..., with closed form ((5-2*sqrt(6))^(2n-1)+(5+2*sqrt(6))^(2n-1)+14)/24 (for n>0). - Bruno Berselli, Dec 12 2013
The terms of this sequence satisfy the Diophantine equation m^2 = k * (3k-1)/2, which is equivalent to (6k-1)^2 - 6*(2*m)^2 = 1. Now, with x=6k-1 and y=2*m, we get the Pell-Fermat equation x^2 - 6*y^2 = 1. The solutions (x,y) of this equation are respectively in A046174 and A046175. The indices m=y/2 of the square numbers which are also pentagonal are the terms of this sequence, the indices k=(x+1)/6 of the pentagonal numbers which are also square are in A046172, and the pentagonal square numbers are in A036353. - Bernard Schott, Mar 10 2019
Also, this sequence is related to A302330 by (sqrt(2) + sqrt(3))^(4*n-2) = A302330(n-1)*5 + a(n)*sqrt(24). - Bruno Berselli, Oct 29 2019

			G.f. = x + 99*x^2 + 9701*x^3 + 950599*x^4 + 93149001*x^5 + ...
99 is a term because 99^2 = 9801 = (1/2) * 81 * (3*81 - 1), so 9801 is the 99th square number, also the 81st pentagonal number, and the second pentagonal square number after 1. - _Bernard Schott_, Mar 10 2019

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 35.

Colin Barker, Table of n, a(n) for n = 1..503
M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (98,-1).

Cf. A036353 (pentagonal square numbers), A046172 (indices of pentagonal numbers that are also square).
Cf. A046174, A046175 (solutions of x^2 - 6*y^2 = 1).
Cf. A302330.

CoefficientList[Series[(1 + x)/(1 - 98* x + x^2), {x, 0, 30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{98, -1}, {1, 99}, 30] (* Harvey P. Dale, Jul 31 2017 *)

{a(n) = subst( poltchebi(n) - poltchebi(n-1), 'x, 49) / 48}; /* Michael Somos, Sep 05 2006 */

Vec(x*(x+1)/(x^2-98*x+1) + O(x^30)) \\ Colin Barker, Jun 23 2015

a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2). - Warut Roonguthai, Jan 05 2001
a(1-n) = -a(n) for all n in Z. - Michael Somos, Sep 05 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),5),5). - Marcos Carreira, Dec 27 2006
a(n) = ((12+5*sqrt(6))/24)*(5+2*sqrt(6))^(2*n)+((12-5*sqrt(6))/24)*(5-2*sqrt(6))^(2*n) for n>=0. - Richard Choulet, Apr 29 2009
a(n+1) = 49*a(n) + 10*sqrt(24*a(n)^2+1) for n > =0 with a(0)=1. - Richard Choulet, Apr 29 2009
a(n) = b such that (-1)^n*Integral_{x=-Pi/2..Pi/2} (cos(2*n-1)*x)/(5-sin(x)) dx = c + b*(log(2)-log(3)). - Francesco Daddi, Aug 01 2011
a(n) = floor((1/24) * sqrt(6) * (sqrt(2) + sqrt(3))^(4n-2)). - Ant King, Nov 07 2011
a(n) = A138288(n)*A054320(n). - Gerry Martens, May 13 2024

A322904 a(n) = Sum_{k=0..n} binomial(2n+1,2k+1)(n^2-1)^(n-k)n^(2*k).

Original entry on oeis.org

1, 1, 181, 38081, 14526601, 8943235489, 8138661470941, 10287228590683393, 17254778510170993681, 37095265466946847758401, 99474891266913130060486021, 325534304813775692747248543681, 1276941308627620432293188401109401, 5914558735952850788377566338591400673

Offset: 0

Seiichi Manyama, Dec 30 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..193
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Diagonal of A188646.

Cf. A253880, A302329, A302330, A302331, A302332.

[&+[Binomial(2*n+1,2*k+1)*(n^2-1)^(n-k)*n^(2*k): k in [0..n]]: n in [0..20]]; // Vincenzo Librandi, Jan 03 2019

a[0] = 1; a[n_] := 1/n ChebyshevT[2n+1, n];
Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Jan 02 2019 *)

{a(n) = sum(k=0, n, binomial(2*n+1, 2*k+1)*(n^2-1)^(n-k)*n^(2*k))}

a(n) = if (n==0, 1, polchebyshev(2*n+1, 1, n)/n); \\ Michel Marcus, Jan 02 2019

For n > 0, a(n) = (1/n) * T_{2*n+1}(n) where T_{n}(x) is a Chebyshev polynomial of the first kind.
For n > 0, a(n) = (1/n) * cosh((2*n+1)*arccosh(n)).
a(n) ~ 4^n * n^(2*n). - Vaclav Kotesovec, Jan 03 2019

cp's OEIS Frontend

A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

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A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

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A046173 Indices of square numbers that are also pentagonal.

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A322904 a(n) = Sum_{k=0..n} binomial(2*n+1,2*k+1)*(n^2-1)^(n-k)*n^(2*k).

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Formula

A322904 a(n) = Sum_{k=0..n} binomial(2n+1,2k+1)(n^2-1)^(n-k)n^(2*k).