A322699 -id:A322699 - cp's OEIS frontend

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1

Offset: 0

Seiichi Manyama, Dec 26 2018

			Square array begins:
   1,  1,   1,    1,      1,       1,         1, ...
   1,  3,  17,   99,    577,    3363,     19601, ...
   1,  5,  49,  485,   4801,   47525,    470449, ...
   1,  7,  97, 1351,  18817,  262087,   3650401, ...
   1,  9, 161, 2889,  51841,  930249,  16692641, ...
   1, 11, 241, 5291, 116161, 2550251,  55989361, ...
   1, 13, 337, 8749, 227137, 5896813, 153090001, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-3 give A000012, A005408, A069129(n+1), A322830.
Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986.
Main diagonal gives A173174.
A(n-1,n) gives A173148(n).
Cf. A322699, A322747.

A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *)

a(n) = 2 * A322699(n) + 1.
A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k).
A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k).
A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1.
A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = 2*n+1.

A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

Original entry on oeis.org

0, 6, 30, 35, 84, 180, 204, 210, 297, 330, 546, 840, 1170, 1189, 1224, 1710, 2310, 2940, 2970, 3036, 3230, 3900, 4914, 6090, 6930, 7134, 7140, 7245, 7440, 8976, 10710, 12654, 14175, 14820, 16296, 16380, 17220, 19866, 22770, 25172, 25944, 29103

Offset: 1

Zak Seidov, May 30 2010

nonn

From Robert G. Wilson v, Jul 24 2010: (Start)
Terms in the i-th row are products contributed with a factor A000217(i):
(1) 0, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, ...
(2) 30, 297, 2940, 29103, 288090, 2851797, 28229880, ...
(3) 84, 1170, 16296, 226974, 3161340, ...
(4) 180, 3230, 57960, 1040050, 18662940, ...
(5) 330, 7245, 159060, 3492075, 76666590, ...
(6) 546, 14175, 368004, 9553929, ...
(7) 840, 25172, 754320, 22604428, ...
(8) 210, 1224, 7134, 41580, 242346, 1412496, 8232630, 47983284, ...
(9) 1710, 64935, 2465820, 93636225, ...
(10) 2310, 96965, 4070220, ...
(11) 3036, 139590, 6418104, ...
(12) 3900, 194922, 9742200, ...
(13) 4914, 265265, 14319396, ...
(14) 6090, 353115, 20474580, ...
(15) 7440, 461160, 28584480, ...
(End)
Numbers m with property that m^2 is a product of two distinct triangular numbers T(i) and T(j) such that i and j are in the same row of the square array A(n, k) defined in A322699. - Onur Ozkan, Mar 17 2023

R. J. Mathar, Table of n, a(n) for n = 1..1000 replacing a b-file of _Robert G. Wilson v_ of 2010.
R. J. Mathar, OEIS A175497, Mar 16 2023
Project Euler, Problem 833. Square Triangle Products.
M. Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT], 2008. Theorem 5.6.

From Robert G. Wilson v, Jul 24 2010: (Start)
A001109 (with the exception of 1), A011945, A075848 and A055112 are all proper subsets.
Many terms are in common with A147779.
Cf. A001108, A132596, A007654, A001108, A132593. (End)
Cf. A152005 (two distinct tetrahedral numbers).

isA175497 := proc(n)
    local i,Ti,Tj;
    if n = 0 then
        return true;
    end if;
    for i from 1 do
        Ti := i*(i+1)/2 ;
        if Ti > n^2 then
            return false;
        else
            Tj := n^2/Ti ;
            if Tj <> Ti and type(Tj,'integer') then
                if isA000217(Tj) then  # code in A000217
                    return true;
                end if;
            end if;
        end if;
    end do:
end proc:
for n from 0 do
    if isA175497(n) then
        printf("%d,\n",n);
    end if;
end do: # R. J. Mathar, May 26 2016

triangularQ[n_] := IntegerQ[Sqrt[8n + 1]];
okQ[n_] := Module[{i, Ti, Tj}, If[n == 0, Return[True]]; For[i = 1, True, i++, Ti = i(i+1)/2; If[Ti > n^2, Return[False], Tj = n^2/Ti; If[Tj != Ti && IntegerQ[Tj], If[ triangularQ[Tj], Return[True]]]]]];
Reap[For[k = 0, k < 30000, k++, If[okQ[k], Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Jun 13 2023, after R. J. Mathar *)

from itertools import count, islice, takewhile
from sympy import divisors
from sympy.ntheory.primetest import is_square
def A175497_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda k:not k or any(map(lambda d: is_square((d<<3)+1) and is_square((k**2//d<<3)+1), takewhile(lambda d:d**2A175497_list = list(islice(A175497_gen(),20)) # Chai Wah Wu, Mar 13 2023

def A175497_list(n):
    def A322699_A(k, n):
        p, q, r, m = 0, k, 4*k*(k+1), 0
        while m < n:
            p, q, r = q, r, (4*k+3)*(r-q) + p
            m += 1
        return p
    def a(k, n, j):
        if n == 0: return 0
        p = A322699_A(k, n)*(A322699_A(k, n)+1)*(2*k+1) - a(k, n-1, 1)
        q = (4*k+2)*p - A322699_A(k, n)*(A322699_A(k, n)+1)//2
        m = 1
        while m < j: p, q = q, (4*k+2)*q - p; m += 1
        return p
    A = set([a(k, 1, 1) for k in range(n+1)])
    k, l, m = 1, 1, 2
    while True:
        x = a(k, l, m)
        if x < max(A):
            A |= {x}
            A  = set(sorted(A)[:n+1])
            m += 1
        else:
            if m == 1 and l == 1:
                if k > n:
                    return sorted(A)
                k += 1
            elif m > 1:
                l += 1; m = 1
            elif l > 1:
                k += 1; l, m = 1, 1
# Onur Ozkan, Mar 15 2023

a(n)^2 = A169836(n). - R. J. Mathar, Mar 12 2023

A322746 a(n) = 1/2 * (-1 + Sum_{k=0..n} binomial(2n,2k)(n+1)^(n-k)n^k).

Original entry on oeis.org

0, 1, 24, 675, 25920, 1275125, 76545000, 5425069447, 443365544448, 41047124680809, 4245890890571000, 485307363135371051, 60742714406414040000, 8262695239025750162653, 1213734518568509516047560, 191478489107270936785743375, 32288451913272713227175006208

Offset: 0

Seiichi Manyama, Dec 25 2018

nonn

			(sqrt(3) + sqrt(2))^2 = 5 + 2*sqrt(6) = sqrt(25) + sqrt(24). So a(2) = 24.

Seiichi Manyama, Table of n, a(n) for n = 0..321
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Main diagonal of A322699.

Cf. A322747.

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

{a(n) = (polchebyshev(n, 1, 2*n+1)-1)/2}

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^n.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^n.
a(n) = (A173174(n) - 1)/2.
a(n) ~ exp(1/2) * 2^(2*n - 2) * n^n. - Vaclav Kotesovec, Dec 25 2018

A173175 a(n) = sinh^2( 2n*arcsinh(sqrt n)).

Original entry on oeis.org

0, 8, 2400, 1825200, 2687489280, 6503780163000, 23436548406180000, 117725514040791821024, 786292024016459316676608, 6739465778247681589030301160, 72110357818535214970387726284000, 942092946853627620313318842336862608, 14758709413836719039368938494112056160000

Offset: 0

Artur Jasinski, Feb 11 2010

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..175
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Cf. A132592, A146311, A146312, A146313, A173115, A173116, A173121, A173127, A173128, A173129, A173130, A173131, A173133, A173134, A173148, A173151, A173170, A173171, A322699.

A173175 := proc(n) sinh(2*n*arcsinh(sqrt(n))) ; %^2 ; expand(%); simplify(%) ; end proc: # R. J. Mathar, Feb 26 2011

Table[Round[N[Sinh[(2 n) ArcSinh[Sqrt[n]]]^2, 100]], {n, 0, 20}]

{a(n) = (polchebyshev(2*n, 1, 2*n+1)-1)/2} \\ Seiichi Manyama, Jan 02 2019

{a(n) = 1/2*(-1+sum(k=0, 2*n, binomial(4*n, 2*k)*(n+1)^(2*n-k)*n^k))} \\ Seiichi Manyama, Jan 02 2019

From Seiichi Manyama, Jan 02 2019: (Start)
a(n) = A322699(n,2*n).
a(n) = (T_{2*n}(2*n+1) - 1)/2 where T_{n}(x) is a Chebyshev polynomial of the first kind.
a(n) = 1/2 * (-1 + Sum_{k=0..2*n} binomial(4*n,2*k)*(n+1)^(2*n-k)*n^k). (End)
a(n) ~ exp(1) * 2^(4*n - 2) * n^(2*n). - Vaclav Kotesovec, Jan 02 2019

a(11)-a(12) from Seiichi Manyama, Jan 02 2019

A322675 a(n) = n * (4*n + 3)^2.

Original entry on oeis.org

0, 49, 242, 675, 1444, 2645, 4374, 6727, 9800, 13689, 18490, 24299, 31212, 39325, 48734, 59535, 71824, 85697, 101250, 118579, 137780, 158949, 182182, 207575, 235224, 265225, 297674, 332667, 370300, 410669, 453870, 499999, 549152, 601425, 656914, 715715, 777924, 843637

Offset: 0

Seiichi Manyama, Dec 23 2018

			(sqrt(2) - sqrt(1))^3 = 5*sqrt(2) - 7 = sqrt(50) - sqrt(49). So a(1) = 49.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Column 3 of A322699.

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^k: A033996(n) (k=2), this sequence (k=3), A322677 (k=4), A322745 (k=5).

PARI
```
{a(n) = n*(4*n+3)^2}
```

concat(0, Vec(x*(49 + 46*x + x^2) / (1 - x)^4 + O(x^40))) \\ Colin Barker, Dec 23 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^3.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^3.
Sum_{n>=1} 1/a(n) = 8/27 + 2*c/3 + Pi/18 - Pi^2/12 - log(2)/3 = 0.027956857336446942649782759291008857522041405948099294509008..., where c is the Catalan constant A006752. - Vaclav Kotesovec, Dec 23 2018
From Colin Barker, Dec 23 2018: (Start)
G.f.: x*(49 + 46*x + x^2) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.
(End)

A322745 a(n) = n * (16n^2+20n+5)^2.

Original entry on oeis.org

0, 1681, 23762, 131043, 465124, 1275125, 2948406, 6041287, 11309768, 19740249, 32580250, 51369131, 77968812, 114594493, 163845374, 228735375, 312723856, 419746337, 554245218, 721200499, 926160500, 1175272581, 1475313862, 1833721943, 2258625624, 2758875625, 3344075306

Offset: 0

Seiichi Manyama, Dec 25 2018

			(sqrt(2) + sqrt(1))^5 = 29*sqrt(2) + 41 = sqrt(1682) + sqrt(1681). So a(1) = 1681.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Column 5 of A322699.

PARI
```
{a(n) = n*(16*n^2+20*n+5)^2}
```

concat(0, Vec(x*(1681 + 13676*x + 13686*x^2 + 1676*x^3 + x^4) / (1 - x)^6 + O(x^30))) \\ Colin Barker, Dec 25 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(n+1) + sqrt(n))^5.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(n+1) - sqrt(n))^5.
From Colin Barker, Dec 25 2018: (Start)
G.f.: x*(1681 + 13676*x + 13686*x^2 + 1676*x^3 + x^4) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5.
(End)

A322707 a(0)=0, a(1)=5 and a(n) = 22*a(n-1) - a(n-2) + 10 for n > 1.

Original entry on oeis.org

0, 5, 120, 2645, 58080, 1275125, 27994680, 614607845, 13493377920, 296239706405, 6503780163000, 142786923879605, 3134808545188320, 68823001070263445, 1510971215000607480, 33172543728943101125, 728284990821747617280, 15989097254349504479045

Offset: 0

Seiichi Manyama, Dec 24 2018

Solutions to X*(X+1)=30*Y^2 with Y=A077421. - R. J. Mathar, Mar 14 2023

			(sqrt(6) + sqrt(5))^2 = 11 + 2*sqrt(30) = sqrt(121) + sqrt(120). So a(2) = 120.

Colin Barker, Table of n, a(n) for n = 0..700
Index entries for linear recurrences with constant coefficients, signature (23,-23,1).

Row 5 of A322699.

Cf. A188930 (sqrt(5)+sqrt(6)).

concat(0, Vec(5*x*(1 + x) / ((1 - x)*(1 - 22*x + x^2)) + O(x^20))) \\ Colin Barker, Dec 24 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(6) + sqrt(5))^n.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(6) - sqrt(5))^n.
a(n) = 23*a(n-1) - 23*a(n-2) + a(n-3) for n > 2.
From Colin Barker, Dec 24 2018: (Start)
G.f.: 5*x*(1 + x) / ((1 - x)*(1 - 22*x + x^2)).
a(n) = ((11+2*sqrt(30))^(-n) * (-1+(11+2*sqrt(30))^n)^2) / 4.
(End)
2*a(n) = A077422(n)-1. - R. J. Mathar, Mar 16 2023

A322708 a(0)=0, a(1)=6 and a(n) = 26*a(n-1) - a(n-2) + 12 for n > 1.

Original entry on oeis.org

0, 6, 168, 4374, 113568, 2948406, 76545000, 1987221606, 51591216768, 1339384414374, 34772403556968, 902743108066806, 23436548406180000, 608447515452613206, 15796198853361763368, 410092722671953234374, 10646614590617422330368, 276401886633381027355206

Offset: 0

Seiichi Manyama, Dec 24 2018

Solutions to X*(X+1)=42*Y^2 with Y=A097309. - R. J. Mathar, Mar 14 2023

			(sqrt(7) + sqrt(6))^2 = 13 + 2*sqrt(42) = sqrt(169) + sqrt(168). So a(2) = 168.

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (27,-27,1).

Row 6 of A322699.

LinearRecurrence[{27,-27,1},{0,6,168},20] (* Harvey P. Dale, Apr 30 2022 *)

concat(0, Vec(6*x*(1 + x) / ((1 - x)*(1 - 26*x + x^2)) + O(x^20))) \\ Colin Barker, Dec 24 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(7) + sqrt(6))^n.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(7) - sqrt(6))^n.
a(n) = 27*a(n-1) - 27*a(n-2) + a(n-3) for n > 2.
From Colin Barker, Dec 24 2018: (Start)
G.f.: 6*x*(1 + x) / ((1 - x)*(1 - 26*x + x^2)).
a(n) = ((13+2*sqrt(42))^(-n) *  (-1+(13+2*sqrt(42))^n)^2) / 4.
(End)
2*a(n) = A097308(n)-1. - R. J. Mathar, Mar 14 2023

A322709 a(0)=0, a(1)=7 and a(n) = 30*a(n-1) - a(n-2) + 14 for n > 1.

Original entry on oeis.org

0, 7, 224, 6727, 201600, 6041287, 181037024, 5425069447, 162571046400, 4871706322567, 145988618630624, 4374786852596167, 131097616959254400, 3928553721925035847, 117725514040791821024, 3527836867501829594887, 105717380511014096025600, 3167993578462921051173127

Offset: 0

Seiichi Manyama, Dec 24 2018

Also numbers k such that 7*A000217(k) is a square. - Metin Sariyar, Nov 16 2019

			(sqrt(8) + sqrt(7))^2 = 15 + 2*sqrt(56) = sqrt(225) + sqrt(224). So a(2) = 224.

Colin Barker, Table of n, a(n) for n = 0..650
Index entries for linear recurrences with constant coefficients, signature (31,-31,1).

Row 7 of A322699.

Cf. A188932 (sqrt(7)+sqrt(8)).

a:=[0,7]; [n le 2 select a[n] else 30*Self(n-1)-Self(n-2)+14: n in [1..18]]; // Marius A. Burtea, Nov 16 2019

R:=PowerSeriesRing(Integers(), 18); [0] cat Coefficients(R!(7*x*(1 + x) / ((1 - x)*(1-30*x + x^2))));  // Marius A. Burtea, Nov 16 2019

LinearRecurrence[{31,-31,1}, {0, 7, 224}, 18] (* Metin Sariyar, Nov 23 2019 *)

concat(0, Vec(7*x*(1 + x) / ((1 - x)*(1 - 30*x + x^2)) + O(x^20))) \\ Colin Barker, Dec 25 2018

sqrt(a(n)+1) + sqrt(a(n)) = (sqrt(8) + sqrt(7))^n.
sqrt(a(n)+1) - sqrt(a(n)) = (sqrt(8) - sqrt(7))^n.
a(n) = 31*a(n-1) - 31*a(n-2) + a(n-3) for n > 2.
From Colin Barker, Dec 25 2018: (Start)
G.f.: 7*x*(1 + x) / ((1 - x)*(1 - 30*x + x^2)).
a(n) = ((15+4*sqrt(14))^(-n) * (-1+(15+4*sqrt(14))^n)^2) / 4.
(End)
E.g.f.: (1/4)*(-2*exp(x) + exp((15-4*sqrt(14))*x) + exp((15+4*sqrt(14))*x)). - Stefano Spezia, Nov 16 2019
2*a(n) = A068203(n)-1. - R. J. Mathar, Mar 16 2023

cp's OEIS Frontend

A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j.

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A175497 Numbers k with the property that k^2 is a product of two distinct triangular numbers.

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A322746 a(n) = 1/2 * (-1 + Sum_{k=0..n} binomial(2*n,2*k)*(n+1)^(n-k)*n^k).

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A173175 a(n) = sinh^2( 2n*arcsinh(sqrt n)).

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A322675 a(n) = n * (4*n + 3)^2.

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A322745 a(n) = n * (16*n^2+20*n+5)^2.

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A322707 a(0)=0, a(1)=5 and a(n) = 22*a(n-1) - a(n-2) + 10 for n > 1.

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A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j.

A322746 a(n) = 1/2 * (-1 + Sum_{k=0..n} binomial(2n,2k)(n+1)^(n-k)n^k).

A322745 a(n) = n * (16n^2+20n+5)^2.