A165677 -id:A165677 - cp's OEIS frontend

A080663 a(n) = 3*n^2 - 1.

2, 11, 26, 47, 74, 107, 146, 191, 242, 299, 362, 431, 506, 587, 674, 767, 866, 971, 1082, 1199, 1322, 1451, 1586, 1727, 1874, 2027, 2186, 2351, 2522, 2699, 2882, 3071, 3266, 3467, 3674, 3887, 4106, 4331, 4562, 4799, 5042, 5291, 5546, 5807, 6074, 6347, 6626

Offset: 1

Cino Hilliard, Mar 01 2003

These numbers cannot be perfect squares. See the Hilliard link for a proof.
2nd elementary symmetric polynomial of n, n + 1 and n + 2: n(n+1) + n(n+2) + (n+1)(n+2). - Zak Seidov, Mar 23 2005
This sequence equals for n >= 2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663(n-1) leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 16 2009
The Diophantine equation x(x+1) + (x+2)(x+3) = (x+y)^2 + (x-y)^2 has solutions x = a(n), y = 3n. - Bruno Berselli, Mar 29 2013
A simpler proof that these numbers can't be perfect squares can easily be constructed using congruences: If the equation x^2 = 3y^2 - 1 has a solution in positive integers, then x^2 = 2 mod 3. Obviously we can't have x = 0 mod 3, and x = 1 mod 3 doesn't work either because then x^2 = 1 mod 3 also. That leaves x = 2 mod 3, but then x^2 = 1 mod 3. - Alonso del Arte, Oct 19 2013
2*a(n+1) is surface area of a rectangular prism with consecutive integer sides: n, n+1, and n+2, (n>0). - Wesley Ivan Hurt, Sep 06 2014
Numbers m such that 3*m+3 is a square. So, these are the numbers m such that the system of equations x=sqrt(m-2yz), y=sqrt(m+1-2xz), z=sqrt(m+2-2xy) admits 3 real positive solutions whose sum is an integer. See the Rechtman link. - Michel Marcus, Jun 06 2020

Ethan D. Bolker, Elementary Number Theory: An Algebraic Approach. Mineola, New York: Dover Publications (1969, reprinted 2007): p. 7, Problem 6.6.
E. Grosswald, Topics from the Theory of Numbers, 1966 p 64 problem 11

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Cino Hilliard, 3n^2 - 1 not square. [Archived copy as of Apr 11 2008 from web.archive.org]
Ana Rechtman, Juin 2020, 1er défi, Images des Mathématiques, CNRS, 2020 (in French).
Leo Tavares, Illustration: Conjoined Trapezoids
Eric Weisstein's World of Mathematics, Symmetric Polynomial.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A007318, A028421, A080663, A121670, A126671, A165674, A165676, A165677, A165678, A165679.

Cf. A005449, A115067.

[3*n^2-1 : n in [1..50]]; // Wesley Ivan Hurt, Sep 04 2014

A080663 := proc(n) return 3*n^2-1: end proc: seq(A080663(n), n=1..50); # Nathaniel Johnston, Oct 16 2013

3*Range[47]^2 - 1 (* Alonso del Arte, Oct 19 2013 *)

list(n) = { for(x=1,n, y = 3*x*x-1; print1(y, ", ") ) } \\ edited by Michel Marcus, Feb 01 2020

Vec(x*(2+5*x-x^2)/(1-x)^3+O(x^66)) \\ Joerg Arndt, Sep 06 2014

a(n) = -Re((1 + n*i)^3) where i=sqrt(-1). - Gary W. Adamson, Aug 14 2006
a(n) = 3*n^2 - 1. - Stephen Crowley, Jul 06 2009
a(n) = a(n-1) - 3*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Oct 16 2009
G.f.: x*(2 + 5*x - x^2)/(1-x)^3. - Joerg Arndt, Sep 06 2014
a(n) = a(n-1) + 6*n - 3 for n > 1. - Vincenzo Librandi, Aug 08 2010
E.g.f.: 1 + exp(x)*(3*x^2 + 3*x - 1). - Stefano Spezia, Feb 01 2020
From Amiram Eldar, Feb 04 2021: (Start)
Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(3))*cot(Pi/sqrt(3)))/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(3))*csc(Pi/sqrt(3)) - 1)/2.
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(3))*csc(Pi/sqrt(3)).
Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(3))*sin(sqrt(2/3)*Pi)/sqrt(2). (End)
a(n) = A005449(n) + A115067(n). - Leo Tavares, May 25 2022
a(n) = (n-1)*n + (n-1)*(n+1) + n*(n+1), for n >= 1. See the Zak Seidov comment above. - Wolfdieter Lang, Aug 15 2024

A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

Original entry on oeis.org

1, 3, 1, 11, 5, 1, 50, 26, 7, 1, 274, 154, 47, 9, 1, 1764, 1044, 342, 74, 11, 1, 13068, 8028, 2754, 638, 107, 13, 1, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1

Offset: 1

Johannes W. Meijer, Oct 05 2009

The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2+6*n+3*n^2)/x^2 - (6+22*n+18*n^2+ 4*n^3)/x^3 + ... ) is discussed in A028421. The formula for the asymptotic expansion leads for n = 1, 2, 3, .., to the left hand columns of the triangle given above.
The recurrence relations of the right hand columns of this triangle lead to Pascal's triangle A007318, their a(n) formulas lead to Wiggen's triangle A028421 and their o.g.f.s lead to Wood's polynomials A126671; cf. A080663, A165676, A165677, A165678 and A165679.
The row sums of this triangle lead to A093344. Surprisingly the e.g.f. of the row sums Egf(x) = (exp(1)*Ei(1,1-x) - exp(1)*Ei(1,1))/(1-x) leads to the exponential integrals in view of the fact that E(x,m=1,n=1) = Ei(n=1,x). We point out that exp(1)*Ei(1,1) = A073003.
The Maple programs generate the coefficients of the triangle given above. The first one makes use of a relation between the triangle coefficients, see the formulas, and the second one makes use of the asymptotic expansions of the E(x,m=2,n).
Amarnath Murthy discovered triangle A093905 which is the reversal of our triangle.
A165675 is an extended version of this triangle. Its reversal is A105954.
Triangle A094587 is generated by the asymptotic expansions of E(x,m=1,n).

A093905 is the reversal of this triangle.
A000254, A001705, A001711, A001716, A001721, A051524, A051545, A051560, A051562, A051564 are the first ten left hand columns.
A080663, n>=2, is the third right hand column.
A165676, A165677, A165678 and A165679 are the next right hand columns, A093344 gives the row sums.
A073003 is Gompertz's constant.
A094587 is generated by the asymptotic expansions of E(x, m=1, n).
Cf. A165675, A105954 (Quet) and A067176 (Bottomley).
Cf. A007318 (Pascal), A028421 (Wiggen), A126671 (Wood).

nmax:=9; for n from 1 to nmax do a(n, n) := 1 od: for n from 2 to nmax do a(n, 1) := n*a(n-1, 1) + (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m = 1..n), n = 1..nmax);
# End program 1
nmax := nmax+1: m:=2; with(combinat): EA := proc(x, m, n) local E, i; E:=0: for i from m-1 to nmax+2 do E := E + sum((-1)^(m+k1+1) * binomial(k1, m-1) * n^(k1-m+1) * stirling1(i, k1), k1=m-1..i) / x^(i-m+1) od: E:= exp(-x)/x^(m) * E: return(E); end: for n1 from 1 to nmax do f(n1-1) := simplify(exp(x) * x^(nmax+3) * EA(x, m, n1)); for m1 from 0 to nmax+2 do b(n1-1, m1) := coeff(f(n1-1), x, nmax+2-m1) od: od: for n1 from 0 to nmax-1 do for m1 from 0 to n1-m+1 do a(n1-m+2, m1+1) := abs(b(m1, n1-m1)) od: od: seq(seq(a(n, m), m = 1..n),n = 1..nmax-1);
# End program 2
# Maple programs revised by Johannes W. Meijer, Sep 22 2012

a(n,m) = (n-m+1)*a(n-1,m) + a(n-1,m-1), for 2 <= m <= n-1, with a(n,n) = 1 and a(n,1) = n*a(n-1,1) + (n-1)!.

a(n,m) = product(i, i= m..n)*sum(1/i, i = m..n).

A165675 Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 5, 1, 24, 50, 26, 7, 1, 120, 274, 154, 47, 9, 1, 720, 1764, 1044, 342, 74, 11, 1, 5040, 13068, 8028, 2754, 638, 107, 13, 1, 40320, 109584, 69264, 24552, 5944, 1066, 146, 15, 1, 362880, 1026576, 663696, 241128, 60216, 11274, 1650, 191, 17, 1

Offset: 0

Johannes W. Meijer, Oct 05 2009

Previous name: Extended triangle related to the asymptotic expansions of the E(x, m = 2, n).
For the definition of the hyperharmonic numbers see the formula section.
This triangle is the same as triangle A165674 except for the extra left-hand column T(n, 0) = n!. The T(n) formulas for the right-hand columns generate the coefficients of this extra left-hand column.
Leroy Quet discovered triangle A105954 which is the reversal of our triangle.
In square format, row k gives the (n-1)-st elementary symmetric function of {k, k+1, k+2,..., k+n}, as in the Mathematica section. - Clark Kimberling, Dec 29 2011

			Triangle T(n, k) begins:
  [0]    1;
  [1]    1,     1;
  [2]    2,     3,    1;
  [3]    6,    11,    5,    1;
  [4]   24,    50,   26,    7,   1;
  [5]  120,   274,  154,   47,   9,   1;
  [6]  720,  1764, 1044,  342,  74,  11,  1;
  [7] 5040, 13068, 8028, 2754, 638, 107, 13, 1;
Seen as an array (the triangle arises when read by descending antidiagonals):
  [0] 1,  1,   2,    6,    24,    120,     720,     5040, ...
  [1] 1,  3,  11,   50,   274,   1764,   13068,   109584, ...
  [2] 1,  5,  26,  154,  1044,   8028,   69264,   663696, ...
  [3] 1,  7,  47,  342,  2754,  24552,  241128,  2592720, ...
  [4] 1,  9,  74,  638,  5944,  60216,  662640,  7893840, ...
  [5] 1, 11, 107, 1066, 11274, 127860, 1557660, 20355120, ...
  [6] 1, 13, 146, 1650, 19524, 245004, 3272688, 46536624, ...
  [7] 1, 15, 191, 2414, 31594, 434568, 6314664, 97053936, ...

A105954 is the reversal of this triangle.
A165674, A138771 and A165680 are related triangles.
A080663 equals the third right hand column.
A000142 equals the first left hand column.
A093345 are the row sums.
Columns include A165676, A165677, A165678 and A165679.

nmax := 8; for n from 0 to nmax do a(n, 0) := n! od: for n from 0 to nmax do a(n, n) := 1 od: for n from 2 to nmax do for m from 1 to n-1 do a(n, m) := (n-m+1)*a(n-1, m) + a(n-1, m-1) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax);
# Johannes W. Meijer, revised Nov 27 2012
# Shows the array format, using hyperharmonic numbers.
H := proc(n, k) option remember; if n = 0 then 1/(k+1)
else add(H(n - 1, j), j = 0..k) fi end:
seq(lprint(seq((k + 1)!*H(n, k), k = 0..7)), n = 0..7);
# Shows the array format, using the hypergeometric formula.
A := (n, k) -> (k+1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1):
seq(lprint(seq(simplify(A(n, k)), k = 0..7)), n = 0..7);
# Peter Luschny, Jul 03 2022

a[n_] := SymmetricPolynomial[n - 1, t[n]]; z = 10;
t[n_] := Table[k - 1, {k, 1, n}]; t1 = Table[a[n], {n, 1, z}]  (* A000142 *)
t[n_] := Table[k,     {k, 1, n}]; t2 = Table[a[n], {n, 1, z}]  (* A000254 *)
t[n_] := Table[k + 1, {k, 1, n}]; t3 = Table[a[n], {n, 1, z}]  (* A001705 *)
t[n_] := Table[k + 2, {k, 1, n}]; t4 = Table[a[n], {n, 1, z}]  (* A001711 *)
t[n_] := Table[k + 3, {k, 1, n}]; t5 = Table[a[n], {n, 1, z}]  (* A001716 *)
t[n_] := Table[k + 4, {k, 1, n}]; t6 = Table[a[n], {n, 1, z}]  (* A001721 *)
t[n_] := Table[k + 5, {k, 1, n}]; t7 = Table[a[n], {n, 1, z}]  (* A051524 *)
t[n_] := Table[k + 6, {k, 1, n}]; t8 = Table[a[n], {n, 1, z}]  (* A051545 *)
t[n_] := Table[k + 7, {k, 1, n}]; t9 = Table[a[n], {n, 1, z}]  (* A051560 *)
t[n_] := Table[k + 8, {k, 1, n}]; t10 = Table[a[n], {n, 1, z}] (* A051562 *)
t[n_] := Table[k + 9, {k, 1, n}]; t11 = Table[a[n], {n, 1, z}] (* A051564 *)
t[n_] := Table[k + 10, {k, 1, n}];t12 = Table[a[n], {n, 1, z}] (* A203147 *)
t = {t1, t2, t3, t4, t5, t6, t7, t8, t9, t10};
TableForm[t]  (* A165675 in square format *)
m[i_, j_] := t[[i]][[j]];
(* A165675 as a sequence *)
Flatten[Table[m[i, n + 1 - i], {n, 1, 10}, {i, 1, n}]]
(* Clark Kimberling, Dec 29 2011 *)
A[n_, k_] := (k + 1)*((n + k)! / n!)*HypergeometricPFQ[{-k, 1, 1}, {2, n + 1}, 1];
Table[A[n, k], {n, 0, 7}, {k, 0, 7}] // TableForm (* Peter Luschny, Jul 03 2022 *)

from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0:
        return [1]
    row = Trow(n - 1) + [1]
    for m in range(n - 1, 0, -1):
        row[m] = (n - m + 1) * row[m] + row[m - 1]
    row[0] *= n
    return row
for n in range(9): print(Trow(n))  # Peter Luschny, Feb 27 2025

The hyperharmonic numbers are H(n, k) = Sum_{j=0..k} H(n - 1, j), with base condition H(0, k) = 1/(k + 1).
T(n, k) = (n - k + 1)*T(n - 1, k) + T(n - 1, k - 1), 1 <= k <= n-1, with T(n, 0) = n! and T(n, n) = 1.
From Peter Luschny, Jul 03 2022: (Start)
The rectangular array is given by:
A(n, k) = (k + 1)!*H(n, k).
A(n, k) = (k + 1)*((n + k)! / n!)*hypergeom([-k, 1, 1], [2, n + 1], 1). (End)
From Werner Schulte, Feb 26 2025: (Start)
T(n, k) = n * T(n-1, k) + (n-1)! / (k-1)! for 0 < k < n.
T(n, k) = (Sum_{i=k..n} 1/i) * n! / (k-1)! for 0 < k <= n.
Matrix inverse M = T^(-1) is given by: M(n, n) = 1, M(n, n-1) = 1 - 2 * n for n > 0, M(n, n-2) = (n-1)^2 for n > 1, and M(i, j) = 0 otherwise. (End)

New name from Peter Luschny, Jul 03 2022

A126671 Triangle read by rows: row n (n>=0) has g.f. Sum_{i=1..n} n!x^i(1+x)^(n-i)/(n+1-i).

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 2, 7, 11, 0, 6, 26, 46, 50, 0, 24, 126, 274, 326, 274, 0, 120, 744, 1956, 2844, 2556, 1764, 0, 720, 5160, 16008, 28092, 30708, 22212, 13068, 0, 5040, 41040, 147120, 304464, 401136, 351504, 212976, 109584, 0, 40320

Offset: 1

N. J. A. Sloane and Carlo Wood (carlo(AT)alinoe.com), Feb 13 2007

The first nonzero column gives the factorial numbers, which are Stirling_1(*,1), the rightmost diagonal gives Stirling_1(*,2), so this triangle may be regarded as interpolating between the first two columns of the Stirling numbers of the first kind.
This is a slice (the right-hand wall) through the infinite square pyramid described in the link. The other three walls give A007318 and A008276 (twice).
The coefficients of the A165674 triangle are generated by the asymptotic expansion of the higher order exponential integral E(x,m=2,n). The a(n) formulas for the coefficients in the right hand columns of this triangle lead to Wiggen's triangle A028421 and their o.g.f.s. lead to the sequence given above. Some right hand columns of the A165674 triangle are A080663, A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 07 2009

			Triangle begins:
0,
0, 1,
0, 1, 3,
0, 2, 7, 11,
0, 6, 26, 46, 50,
0, 24, 126, 274, 326, 274,
0, 120, 744, 1956, 2844, 2556, 1764,
0, 720, 5160, 16008, 28092, 30708, 22212, 13068,
0, 5040, 41040, 147120, 304464, 401136, 351504, 212976, 109584,
0, 40320, 367920, 1498320, 3582000, 5562576, 5868144, 4292496, 2239344, 1026576, ...

N. J. A. Sloane, Notes on Carlo Wood's Polynomials

Columns give A000142, A108217, A126672; diagonals give A000254, A067318, A126673. Row sums give A126674. Alternating row sums give A000142.
See A126682 for the full pyramid of coefficients of the underlying polynomials.
Cf. A165674, A028421, A080663, A165676, A165677, A165678 and A165679. - Johannes W. Meijer, Oct 07 2009

for n from 1 to 15 do t1:=add( n!*x^i*(1+x)^(n-i)/(n+1-i), i=1..n); series(t1,x,100); lprint(seriestolist(%)); od:

Join[{{0}}, Reap[For[n = 1, n <= 15, n++, t1 = Sum[n!*x^i*(1+x)^(n-i)/(n+1-i), {i, 1, n}]; se = Series[t1, {x, 0, 100}]; Sow[CoefficientList[se, x]]]][[2, 1]]] // Flatten (* Jean-François Alcover, Jan 07 2014, after Maple *)

Recurrence: T(n,0) = 0; for n>=0, i>=1, T(n+1,i) = (n+1)*T(n,i) + n!*binomial(n,i).

E.g.f.: x*log(1-(1+x)*y)/(x*y-1)/(1+x). - Vladeta Jovovic, Feb 13 2007

cp's OEIS Frontend

A080663 a(n) = 3*n^2 - 1.

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A165674 Triangle generated by the asymptotic expansions of the E(x,m=2,n).

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A165675 Triangle read by rows. T(n, k) = (n - k + 1)! * H(k, n - k), where H are the hyperharmonic numbers. For 0 <= k <= n.

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A126671 Triangle read by rows: row n (n>=0) has g.f. Sum_{i=1..n} n!*x^i*(1+x)^(n-i)/(n+1-i).

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A126671 Triangle read by rows: row n (n>=0) has g.f. Sum_{i=1..n} n!x^i(1+x)^(n-i)/(n+1-i).